home  /  The World History  /  Examples of quadratic equations with imaginary roots. Square root: calculation formulas. Formula for finding the roots of a quadratic equation. What is a quadratic equation

Examples of quadratic equations with imaginary roots. Square root: calculation formulas. Formula for finding the roots of a quadratic equation. What is a quadratic equation

The World History
09.02.2021

We continue to study the topic “ solving equations" We have already become acquainted with linear equations and are moving on to getting acquainted with quadratic equations.

First we will look at what a quadratic equation is and how it is written in general view, and give related definitions. After this, we will use examples to examine in detail how incomplete quadratic equations are solved. Let's move on to the solution complete equations, we get the root formula, get acquainted with the discriminant of the quadratic equation and consider the solutions typical examples. Finally, let's trace the connections between the roots and coefficients.

Page navigation.

What is a quadratic equation? Their types

First you need to clearly understand what a quadratic equation is. Therefore, it is logical to start a conversation about quadratic equations with the definition of a quadratic equation, as well as related definitions. After this, you can consider the main types quadratic equations: reduced and unreduced, as well as complete and incomplete equations.

Definition and examples of quadratic equations

Definition.

Quadratic equation is an equation of the form a x 2 +b x+c=0, where x is a variable, a, b and c are some numbers, and a is non-zero.

Let's say right away that quadratic equations are often called equations of the second degree. This is due to the fact that the quadratic equation is algebraic equation second degree.

The stated definition allows us to give examples of quadratic equations. So 2 x 2 +6 x+1=0, 0.2 x 2 +2.5 x+0.03=0, etc. These are quadratic equations.

Definition.

Numbers a, b and c are called coefficients of the quadratic equation a·x 2 +b·x+c=0, and coefficient a is called the first, or the highest, or the coefficient of x 2, b is the second coefficient, or the coefficient of x, and c is the free term.

For example, let's take a quadratic equation of the form 5 x 2 −2 x −3=0, here the leading coefficient is 5, the second coefficient is equal to −2, and the free term is equal to −3. Please note that when the coefficients b and/or c are negative, as in the example just given, the short form of the quadratic equation is 5 x 2 −2 x−3=0 , rather than 5 x 2 +(−2 )·x+(−3)=0 .

It is worth noting that when the coefficients a and/or b are equal to 1 or −1, they are usually not explicitly present in the quadratic equation, which is due to the peculiarities of writing such . For example, in the quadratic equation y 2 −y+3=0 the leading coefficient is one, and the coefficient of y is equal to −1.

Reduced and unreduced quadratic equations

Depending on the value of the leading coefficient, reduced and unreduced quadratic equations are distinguished. Let us give the corresponding definitions.

Definition.

A quadratic equation in which the leading coefficient is 1 is called given quadratic equation. Otherwise the quadratic equation is untouched.

According to this definition, quadratic equations x 2 −3·x+1=0, x 2 −x−2/3=0, etc. – given, in each of them the first coefficient is equal to one. A 5 x 2 −x−1=0, etc. - unreduced quadratic equations, their leading coefficients are different from 1.

From any unreduced quadratic equation, by dividing both sides by the leading coefficient, you can go to the reduced one. This action is an equivalent transformation, that is, the reduced quadratic equation obtained in this way has the same roots as the original unreduced quadratic equation, or, like it, has no roots.

Let us look at an example of how the transition from an unreduced quadratic equation to a reduced one is performed.

Example.

From the equation 3 x 2 +12 x−7=0, go to the corresponding reduced quadratic equation.

Solution.

We just need to divide both sides of the original equation by the leading coefficient 3, it is non-zero, so we can perform this action. We have (3 x 2 +12 x−7):3=0:3, which is the same, (3 x 2):3+(12 x):3−7:3=0, and then (3:3) x 2 +(12:3) x−7:3=0, from where . This is how we obtained the reduced quadratic equation, which is equivalent to the original one.

Answer:

Complete and incomplete quadratic equations

The definition of a quadratic equation contains the condition a≠0. This condition is necessary so that the equation a x 2 + b x + c = 0 is quadratic, since when a = 0 it actually becomes a linear equation of the form b x + c = 0.

As for the coefficients b and c, they can be equal to zero, both individually and together. In these cases, the quadratic equation is called incomplete.

Definition.

The quadratic equation a x 2 +b x+c=0 is called incomplete, if at least one of the coefficients b, c equal to zero.

In its turn

Definition.

Complete quadratic equation is an equation in which all coefficients are different from zero.

Such names were not given by chance. This will become clear from the following discussions.

If the coefficient b is zero, then the quadratic equation takes the form a·x 2 +0·x+c=0, and it is equivalent to the equation a·x 2 +c=0. If c=0, that is, the quadratic equation has the form a·x 2 +b·x+0=0, then it can be rewritten as a·x 2 +b·x=0. And with b=0 and c=0 we get the quadratic equation a·x 2 =0. The resulting equations differ from the complete quadratic equation in that their left-hand sides do not contain either a term with the variable x, or a free term, or both. Hence their name - incomplete quadratic equations.

So the equations x 2 +x+1=0 and −2 x 2 −5 x+0.2=0 are examples of complete quadratic equations, and x 2 =0, −2 x 2 =0, 5 x 2 +3=0 , −x 2 −5 x=0 are incomplete quadratic equations.

Solving incomplete quadratic equations

From the information in the previous paragraph it follows that there is three types of incomplete quadratic equations:

  • a·x 2 =0, the coefficients b=0 and c=0 correspond to it;
  • a x 2 +c=0 when b=0 ;
  • and a·x 2 +b·x=0 when c=0.

Let us examine in order how incomplete quadratic equations of each of these types are solved.

a x 2 =0

Let's start with solving incomplete quadratic equations in which the coefficients b and c are equal to zero, that is, with equations of the form a x 2 =0. The equation a·x 2 =0 is equivalent to the equation x 2 =0, which is obtained from the original by dividing both parts by a non-zero number a. Obviously, the root of the equation x 2 =0 is zero, since 0 2 =0. This equation has no other roots, which is explained by the fact that for any non-zero number p the inequality p 2 >0 holds, which means that for p≠0 the equality p 2 =0 is never achieved.

So, the incomplete quadratic equation a·x 2 =0 has a single root x=0.

As an example, we give the solution to the incomplete quadratic equation −4 x 2 =0. It is equivalent to the equation x 2 =0, its only root is x=0, therefore, the original equation has a single root zero.

A short solution in this case can be written as follows:
−4 x 2 =0 ,
x 2 =0,
x=0 .

a x 2 +c=0

Now let's look at how incomplete quadratic equations are solved in which the coefficient b is zero and c≠0, that is, equations of the form a x 2 +c=0. We know that moving a term from one side of the equation to the other with the opposite sign, as well as dividing both sides of the equation by a non-zero number, gives an equivalent equation. Therefore, we can carry out the following equivalent transformations of the incomplete quadratic equation a x 2 +c=0:

  • move c to the right side, which gives the equation a x 2 =−c,
  • and divide both sides by a, we get .

The resulting equation allows us to draw conclusions about its roots. Depending on the values ​​of a and c, the value of the expression can be negative (for example, if a=1 and c=2, then ) or positive (for example, if a=−2 and c=6, then ), it is not zero , since by condition c≠0. Let's look at the cases separately.

If , then the equation has no roots. This statement follows from the fact that the square of any number is a non-negative number. It follows from this that when , then for any number p the equality cannot be true.

If , then the situation with the roots of the equation is different. In this case, if we remember about , then the root of the equation immediately becomes obvious; it is the number, since . It’s easy to guess that the number is also the root of the equation, indeed, . This equation has no other roots, which can be shown, for example, by contradiction. Let's do it.

Let us denote the roots of the equation just announced as x 1 and −x 1 . Suppose that the equation has one more root x 2, different from the indicated roots x 1 and −x 1. It is known that substituting its roots into an equation instead of x turns the equation into a correct numerical equality. For x 1 and −x 1 we have , and for x 2 we have . The properties of numerical equalities allow us to perform term-by-term subtraction of correct numerical equalities, so subtracting the corresponding parts of the equalities gives x 1 2 −x 2 2 =0. The properties of operations with numbers allow us to rewrite the resulting equality as (x 1 −x 2)·(x 1 +x 2)=0. We know that the product of two numbers is equal to zero if and only if at least one of them is equal to zero. Therefore, from the resulting equality it follows that x 1 −x 2 =0 and/or x 1 +x 2 =0, which is the same, x 2 =x 1 and/or x 2 =−x 1. So we came to a contradiction, since at the beginning we said that the root of the equation x 2 is different from x 1 and −x 1. This proves that the equation has no roots other than and .

Let us summarize the information in this paragraph. The incomplete quadratic equation a x 2 +c=0 is equivalent to the equation that

  • has no roots if ,
  • has two roots and , if .

Let's consider examples of solving incomplete quadratic equations of the form a·x 2 +c=0.

Let's start with the quadratic equation 9 x 2 +7=0. After moving the free term to the right side of the equation, it will take the form 9 x 2 =−7. Dividing both sides of the resulting equation by 9, we arrive at . Since the right side has a negative number, this equation has no roots, therefore, the original incomplete quadratic equation 9 x 2 +7 = 0 has no roots.

Let's solve another incomplete quadratic equation −x 2 +9=0. We move the nine to the right side: −x 2 =−9. Now we divide both sides by −1, we get x 2 =9. On the right side there is a positive number, from which we conclude that or . Then we write down the final answer: the incomplete quadratic equation −x 2 +9=0 has two roots x=3 or x=−3.

a x 2 +b x=0

It remains to deal with the solution of the last type of incomplete quadratic equations for c=0. Incomplete quadratic equations of the form a x 2 + b x = 0 allow you to solve factorization method. Obviously, we can, located on the left side of the equation, for which it is enough to take it out of brackets common multiplier x. This allows us to move from the original incomplete quadratic equation to an equivalent equation of the form x·(a·x+b)=0. And this equation is equivalent to a set of two equations x=0 and a·x+b=0, the latter of which is linear and has a root x=−b/a.

So, the incomplete quadratic equation a·x 2 +b·x=0 has two roots x=0 and x=−b/a.

To consolidate the material, we will analyze the solution to a specific example.

Example.

Solve the equation.

Solution.

Taking x out of brackets gives the equation . It is equivalent to two equations x=0 and . Solving what we got linear equation: , and dividing the mixed number by common fraction, we find . Therefore, the roots of the original equation are x=0 and .

After gaining the necessary practice, solutions to such equations can be written briefly:

Answer:

x=0 , .

Discriminant, formula for the roots of a quadratic equation

To solve quadratic equations, there is a root formula. Let's write it down formula for the roots of a quadratic equation: , Where D=b 2 −4 a c- so-called discriminant of a quadratic equation. The entry essentially means that .

It is useful to know how the root formula was derived and how it is used in finding the roots of quadratic equations. Let's figure this out.

Derivation of the formula for the roots of a quadratic equation

Let us need to solve the quadratic equation a·x 2 +b·x+c=0. Let's perform some equivalent transformations:

  • We can divide both sides of this equation by a non-zero number a, resulting in the following quadratic equation.
  • Now select a complete square on its left side: . After this, the equation will take the form .
  • At this stage, it is possible to transfer the last two terms to the right side with the opposite sign, we have .
  • And let’s also transform the expression on the right side: .

As a result, we arrive at an equation that is equivalent to the original quadratic equation a·x 2 +b·x+c=0.

We have already solved equations similar in form in the previous paragraphs, when we examined. This allows us to draw the following conclusions regarding the roots of the equation:

  • if , then the equation has no real solutions;
  • if , then the equation has the form , therefore, , from which its only root is visible;
  • if , then or , which is the same as or , that is, the equation has two roots.

Thus, the presence or absence of roots of the equation, and therefore the original quadratic equation, depends on the sign of the expression on the right side. In turn, the sign of this expression is determined by the sign of the numerator, since the denominator 4·a 2 is always positive, that is, by the sign of the expression b 2 −4·a·c. This expression b 2 −4 a c was called discriminant of a quadratic equation and designated by the letter D. From here the essence of the discriminant is clear - based on its value and sign, they conclude whether the quadratic equation has real roots, and if so, what is their number - one or two.

Let's return to the equation and rewrite it using the discriminant notation: . And we draw conclusions:

  • if D<0 , то это уравнение не имеет действительных корней;
  • if D=0, then this equation has a single root;
  • finally, if D>0, then the equation has two roots or, which can be rewritten in the form or, and after expanding and reducing the fractions to common denominator we receive .

So we derived the formulas for the roots of the quadratic equation, they look like , where the discriminant D is calculated by the formula D=b 2 −4·a·c.

With their help, with a positive discriminant, you can calculate both real roots of a quadratic equation. When the discriminant is equal to zero, both formulas give the same value of the root, corresponding to a unique solution to the quadratic equation. And with a negative discriminant, when trying to use the formula for the roots of a quadratic equation, we are faced with extracting the square root of negative number, which takes us beyond the scope of the school curriculum. With a negative discriminant, the quadratic equation has no real roots, but has a pair complex conjugate roots, which can be found using the same root formulas we obtained.

Algorithm for solving quadratic equations using root formulas

In practice, when solving quadratic equations, you can immediately use the root formula to calculate their values. But this is more related to finding complex roots.

However, in a school algebra course we usually talk not about complex, but about real roots of a quadratic equation. In this case, it is advisable, before using the formulas for the roots of a quadratic equation, to first find the discriminant, make sure that it is non-negative (otherwise, we can conclude that the equation does not have real roots), and only then calculate the values ​​of the roots.

The above reasoning allows us to write algorithm for solving a quadratic equation. To solve the quadratic equation a x 2 +b x+c=0, you need to:

  • using the discriminant formula D=b 2 −4·a·c, calculate its value;
  • conclude that a quadratic equation has no real roots if the discriminant is negative;
  • calculate the only root of the equation using the formula if D=0;
  • find two real roots of a quadratic equation using the root formula if the discriminant is positive.

Here we just note that if the discriminant is equal to zero, you can also use the formula; it will give the same value as .

You can move on to examples of using the algorithm for solving quadratic equations.

Examples of solving quadratic equations

Let's consider solutions to three quadratic equations with a positive, negative and zero discriminant. Having dealt with their solution, by analogy it will be possible to solve any other quadratic equation. Let's begin.

Example.

Find the roots of the equation x 2 +2·x−6=0.

Solution.

In this case, we have the following coefficients of the quadratic equation: a=1, b=2 and c=−6. According to the algorithm, you first need to calculate the discriminant; to do this, we substitute the indicated a, b and c into the discriminant formula, we have D=b 2 −4·a·c=2 2 −4·1·(−6)=4+24=28. Since 28>0, that is, the discriminant is greater than zero, the quadratic equation has two real roots. Let's find them using the root formula, we get , here you can simplify the resulting expressions by doing moving the multiplier beyond the root sign followed by reduction of the fraction:

Answer:

Let's move on to the next typical example.

Example.

Solve the quadratic equation −4 x 2 +28 x−49=0 .

Solution.

We start by finding the discriminant: D=28 2 −4·(−4)·(−49)=784−784=0. Therefore, this quadratic equation has a single root, which we find as , that is,

Answer:

x=3.5.

It remains to consider solving quadratic equations with a negative discriminant.

Example.

Solve the equation 5·y 2 +6·y+2=0.

Solution.

Here are the coefficients of the quadratic equation: a=5, b=6 and c=2. We substitute these values ​​into the discriminant formula, we have D=b 2 −4·a·c=6 2 −4·5·2=36−40=−4. The discriminant is negative, therefore, this quadratic equation has no real roots.

If you need to indicate complex roots, then we apply the well-known formula for the roots of a quadratic equation, and perform actions with complex numbers :

Answer:

there are no real roots, complex roots are: .

Let us note once again that if the discriminant of a quadratic equation is negative, then in school they usually immediately write down an answer in which they indicate that there are no real roots, and complex roots are not found.

Root formula for even second coefficients

The formula for the roots of a quadratic equation, where D=b 2 −4·a·c allows you to obtain a formula of a more compact form, allowing you to solve quadratic equations with an even coefficient for x (or simply with a coefficient having the form 2·n, for example, or 14· ln5=2·7·ln5 ). Let's get her out.

Let's say we need to solve a quadratic equation of the form a x 2 +2 n x+c=0. Let's find its roots using the formula we know. To do this, we calculate the discriminant D=(2 n) 2 −4 a c=4 n 2 −4 a c=4 (n 2 −a c), and then we use the root formula:

Let us denote the expression n 2 −a c as D 1 (sometimes it is denoted D "). Then the formula for the roots of the quadratic equation under consideration with the second coefficient 2 n will take the form , where D 1 =n 2 −a·c.

It is easy to see that D=4·D 1, or D 1 =D/4. In other words, D 1 is the fourth part of the discriminant. It is clear that the sign of D 1 is the same as the sign of D . That is, the sign D 1 is also an indicator of the presence or absence of roots of a quadratic equation.

So, to solve a quadratic equation with a second coefficient 2·n, you need

  • Calculate D 1 =n 2 −a·c ;
  • If D 1<0 , то сделать вывод, что действительных корней нет;
  • If D 1 =0, then calculate the only root of the equation using the formula;
  • If D 1 >0, then find two real roots using the formula.

Let's consider solving the example using the root formula obtained in this paragraph.

Example.

Solve the quadratic equation 5 x 2 −6 x −32=0 .

Solution.

The second coefficient of this equation can be represented as 2·(−3) . That is, you can rewrite the original quadratic equation in the form 5 x 2 +2 (−3) x−32=0, here a=5, n=−3 and c=−32, and calculate the fourth part of the discriminant: D 1 =n 2 −a·c=(−3) 2 −5·(−32)=9+160=169. Since its value is positive, the equation has two real roots. Let's find them using the appropriate root formula:

Note that it was possible to use the usual formula for the roots of a quadratic equation, but in this case more computational work would have to be performed.

Answer:

Simplifying the form of quadratic equations

Sometimes, before starting to calculate the roots of a quadratic equation using formulas, it doesn’t hurt to ask the question: “Is it possible to simplify the form of this equation?” Agree that in terms of calculations it will be easier to solve the quadratic equation 11 x 2 −4 x−6=0 than 1100 x 2 −400 x−600=0.

Typically, simplifying the form of a quadratic equation is achieved by multiplying or dividing both sides by a certain number. For example, in the previous paragraph it was possible to simplify the equation 1100 x 2 −400 x −600=0 by dividing both sides by 100.

A similar transformation is carried out with quadratic equations, the coefficients of which are not . In this case, we usually divide both sides of the equation by absolute values its coefficients. For example, let's take the quadratic equation 12 x 2 −42 x+48=0. absolute values ​​of its coefficients: GCD(12, 42, 48)= GCD(GCD(12, 42), 48)= GCD(6, 48)=6. Dividing both sides of the original quadratic equation by 6, we arrive at the equivalent quadratic equation 2 x 2 −7 x+8=0.

And multiplying both sides of a quadratic equation is usually done to get rid of fractional coefficients. In this case, multiplication is carried out by the denominators of its coefficients. For example, if both sides of the quadratic equation are multiplied by LCM(6, 3, 1)=6, then it will take the simpler form x 2 +4·x−18=0.

In conclusion of this point, we note that they almost always get rid of the minus at the highest coefficient of a quadratic equation by changing the signs of all terms, which corresponds to multiplying (or dividing) both sides by −1. For example, usually one moves from the quadratic equation −2 x 2 −3 x+7=0 to the solution 2 x 2 +3 x−7=0 .

Relationship between roots and coefficients of a quadratic equation

The formula for the roots of a quadratic equation expresses the roots of the equation through its coefficients. Based on the root formula, you can obtain other relationships between roots and coefficients.

The most well-known and applicable formulas from Vieta’s theorem are of the form and . In particular, for the given quadratic equation, the sum of the roots is equal to the second coefficient with the opposite sign, and the product of the roots is equal to the free term. For example, by looking at the form of the quadratic equation 3 x 2 −7 x + 22 = 0, we can immediately say that the sum of its roots is equal to 7/3, and the product of the roots is equal to 22/3.

Using the already written formulas, you can obtain a number of other connections between the roots and coefficients of the quadratic equation. For example, you can express the sum of the squares of the roots of a quadratic equation through its coefficients: .

Bibliography.

  • Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
  • Mordkovich A. G. Algebra. 8th grade. At 2 p.m. Part 1. Textbook for students educational institutions/ A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.

Quadratic equations are studied in 8th grade, so there is nothing complicated here. The ability to solve them is absolutely necessary.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a ≠ 0.

Before studying specific solution methods, note that all quadratic equations can be divided into three classes:

  1. Have no roots;
  2. Have exactly one root;
  3. They have two different roots.

This is important difference quadratic equations from linear ones, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac.

You need to know this formula by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant you can determine how many roots a quadratic equation has. Namely:

  1. If D< 0, корней нет;
  2. If D = 0, there is exactly one root;
  3. If D > 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people believe. Take a look at the examples and you will understand everything yourself:

Task. How many roots do quadratic equations have:

  1. x 2 − 8x + 12 = 0;
  2. 5x 2 + 3x + 7 = 0;
  3. x 2 − 6x + 9 = 0.

Let's write out the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16

So the discriminant is positive, so the equation has two different roots. We analyze the second equation in a similar way:
a = 5; b = 3; c = 7;
D = 3 2 − 4 5 7 = 9 − 140 = −131.

The discriminant is negative, there are no roots. The last equation left is:
a = 1; b = −6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.

The discriminant is zero - the root will be one.

Please note that coefficients have been written down for each equation. Yes, it’s long, yes, it’s tedious, but you won’t mix up the odds and make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you get the hang of it, after a while you won’t need to write down all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not that much.

Roots of a quadratic equation

Now let's move on to the solution itself. If the discriminant D > 0, the roots can be found using the formulas:

Basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you will get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

  1. x 2 − 2x − 3 = 0;
  2. 15 − 2x − x 2 = 0;
  3. x 2 + 12x + 36 = 0.

First equation:
x 2 − 2x − 3 = 0 ⇒ a = 1; b = −2; c = −3;
D = (−2) 2 − 4 1 (−3) = 16.

D > 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 · (−1) · 15 = 64.

D > 0 ⇒ the equation again has two roots. Let's find them

\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and can count, there will be no problems. Most often, errors occur when substituting negative coefficients into the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon you will get rid of errors.

Incomplete quadratic equations

It happens that a quadratic equation is slightly different from what is given in the definition. For example:

  1. x 2 + 9x = 0;
  2. x 2 − 16 = 0.

It is easy to notice that these equations are missing one of the terms. Such quadratic equations are even easier to solve than standard ones: they don’t even require calculating the discriminant. So, let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.

Of course, it’s completely possible Hard case, when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.

Let's consider the remaining cases. Let b = 0, then we obtain an incomplete quadratic equation of the form ax 2 + c = 0. Let us transform it a little:

Since the arithmetic square root exists only of a non-negative number, the last equality makes sense only for (−c /a) ≥ 0. Conclusion:

  1. If in an incomplete quadratic equation of the form ax 2 + c = 0 the inequality (−c /a) ≥ 0 is satisfied, there will be two roots. The formula is given above;
  2. If (−c /a)< 0, корней нет.

As you can see, a discriminant was not required—there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c /a) ≥ 0. It is enough to express the value x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If it is negative, there will be no roots at all.

Now let's look at equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor the polynomial:

Taking the common factor out of brackets

The product is zero when at least one of the factors is zero. This is where the roots come from. In conclusion, let’s look at a few of these equations:

Task. Solve quadratic equations:

  1. x 2 − 7x = 0;
  2. 5x 2 + 30 = 0;
  3. 4x 2 − 9 = 0.

x 2 − 7x = 0 ⇒ x · (x − 7) = 0 ⇒ x 1 = 0; x 2 = −(−7)/1 = 7.

5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, because a square cannot be equal to a negative number.

4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 = −1.5.

Video tutorial 2: Solving Quadratic Equations

Lecture: Quadratic equations


The equation

The equation- this is a kind of equality in the expressions of which there is a variable.

Solve the equation- means finding a number instead of a variable that will bring it into correct equality.

An equation may have one solution, several, or none at all.

To solve any equation, it should be simplified as much as possible to the form:

Linear: a*x = b;

Square: a*x 2 + b*x + c = 0.

That is, any equations must be converted to standard form before solving.

Any equation can be solved in two ways: analytical and graphical.

On the graph, the solution to the equation is considered to be the points at which the graph intersects the OX axis.

Quadratic equations


An equation can be called quadratic if, when simplified, it takes the form:

a*x 2 + b*x + c = 0.

Wherein a, b, c are coefficients of the equation that differ from zero. A "X"- root of the equation. It is believed that a quadratic equation has two roots or may not have a solution at all. The resulting roots may be the same.

"A"- the coefficient that stands before the squared root.

"b"- stands before the unknown in the first degree.

"With" is the free term of the equation.

If, for example, we have an equation of the form:

2x 2 -5x+3=0

In it, “2” is the coefficient of the leading term of the equation, “-5” is the second coefficient, and “3” is the free term.

Solving a quadratic equation

There are a huge variety of ways to solve a quadratic equation. However, in a school mathematics course, the solution is studied using Vieta’s theorem, as well as using a discriminant.

Discriminant solution:

When solving with this method it is necessary to calculate the discriminant using the formula:

If during your calculations you find that the discriminant is less than zero, this means that this equation has no solutions.

If the discriminant is zero, then the equation has two identical solutions. In this case, the polynomial can be collapsed using the abbreviated multiplication formula into the square of the sum or difference. Then solve it as a linear equation. Or use the formula:

If the discriminant is greater than zero, then you must use the following method:

Vieta's theorem


If the equation is given, that is, the coefficient of the leading term is equal to one, then you can use Vieta's theorem.

So let's assume the equation is:

The roots of the equation are found as follows:

Incomplete quadratic equation

There are several options for obtaining an incomplete quadratic equation, the form of which depends on the presence of coefficients.

1. If the second and third coefficients are zero (b = 0, c = 0), then the quadratic equation will look like:

This equation will have a unique solution. The equality will be true only if the solution to the equation is zero.

Continuing the topic “Solving Equations,” the material in this article will introduce you to quadratic equations.

Let's look at everything in detail: the essence and notation of a quadratic equation, define the accompanying terms, analyze the scheme for solving incomplete and complete equations, get acquainted with the formula of roots and the discriminant, establish connections between the roots and coefficients, and of course we will give a visual solution to practical examples.

Quadratic equation, its types

Definition 1

Quadratic equation is an equation written as a x 2 + b x + c = 0, Where x– variable, a , b and c– some numbers, while a is not zero.

Often, quadratic equations are also called equations of the second degree, since in essence a quadratic equation is algebraic equation second degree.

Let's give an example to illustrate the given definition: 9 x 2 + 16 x + 2 = 0 ; 7, 5 x 2 + 3, 1 x + 0, 11 = 0, etc. These are quadratic equations.

Definition 2

Numbers a, b and c are the coefficients of the quadratic equation a x 2 + b x + c = 0, while the coefficient a is called the first, or senior, or coefficient at x 2, b - the second coefficient, or coefficient at x, A c called a free member.

For example, in the quadratic equation 6 x 2 − 2 x − 11 = 0 the leading coefficient is 6, the second coefficient is − 2 , and the free term is equal to − 11 . Let us pay attention to the fact that when the coefficients b and/or c are negative, then a short form of the form is used 6 x 2 − 2 x − 11 = 0, but not 6 x 2 + (− 2) x + (− 11) = 0.

Let us also clarify this aspect: if the coefficients a and/or b equal 1 or − 1 , then they may not take an explicit part in writing the quadratic equation, which is explained by the peculiarities of writing the indicated numerical coefficients. For example, in the quadratic equation y 2 − y + 7 = 0 the leading coefficient is 1, and the second coefficient is − 1 .

Reduced and unreduced quadratic equations

Based on the value of the first coefficient, quadratic equations are divided into reduced and unreduced.

Definition 3

Reduced quadratic equation is a quadratic equation where the leading coefficient is 1. For other values ​​of the leading coefficient, the quadratic equation is unreduced.

Let's give examples: quadratic equations x 2 − 4 · x + 3 = 0, x 2 − x − 4 5 = 0 are reduced, in each of which the leading coefficient is 1.

9 x 2 − x − 2 = 0- unreduced quadratic equation, where the first coefficient is different from 1 .

Any unreduced quadratic equation can be converted into a reduced equation by dividing both sides by the first coefficient (equivalent transformation). The transformed equation will have the same roots as the given unreduced equation or will also have no roots at all.

Consideration of a specific example will allow us to clearly demonstrate the transition from an unreduced quadratic equation to a reduced one.

Example 1

Given the equation 6 x 2 + 18 x − 7 = 0 . It is necessary to convert the original equation into the reduced form.

Solution

According to the above scheme, we divide both sides of the original equation by the leading coefficient 6. Then we get: (6 x 2 + 18 x − 7) : 3 = 0: 3, and this is the same as: (6 x 2) : 3 + (18 x) : 3 − 7: 3 = 0 and further: (6: 6) x 2 + (18: 6) x − 7: 6 = 0. From here: x 2 + 3 x - 1 1 6 = 0 . Thus, an equation equivalent to the given one is obtained.

Answer: x 2 + 3 x - 1 1 6 = 0 .

Complete and incomplete quadratic equations

Let's turn to the definition of a quadratic equation. In it we specified that a ≠ 0. A similar condition is necessary for the equation a x 2 + b x + c = 0 was precisely square, since at a = 0 it essentially transforms into a linear equation b x + c = 0.

In the case when the coefficients b And c are equal to zero (which is possible, both individually and jointly), the quadratic equation is called incomplete.

Definition 4

Incomplete quadratic equation- such a quadratic equation a x 2 + b x + c = 0, where at least one of the coefficients b And c(or both) is zero.

Complete quadratic equation– a quadratic equation in which all numerical coefficients are not equal to zero.

Let's discuss why the types of quadratic equations are given exactly these names.

When b = 0, the quadratic equation takes the form a x 2 + 0 x + c = 0, which is the same as a x 2 + c = 0. At c = 0 the quadratic equation is written as a x 2 + b x + 0 = 0, which is equivalent a x 2 + b x = 0. At b = 0 And c = 0 the equation will take the form a x 2 = 0. The equations that we obtained differ from the complete quadratic equation in that their left-hand sides do not contain either a term with the variable x, or a free term, or both. Actually, this fact gave the name to this type of equation – incomplete.

For example, x 2 + 3 x + 4 = 0 and − 7 x 2 − 2 x + 1, 3 = 0 are complete quadratic equations; x 2 = 0, − 5 x 2 = 0; 11 x 2 + 2 = 0, − x 2 − 6 x = 0 – incomplete quadratic equations.

Solving incomplete quadratic equations

The definition given above makes it possible to distinguish the following types of incomplete quadratic equations:

  • a x 2 = 0, this equation corresponds to the coefficients b = 0 and c = 0 ;
  • a · x 2 + c = 0 at b = 0 ;
  • a · x 2 + b · x = 0 at c = 0.

Let us consider sequentially the solution of each type of incomplete quadratic equation.

Solution of the equation a x 2 =0

As mentioned above, this equation corresponds to the coefficients b And c, equal to zero. The equation a x 2 = 0 can be converted into an equivalent equation x 2 = 0, which we get by dividing both sides of the original equation by the number a, not equal to zero. The obvious fact is that the root of the equation x 2 = 0 this is zero because 0 2 = 0 . This equation has no other roots, which can be explained by the properties of the degree: for any number p, Not equal to zero, the inequality is true p 2 > 0, from which it follows that when p ≠ 0 equality p 2 = 0 will never be achieved.

Definition 5

Thus, for the incomplete quadratic equation a x 2 = 0 there is a unique root x = 0.

Example 2

For example, let’s solve an incomplete quadratic equation − 3 x 2 = 0. It is equivalent to the equation x 2 = 0, its only root is x = 0, then the original equation has a single root - zero.

Briefly, the solution is written as follows:

− 3 x 2 = 0, x 2 = 0, x = 0.

Solving the equation a x 2 + c = 0

Next in line is the solution of incomplete quadratic equations, where b = 0, c ≠ 0, that is, equations of the form a x 2 + c = 0. Let's transform this equation by moving a term from one side of the equation to the other, changing the sign to the opposite one and dividing both sides of the equation by a number that is not equal to zero:

  • transfer c to the right hand side, which gives the equation a x 2 = − c;
  • divide both sides of the equation by a, we end up with x = - c a .

Our transformations are equivalent; accordingly, the resulting equation is also equivalent to the original one, and this fact makes it possible to draw conclusions about the roots of the equation. From what the values ​​are a And c the value of the expression - c a depends: it can have a minus sign (for example, if a = 1 And c = 2, then - c a = - 2 1 = - 2) or a plus sign (for example, if a = − 2 And c = 6, then - c a = - 6 - 2 = 3); it is not zero because c ≠ 0. Let us dwell in more detail on situations when - c a< 0 и - c a > 0 .

In the case when - c a< 0 , уравнение x 2 = - c a не будет иметь корней. Утверждая это, мы опираемся на то, что квадратом любого числа является число неотрицательное. Из сказанного следует, что при - c a < 0 ни для какого числа p the equality p 2 = - c a cannot be true.

Everything is different when - c a > 0: remember the square root, and it will become obvious that the root of the equation x 2 = - c a will be the number - c a, since - c a 2 = - c a. It is not difficult to understand that the number - - c a is also the root of the equation x 2 = - c a: indeed, - - c a 2 = - c a.

The equation will have no other roots. We can demonstrate this using the method of contradiction. To begin with, let us define the notations for the roots found above as x 1 And − x 1. Let us assume that the equation x 2 = - c a also has a root x 2, which is different from the roots x 1 And − x 1. We know that by substituting into the equation x its roots, we transform the equation into a fair numerical equality.

For x 1 And − x 1 we write: x 1 2 = - c a , and for x 2- x 2 2 = - c a . Based on the properties of numerical equalities, we subtract one correct equality term by term from another, which will give us: x 1 2 − x 2 2 = 0. We use the properties of operations with numbers to rewrite the last equality as (x 1 − x 2) · (x 1 + x 2) = 0. It is known that the product of two numbers is zero if and only if at least one of the numbers is zero. From the above it follows that x 1 − x 2 = 0 and/or x 1 + x 2 = 0, which is the same x 2 = x 1 and/or x 2 = − x 1. An obvious contradiction arose, because at first it was agreed that the root of the equation x 2 differs from x 1 And − x 1. So, we have proven that the equation has no roots other than x = - c a and x = - - c a.

Let us summarize all the arguments above.

Definition 6

Incomplete quadratic equation a x 2 + c = 0 is equivalent to the equation x 2 = - c a, which:

  • will have no roots at - c a< 0 ;
  • will have two roots x = - c a and x = - - c a for - c a > 0.

Let us give examples of solving the equations a x 2 + c = 0.

Example 3

Given a quadratic equation 9 x 2 + 7 = 0. It is necessary to find a solution.

Solution

Let's move the free term to the right side of the equation, then the equation will take the form 9 x 2 = − 7.
Let us divide both sides of the resulting equation by 9 , we arrive at x 2 = - 7 9 . On the right side we see a number with a minus sign, which means: y given equation no roots. Then the original incomplete quadratic equation 9 x 2 + 7 = 0 will have no roots.

Answer: the equation 9 x 2 + 7 = 0 has no roots.

Example 4

The equation needs to be solved − x 2 + 36 = 0.

Solution

Let's move 36 to the right side: − x 2 = − 36.
Let's divide both parts by − 1 , we get x 2 = 36. On the right side there is a positive number, from which we can conclude that x = 36 or x = - 36 .
Let's extract the root and write down the final result: incomplete quadratic equation − x 2 + 36 = 0 has two roots x=6 or x = − 6.

Answer: x=6 or x = − 6.

Solution of the equation a x 2 +b x=0

Let us analyze the third type of incomplete quadratic equations, when c = 0. To find a solution to an incomplete quadratic equation a x 2 + b x = 0, we will use the factorization method. Let's factorize the polynomial that is on the left side of the equation, taking the common factor out of brackets x. This step will make it possible to transform the original incomplete quadratic equation into its equivalent x (a x + b) = 0. And this equation, in turn, is equivalent to a set of equations x = 0 And a x + b = 0. The equation a x + b = 0 linear, and its root: x = − b a.

Definition 7

Thus, the incomplete quadratic equation a x 2 + b x = 0 will have two roots x = 0 And x = − b a.

Let's reinforce the material with an example.

Example 5

It is necessary to find a solution to the equation 2 3 · x 2 - 2 2 7 · x = 0.

Solution

We'll take it out x outside the brackets we get the equation x · 2 3 · x - 2 2 7 = 0 . This equation is equivalent to the equations x = 0 and 2 3 x - 2 2 7 = 0. Now you should solve the resulting linear equation: 2 3 · x = 2 2 7, x = 2 2 7 2 3.

Briefly write the solution to the equation as follows:

2 3 x 2 - 2 2 7 x = 0 x 2 3 x - 2 2 7 = 0

x = 0 or 2 3 x - 2 2 7 = 0

x = 0 or x = 3 3 7

Answer: x = 0, x = 3 3 7.

Discriminant, formula for the roots of a quadratic equation

To find solutions to quadratic equations, there is a root formula:

Definition 8

x = - b ± D 2 · a, where D = b 2 − 4 a c– the so-called discriminant of a quadratic equation.

Writing x = - b ± D 2 · a essentially means that x 1 = - b + D 2 · a, x 2 = - b - D 2 · a.

It would be useful to understand how this formula was derived and how to apply it.

Derivation of the formula for the roots of a quadratic equation

Let us be faced with the task of solving a quadratic equation a x 2 + b x + c = 0. Let us carry out a number of equivalent transformations:

  • divide both sides of the equation by a number a, different from zero, we obtain the following quadratic equation: x 2 + b a · x + c a = 0 ;
  • Let's select the complete square on the left side of the resulting equation:
    x 2 + b a · x + c a = x 2 + 2 · b 2 · a · x + b 2 · a 2 - b 2 · a 2 + c a = = x + b 2 · a 2 - b 2 · a 2 + c a
    After this, the equation will take the form: x + b 2 · a 2 - b 2 · a 2 + c a = 0;
  • Now it is possible to transfer the last two terms to the right side, changing the sign to the opposite, after which we get: x + b 2 · a 2 = b 2 · a 2 - c a ;
  • Finally, we transform the expression written on the right side of the last equality:
    b 2 · a 2 - c a = b 2 4 · a 2 - c a = b 2 4 · a 2 - 4 · a · c 4 · a 2 = b 2 - 4 · a · c 4 · a 2 .

Thus, we arrive at the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 , equivalent to the original equation a x 2 + b x + c = 0.

We examined the solution of such equations in the previous paragraphs (solving incomplete quadratic equations). The experience already gained makes it possible to draw a conclusion regarding the roots of the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2:

  • with b 2 - 4 a c 4 a 2< 0 уравнение не имеет действительных решений;
  • when b 2 - 4 · a · c 4 · a 2 = 0 the equation is x + b 2 · a 2 = 0, then x + b 2 · a = 0.

From here the only root x = - b 2 · a is obvious;

  • for b 2 - 4 · a · c 4 · a 2 > 0, the following will be true: x + b 2 · a = b 2 - 4 · a · c 4 · a 2 or x = b 2 · a - b 2 - 4 · a · c 4 · a 2 , which is the same as x + - b 2 · a = b 2 - 4 · a · c 4 · a 2 or x = - b 2 · a - b 2 - 4 · a · c 4 · a 2 , i.e. the equation has two roots.

It is possible to conclude that the presence or absence of roots of the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 (and therefore the original equation) depends on the sign of the expression b 2 - 4 · a · c 4 · a 2 written on the right side. And the sign of this expression is given by the sign of the numerator, (denominator 4 a 2 will always be positive), that is, the sign of the expression b 2 − 4 a c. This expression b 2 − 4 a c the name is given - the discriminant of the quadratic equation and the letter D is defined as its designation. Here you can write down the essence of the discriminant - based on its value and sign, they can conclude whether the quadratic equation will have real roots, and, if so, what is the number of roots - one or two.

Let's return to the equation x + b 2 · a 2 = b 2 - 4 · a · c 4 · a 2 . Let's rewrite it using discriminant notation: x + b 2 · a 2 = D 4 · a 2 .

Let us formulate our conclusions again:

Definition 9

  • at D< 0 the equation has no real roots;
  • at D=0 the equation has a single root x = - b 2 · a ;
  • at D > 0 the equation has two roots: x = - b 2 · a + D 4 · a 2 or x = - b 2 · a - D 4 · a 2. Based on the properties of radicals, these roots can be written in the form: x = - b 2 · a + D 2 · a or - b 2 · a - D 2 · a. And, when we open the modules and bring the fractions to a common denominator, we get: x = - b + D 2 · a, x = - b - D 2 · a.

So, the result of our reasoning was the derivation of the formula for the roots of a quadratic equation:

x = - b + D 2 a, x = - b - D 2 a, discriminant D calculated by the formula D = b 2 − 4 a c.

These formulas make it possible to determine both real roots when the discriminant is greater than zero. When the discriminant is zero, applying both formulas will give the same root as the only solution to the quadratic equation. In the case where the discriminant is negative, if we try to use the quadratic root formula, we will be faced with the need to take the square root of a negative number, which will take us beyond the scope of real numbers. With a negative discriminant, the quadratic equation will not have real roots, but a pair of complex conjugate roots is possible, determined by the same root formulas we obtained.

Algorithm for solving quadratic equations using root formulas

It is possible to solve a quadratic equation by immediately using the root formula, but this is generally done when it is necessary to find complex roots.

In the majority of cases, it usually means searching not for complex, but for real roots of a quadratic equation. Then it is optimal, before using the formulas for the roots of a quadratic equation, to first determine the discriminant and make sure that it is not negative (otherwise we will conclude that the equation has no real roots), and then proceed to calculate the value of the roots.

The reasoning above makes it possible to formulate an algorithm for solving a quadratic equation.

Definition 10

To solve a quadratic equation a x 2 + b x + c = 0, necessary:

  • according to the formula D = b 2 − 4 a c find the discriminant value;
  • at D< 0 сделать вывод об отсутствии у квадратного уравнения действительных корней;
  • for D = 0, find the only root of the equation using the formula x = - b 2 · a ;
  • for D > 0, determine two real roots of the quadratic equation using the formula x = - b ± D 2 · a.

Note that when the discriminant is zero, you can use the formula x = - b ± D 2 · a, it will give the same result as the formula x = - b 2 · a.

Let's look at examples.

Examples of solving quadratic equations

Let us give solutions to examples for different values ​​of the discriminant.

Example 6

We need to find the roots of the equation x 2 + 2 x − 6 = 0.

Solution

Let's write down the numerical coefficients of the quadratic equation: a = 1, b = 2 and c = − 6. Next we proceed according to the algorithm, i.e. Let's start calculating the discriminant, for which we will substitute the coefficients a, b And c into the discriminant formula: D = b 2 − 4 · a · c = 2 2 − 4 · 1 · (− 6) = 4 + 24 = 28 .

So we get D > 0, which means that the original equation will have two real roots.
To find them, we use the root formula x = - b ± D 2 · a and, substituting the corresponding values, we get: x = - 2 ± 28 2 · 1. Let us simplify the resulting expression by taking the factor out of the root sign and then reducing the fraction:

x = - 2 ± 2 7 2

x = - 2 + 2 7 2 or x = - 2 - 2 7 2

x = - 1 + 7 or x = - 1 - 7

Answer: x = - 1 + 7 ​​​​​​, x = - 1 - 7 .

Example 7

Need to solve a quadratic equation − 4 x 2 + 28 x − 49 = 0.

Solution

Let's define the discriminant: D = 28 2 − 4 · (− 4) · (− 49) = 784 − 784 = 0. With this value of the discriminant, the original equation will have only one root, determined by the formula x = - b 2 · a.

x = - 28 2 (- 4) x = 3.5

Answer: x = 3.5.

Example 8

The equation needs to be solved 5 y 2 + 6 y + 2 = 0

Solution

The numerical coefficients of this equation will be: a = 5, b = 6 and c = 2. We use these values ​​to find the discriminant: D = b 2 − 4 · a · c = 6 2 − 4 · 5 · 2 = 36 − 40 = − 4 . The calculated discriminant is negative, so the original quadratic equation has no real roots.

In the case when the task is to indicate complex roots, we apply the root formula, performing actions with complex numbers:

x = - 6 ± - 4 2 5,

x = - 6 + 2 i 10 or x = - 6 - 2 i 10,

x = - 3 5 + 1 5 · i or x = - 3 5 - 1 5 · i.

Answer: there are no real roots; the complex roots are as follows: - 3 5 + 1 5 · i, - 3 5 - 1 5 · i.

IN school curriculum There is no standard requirement to look for complex roots, therefore, if during the solution the discriminant is determined to be negative, the answer is immediately written down that there are no real roots.

Root formula for even second coefficients

The root formula x = - b ± D 2 · a (D = b 2 − 4 · a · c) makes it possible to obtain another formula, more compact, allowing one to find solutions to quadratic equations with an even coefficient for x (or with a coefficient of the form 2 · n, for example, 2 3 or 14 ln 5 = 2 7 ln 5). Let us show how this formula is derived.

Let us be faced with the task of finding a solution to the quadratic equation a · x 2 + 2 · n · x + c = 0 . We proceed according to the algorithm: we determine the discriminant D = (2 n) 2 − 4 a c = 4 n 2 − 4 a c = 4 (n 2 − a c), and then use the root formula:

x = - 2 n ± D 2 a, x = - 2 n ± 4 n 2 - a c 2 a, x = - 2 n ± 2 n 2 - a c 2 a, x = - n ± n 2 - a · c a .

Let the expression n 2 − a · c be denoted as D 1 (sometimes it is denoted D "). Then the formula for the roots of the quadratic equation under consideration with the second coefficient 2 · n will take the form:

x = - n ± D 1 a, where D 1 = n 2 − a · c.

It is easy to see that D = 4 · D 1, or D 1 = D 4. In other words, D 1 is a quarter of the discriminant. Obviously, the sign of D 1 is the same as the sign of D, which means the sign of D 1 can also serve as an indicator of the presence or absence of roots of a quadratic equation.

Definition 11

Thus, to find a solution to a quadratic equation with a second coefficient of 2 n, it is necessary:

  • find D 1 = n 2 − a · c ;
  • at D 1< 0 сделать вывод, что действительных корней нет;
  • when D 1 = 0, determine the only root of the equation using the formula x = - n a;
  • for D 1 > 0, determine two real roots using the formula x = - n ± D 1 a.

Example 9

It is necessary to solve the quadratic equation 5 x 2 − 6 x − 32 = 0.

Solution

We can represent the second coefficient of the given equation as 2 · (− 3) . Then we rewrite the given quadratic equation as 5 x 2 + 2 (− 3) x − 32 = 0, where a = 5, n = − 3 and c = − 32.

Let's calculate the fourth part of the discriminant: D 1 = n 2 − a · c = (− 3) 2 − 5 · (− 32) = 9 + 160 = 169. The resulting value is positive, which means that the equation has two real roots. Let us determine them using the corresponding root formula:

x = - n ± D 1 a, x = - - 3 ± 169 5, x = 3 ± 13 5,

x = 3 + 13 5 or x = 3 - 13 5

x = 3 1 5 or x = - 2

It would be possible to carry out calculations using the usual formula for the roots of a quadratic equation, but in this case the solution would be more cumbersome.

Answer: x = 3 1 5 or x = - 2 .

Simplifying the form of quadratic equations

Sometimes it is possible to optimize the form of the original equation, which will simplify the process of calculating the roots.

For example, the quadratic equation 12 x 2 − 4 x − 7 = 0 is clearly more convenient to solve than 1200 x 2 − 400 x − 700 = 0.

More often, simplification of the form of a quadratic equation is carried out by multiplying or dividing its both sides by a certain number. For example, above we showed a simplified representation of the equation 1200 x 2 − 400 x − 700 = 0, obtained by dividing both sides by 100.

Such a transformation is possible when the coefficients of the quadratic equation are not mutually prime numbers. Then we usually divide both sides of the equation by the largest common divisor absolute values ​​of its coefficients.

As an example, we use the quadratic equation 12 x 2 − 42 x + 48 = 0. Let us determine the GCD of the absolute values ​​of its coefficients: GCD (12, 42, 48) = GCD(GCD (12, 42), 48) = GCD (6, 48) = 6. Let us divide both sides of the original quadratic equation by 6 and obtain the equivalent quadratic equation 2 x 2 − 7 x + 8 = 0.

By multiplying both sides of a quadratic equation, you usually get rid of fractional coefficients. In this case, they multiply by the least common multiple of the denominators of its coefficients. For example, if each part of the quadratic equation 1 6 x 2 + 2 3 x - 3 = 0 is multiplied with LCM (6, 3, 1) = 6, then it will become written in more in simple form x 2 + 4 x − 18 = 0 .

Finally, we note that we almost always get rid of the minus at the first coefficient of a quadratic equation by changing the signs of each term of the equation, which is achieved by multiplying (or dividing) both sides by − 1. For example, from the quadratic equation − 2 x 2 − 3 x + 7 = 0, you can go to its simplified version 2 x 2 + 3 x − 7 = 0.

Relationship between roots and coefficients

The formula for the roots of quadratic equations, already known to us, x = - b ± D 2 · a, expresses the roots of the equation through its numerical coefficients. Based on this formula, we have the opportunity to specify other dependencies between the roots and coefficients.

The most famous and applicable formulas are Vieta’s theorem:

x 1 + x 2 = - b a and x 2 = c a.

In particular, for the given quadratic equation, the sum of the roots is the second coefficient with the opposite sign, and the product of the roots is equal to the free term. For example, by looking at the form of the quadratic equation 3 x 2 − 7 x + 22 = 0, it is possible to immediately determine that the sum of its roots is 7 3 and the product of the roots is 22 3.

You can also find a number of other connections between the roots and coefficients of a quadratic equation. For example, the sum of the squares of the roots of a quadratic equation can be expressed in terms of coefficients:

x 1 2 + x 2 2 = (x 1 + x 2) 2 - 2 x 1 x 2 = - b a 2 - 2 c a = b 2 a 2 - 2 c a = b 2 - 2 a c a 2.

If you notice an error in the text, please highlight it and press Ctrl+Enter

I came up with such a cool theorem for them,
and they decide through the discriminant:-(((
(c) Francois Viet “Non-existent statements”

Root formula, or the long way

Everyone who attended even the slightest bit of mathematics lessons in the 8th grade knows the formula for the roots of a quadratic equation. The solution using the root formula is often called in common parlance “the solution through the discriminant.” Let us briefly recall the formula for roots.

[You can also view the contents of this article at video format ]

The quadratic equation has the form ax 2 +bx+c= 0, where a, b, c- some numbers. For example, in Eq. 2x 2 + 3x – 5 = 0 these numbers are equal: a = 2, b = 3. c= -5. Before solving any quadratic equation, you need to “see” these numbers and understand what they equal.

Next, the so-called discriminant is calculated using the formula D=b^2-4ac. In our case D = 3^2 – 4 \cdot 2 \cdot (-5) = 9 + 40 = 49. Then the root is extracted from the discriminant: \sqrt(D) = \sqrt(49) = 7 .

After the discriminant has been calculated, the root formula is used: x_1=\frac(-b-\sqrt(D))(2a); x_2=\frac(-b+\sqrt(D))(2a):

x_1=\frac(-3-7)(2 \cdot 2)=\frac(-10)(4)=-2.5
x_2= \frac(-3+7)(2 \cdot 2)=\frac(4)(4)=1

And thus, the equation is solved. It has two roots: 1 and -2.5.

But this equation, like many others proposed in school textbooks/problems, could be solved much more in a fast way, if you know a couple of life hacks. And we’re not just talking about Vieta’s theorem, although it is a useful tool.

Life hack first. If a + b + c= 0, then x_1=1, x_2=\frac(c)(a) .

It only applies if all three coefficients in a quadratic equation are a, b, c when added they give 0. For example, we had the equation 2x 2 + 3x – 5 = 0 . Adding all three coefficients, we get 2 + 3 – 5, which is equal to 0. In this case, you can not count the discriminant and not apply the root formula. Instead, you can immediately write that

x_1=1,
x_2=\frac(c)(a)=\frac(-5)(2)=-2.5

(note that we got the same result in the roots formula).

People often ask whether x_1=1 will always work? Yes, whenever a + b + c = 0.

Life hack second. If a + c = b, then x_1=-1, x_2=-\frac(c)(a) .

Let the equation be given 5x 2 + 6x + 1 = 0 . In him a = 5, b = 6, c= 1. If we add up the “extreme” coefficients a And c, we get 5+1 = 6, which is exactly equal to the “average” coefficient b. This means we can do without a discriminant! We immediately write down:

x_1=-1,
x_2=-\frac(c)(a)=\frac(-1)(5)=-0.2

Life hack third(the theorem inverse to Vieta’s theorem). If a= 1, then

Consider the equation x 2 – 12x+ 35 = 0. It contains a = 1, b = -12, c = 35. It does not fit either the first or the second life hack - the conditions are not met. If it fit the first or second, then we would do without Vieta’s theorem.

The very use of Vieta's theorem implies an understanding of some useful techniques.

First appointment. Don’t be shy about writing down the view system itself \begin(cases) x_1+x_2 = -b \\ x_1 \cdot x_2 = c \end(cases), which is obtained by using Vieta’s theorem. There is no need to try at all costs to solve the equation absolutely orally, without written notes, as “advanced users” do.

For our equation x 2 – 12x+ 35 = 0 this system has the form

\begin(cases) x_1+x_2 = 12 \\ x_1 \cdot x_2 = 35 \end(cases)

Now we need to verbally select the numbers x_1 and x_2 that satisfy our system, i.e. the total is 12, and when multiplied it is 35.

So, second appointment is that you need to start the selection not with the sum, but with the product. Let's look at the second equation of the system and ask ourselves: what numbers, when multiplied, give 35? If everything is in order with the multiplication table, then the answer immediately comes to mind: 7 and 5. And only now let’s substitute these numbers into the first equation: we will have 7 + 5 = 12, which is true equality. So, the numbers 7 and 5 satisfy both equations, so we immediately write:

x_1 = 7, x_2 = 5

Third reception is that if the numbers cannot be found quickly (within 15-20 seconds), then, regardless of the reason, you need to calculate the discriminant and use the root formula. Why? Because the roots may not be found if the equation does not have them at all (the discriminant is negative), or the roots are numbers that are not integers.

Training exercises for solving quadratic equations

Practice! Try solving the following equations. Look at each equation in the following order:

  • if the equation fits the first life hack (when a + b + c = 0), then we solve it with its help;
  • if the equation fits the second life hack (when a + c = b), then we solve it with its help;
  • if the equation fits the third life hack (Viete’s theorem), we solve it with its help;
  • and only in the most extreme case - if nothing fits and/or it was not possible to solve using Vieta’s theorem - we calculate the discriminant. Again: discriminant - last but not least!
  1. Solve the equation x 2 + 3x + 2 = 0
    View solution and answer

    See lifehack two
    In this equation, a = 1, b = 3, c = 2. Thus, a + c = b, whence x_1=-1, x_2 = -\frac(c)(a) = -\frac(2)(1)=-2.
    Answer: -1, -2.

  2. Solve the equation x 2 + 8x – 9 = 0
    View solution and answer

    See life hack first
    In this equation, a = 1, b = 8, c = -9. Thus, a + b + c = 0, whence x_1=1, x_2 = \frac(c)(a) = \frac(-9)(1)=-9.
    Answer: 1, -9.

  3. Solve the equation 15x 2 – 11x + 2 = 0
    View solution and answer

    This equation (the only one from the entire list) does not fall under any of the life hacks, so we will solve it using the root formula:
    D=b^2-4ac = (-11)^2 – 4 \cdot 15 \cdot 2 = 121 – 120 = 1.x_1=\frac(11-1)(2 \cdot 15)=\frac(10)(30)=\frac(1)(3)x_2= \frac(11+1)(2 \cdot 15)=\frac(12)(30)=\frac(2)(5) Answer: \frac(1)(3), \frac(2)(5).

  4. Solve the equation x 2 + 9x + 20 = 0
    View solution and answer


    \begin(cases) x_1+x_2 = -9 \\ x_1 \cdot x_2 = 20 \end(cases)
    By selection we establish that x_1 = -4, x_2 = -5.
    Answer: -4, -5.

  5. Solve the equation x 2 – 7x – 30 = 0
    View solution and answer

    See life hack three (Vieta’s theorem)
    In this equation a = 1, so we can write that \begin(cases) x_1+x_2 = 7 \\ x_1 \cdot x_2 = -30 \end(cases)
    By selection we establish that x_1 = 10, x_2 = -3.
    Answer: 10, -3.

  6. Solve the equation x 2 – 19x + 18 = 0
    View solution and answer

    See life hack first
    In this equation, a = 1, b = -19, c = 18. Thus, a + b + c = 0, whence x_1=1, x_2 = \frac(c)(a) = \frac(18)(1)=18.
    Answer: 1, 18.

  7. Solve the equation x 2 + 7x + 6 = 0
    View solution and answer

    See lifehack two
    In this equation, a = 1, b = 7, c = 6. Thus, a + c = b, whence x_1=-1, x_2 = -\frac(c)(a) = -\frac(6)(1)=-6.
    Answer: -1, -6.

  8. Solve the equation x 2 – 8x + 12 = 0
    View solution and answer

    See life hack three (Vieta’s theorem)
    In this equation a = 1, so we can write that \begin(cases) x_1+x_2 = 8 \\ x_1 \cdot x_2 = 12 \end(cases)
    By selection we establish that x_1 = 6, x_2 = 2.
    Answer: 6, 2.

  9. Solve the equation x 2 – x – 6 = 0
    View solution and answer

    See life hack three (Vieta’s theorem)
    In this equation a = 1, so we can write that \begin(cases) x_1+x_2 = 1 \\ x_1 \cdot x_2 = -6 \end(cases)
    By selection we establish that x_1 = 3, x_2 = -2.
    Answer: 3, -2.

  10. Solve the equation x 2 – 15x – 16 = 0
    View solution and answer

    See lifehack two
    In this equation, a = 1, b = -15, c = -16. Thus, a + c = b, whence x_1=-1, x_2 = -\frac(c)(a) = -\frac(-16)(1)=16.
    Answer: -1, 16.

  11. Solve the equation x 2 + 11x – 12 = 0
    View solution and answer

    See life hack first
    In this equation, a = 1, b = 11, c = -12. Thus, a + b + c = 0, whence x_1=1, x_2 = \frac(c)(a) = \frac(-12)(1)=-12.
    Answer: 1, -12.