Discriminant from 1681. Solution of square equations, root formula, examples

I hope that studying this article, you will learn to find the roots of a complete square equation.

With the help of discriminant, only complete square equations are solved, for solve square equations, other methods that you find in the article "Decision of incomplete square equations" are used.

What square equations are called full? it equations of the form ah 2 + b x + c \u003d 0where the coefficients a, b and are not equal to zero. So, to solve a complete square equation, it is necessary to calculate the discriminant D.

D \u003d b 2 - 4As.

Depending on what kind of importance is discriminant, we will write the answer.

If discriminant is a negative number (D< 0),то корней нет.

If the discriminant is zero, x \u003d (-b) / 2a. When the discriminant is a positive number (D\u003e 0),

then x 1 \u003d (-b - √d) / 2a, and x 2 \u003d (-b + √d) / 2a.

For example. Solve equation x 2 - 4x + 4 \u003d 0.

D \u003d 4 2 - 4 · 4 \u003d 0

x \u003d (- (-4)) / 2 \u003d 2

Answer: 2

Solve equation 2. x 2 + x + 3 \u003d 0.

D \u003d 1 2 - 4 · 2 · 3 \u003d - 23

Answer: No roots.

Solve equation 2. x 2 + 5x - 7 \u003d 0.

D \u003d 5 2 - 4 · 2 · (-7) \u003d 81

x 1 \u003d (-5 - √81) / (2 · 2) \u003d (-5 - 9) / 4 \u003d - 3.5

x 2 \u003d (-5 + √81) / (2 · 2) \u003d (-5 + 9) / 4 \u003d 1

Answer: - 3.5; one.

So let's imagine the solution of complete square equations by the scheme in Figure1.

According to these formulas, you can solve any complete square equation. You only need to carefully monitor the equation was recorded by a polynomial of a standard type.

but x 2 + BX + C, Otherwise you can make an error. For example, in the record of the equation x + 3 + 2x 2 \u003d 0, it is erroneously can be solved that

a \u003d 1, b \u003d 3 and c \u003d 2. then

D \u003d 3 2 - 4 · 1 · 2 \u003d 1 and then the equation has two roots. And this is incorrect. (See the solution of Example 2 above).

Therefore, if the equation is not written not to a polynomial of a standard species, at first a complete square equation should be recorded by a polynomial of a standard species (in the first place should be unrocked with the greatest indicator, that is but x 2 then with smaller – bX.and then free dick from.

When solving a given square equation and a square equation with an even coefficient, with the second term, other formulas can be used. Let's get acquainted with these formulas. If in a complete square equation in the second term, the coefficient will be even (b \u003d 2k), then the equation according to the formulas in the figure 2 can be solved.

The full square equation is called the above, if the coefficient is x 2 equal to one and the equation will take the form x 2 + px + q \u003d 0. Such an equation can be given to solve, or is obtained by dividing all coefficients to the coefficient equation butstanding for x 2 .

Figure 3 shows the scheme of solving the above square
equations. Consider on the example the application of the formulas considered in this article.

Example. Solve equation

3x 2 + 6x - 6 \u003d 0.

Let's decide this equation using the formulas shown in the Figure 1 scheme.

D \u003d 6 2 - 4 · 3 · (- 6) \u003d 36 + 72 \u003d 108

√D \u003d √108 \u003d √ (36 · 3) \u003d 6√3

x 1 \u003d (-6 - 6√3) / (2 · 3) \u003d (6 (-1- √ (3))) / 6 \u003d -1 - √3

x 2 \u003d (-6 + 6√3) / (2 · 3) \u003d (6 (-1+ √ (3))) / 6 \u003d -1 + √3

Answer: -1 - √3; -1 + √3

It can be seen that the coefficient at x in this equation is an even number, that is, b \u003d 6 or b \u003d 2k, from where k \u003d 3. Then we try to solve the equation according to the formulas shown in the diagram D 1 \u003d 3 2 - 3 · (- 6 ) \u003d 9 + 18 \u003d 27

√ (D 1) \u003d √27 \u003d √ (9 · 3) \u003d 3√3

x 1 \u003d (-3 - 3√3) / 3 \u003d (3 (-1 - √ (3))) / 3 \u003d - 1 - √3

x 2 \u003d (-3 + 3√3) / 3 \u003d (3 (-1 + √ (3))) / 3 \u003d - 1 + √3

Answer: -1 - √3; -1 + √3. Noticed that all coefficients in this square equation are divided into 3 and by performing division, we obtain the reduced square equation x 2 + 2x - 2 \u003d 0 by solving this equation using formulas for the specified square
equations Figure 3.

D 2 \u003d 2 2 - 4 · (- 2) \u003d 4 + 8 \u003d 12

√ (D 2) \u003d √12 \u003d √ (4 · 3) \u003d 2√3

x 1 \u003d (-2 - 2√3) / 2 \u003d (2 (-1 - √ (3))) / 2 \u003d - 1 - √3

x 2 \u003d (-2 + 2√3) / 2 \u003d (2 (-1+ √ (3))) / 2 \u003d - 1 + √3

Answer: -1 - √3; -1 + √3.

As we see, when solving this equation on various formulas, we received the same answer. Therefore, it is well aware of the formulas shown in the Figure 1 scheme, you can always solve any complete square equation.

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The challenges per square equation are studied in the school program and in universities. Beneath them understand the equations of the form a * x ^ 2 + b * x + c \u003d 0, where x - variable, a, b, c - constants; A.<>0. The task is to find the roots of the equation.

Geometric meaning of the square equation

The graph of the function, which is represented by the square equation is parabola. Solutions (roots) of the square equation are the points of intersection of the parabola with the abscissa axis (x). It follows from this that there are three possible cases:
1) Parabola has no intersection points with an abscissa axis. This means that it is in the upper plane with branches up or bottom with branches down. In such cases, the square equation has no valid roots (has two complex roots).

2) Parabola has one intersection point with the axis oh. Such a point is called the pearabol vertex, and the square equation in it acquires its minimum or maximum value. In this case, the square equation has one valid root (or two identical root).

3) The last case in practice is interesting more - there are two points of intersection of parabola with the abscissa axis. This means that there are two valid equation root.

Based on the analysis of coefficients in the degrees of variables, it is possible to make interesting conclusions about the placement of parabola.

1) If the coefficient is the more zero, the parabola is directed upwards, if negative - the parabola branches are directed down.

2) If the coefficient b is greater than zero, then the top of the parabola lies in the left half-plane, if it takes a negative value - then in the right.

Output of the formula for solving a square equation

We transfer the constant from the square equation

per sign of equality, we get expression

Multiply both parts on 4a

To get the left of the full square add in both parts b ^ 2 and implement the transformation

From here to find

The formula of the discriminant and the roots of the square equation

The discriminant is called the value of the conditioned expression, it is positive, the equation has two valid roots calculated by the formula At a zero discriminant, the square equation has one solution (two coinciding root), which can be easily obtained from the above formula for d \u003d 0 with a negative discriminant of the equation of valid roots. However, to maintain the solutions of the square equation in the complex plane, and their value is calculated by the formula

Vieta theorem

Consider two roots of the square equation and construct on their basis the square equation. The record itself is easily followed by the Vieta Theorem itself: if we have a square equation of type the sum of its roots is equal to the P coefficient, taken with the opposite sign, and the product of the equation's roots is equal to the free term Q. The formulae record of the above will have seen in the classic equation of a constant A is different from zero, then all equation should be divided into it, and then apply the theorem of the Vieta.

Schedule of a square equation for multipliers

Let the task: decompose the square equation on multipliers. To fulfill it, we first solve the equation (we find the roots). Further, the roots found substituted in the decomposition formula of the square equation this task will be allowed.

Square equation

Task 1. Find the roots of the square equation

x ^ 2-26x + 120 \u003d 0.

Solution: We write the coefficients and substitute in the formula of the discriminant

The root of this value is 14, it is easy to find it with a calculator, or remember with frequent use, however, for convenience, at the end of the article, I will give you a list of numbers squares that can often meet with such tasks.
The foundation is substituted in the root formula

And get

Task 2. Solve equation

2x 2 + x-3 \u003d 0.

Solution: we have a complete square equation, we write out the coefficients and find the discriminant


According to famous formulas we find the roots of the square equation

Task 3. Solve equation

9x 2 -12x + 4 \u003d 0.

Solution: We have a complete square equation. Determine discriminant

We received a case when the roots coincide. Find the values \u200b\u200bof the roots by the formula

Task 4. Solve equation

x ^ 2 + x-6 \u003d 0.

Solution: In cases where there are small coefficients at x it is advisable to apply the theorem of the Vieta. According to her, we get two equations

From the second condition, we get that the work should be equal to -6. This means that one of the roots is negative. We have the following possible pair of solutions (-3; 2), (3; -2). Taking into account the first condition, the second pair of solutions reject.
Root equations are equal

Task 5. Find the lengths of the side of the rectangle, if its perimeter is 18 cm, and the area is 77 cm 2.

Solution: Half of the perimeter of the rectangle is equal to the sum of neighboring sides. Denote by x - most side, then 18-x is a smaller side. The area of \u200b\u200bthe rectangle is equal to the product of these lengths:
x (18-x) \u003d 77;
or
x 2 -18x + 77 \u003d 0.
We find the discriminant of the equation

Calculate the roots of the equation

If a x \u003d 11,that 18h \u003d 7, On the contrary, it is also true (if x \u003d 7, then 21-x \u003d 9).

Task 6. Square square 10x 2 -11x + 3 \u003d 0 Equations for multipliers.

Solution: Calculate the roots of the equation, for this we find discriminant

We substitute the value found in the root formula and calculate

Apply the decomposition formula of the square equation along the roots

The layout of the bracket will receive identity.

Square equation with parameter

Example 1. Under what values \u200b\u200bof the parameter but , Equation (A-3) x 2 + (3-A) x-1/4 \u003d 0 has one root?

Solution: a direct substitution of the value A \u003d 3 we see that it has no solution. Next, we use that at zero discriminant, the equation has one root of multiplicity 2. Drink discriminant

simplify it and equate to zero

Received a square equation on the parameter A, the solution of which is easy to obtain on the Vieta theorem. The amount of the roots is 7, and their work 12. Simple bust by installing that the numbers 3.4 will be rooted equations. Since the solution A \u003d 3, we already rejected at the beginning of the calculations, the only right will be - a \u003d 4.Thus, when A \u003d 4, the equation has one root.

Example 2. Under what values \u200b\u200bof the parameter but , the equation a (a + 3) x ^ 2 + (2a + 6) x-3a-9 \u003d 0has more than one root?

Solution: Consider first singular points, they will be values \u200b\u200ba \u003d 0 and a \u003d -3. When A \u003d 0, the equation will be simplified to the form 6x-9 \u003d 0; x \u003d 3/2 and there will be one root. When A \u003d -3, we obtain the identity 0 \u003d 0.
Calculate discriminant

and find values \u200b\u200band in which it is positive

From the first condition we will get a\u003e 3. For the second we find the discriminant and roots of the equation


We define the gaps where the function takes positive values. Figure point a \u003d 0 Get 3>0 . So, beyond the interval (-3; 1/3) the function is negative. Don't forget about the point a \u003d 0,this should be excluded because the initial equation in it has one root.
As a result, we obtain two intervals that satisfy the condition of the task

There will be many similar tasks in practice, try to deal with the tasks yourself and do not forget to consider the conditions that are mutually exclusive. Well read the formula for solving square equations, they are often needed when calculating in different tasks and sciences.

For example, for three-shots \\ (3x ^ 2 + 2x-7 \\), the discriminant will be equal to \\ (2 ^ 2-4 \\ Cdot3 \\ Cdot (-7) \u003d 4 + 84 \u003d 88 \\). And for three-shots \\ (x ^ 2-5x + 11 \\), it will be equal to \\ ((- 5) ^ 2-4 \\ cdot1 \\ cdot11 \u003d 25-44 \u003d -19 \\).

The discriminant is indicated by the letter \\ (D \\) and is often used when solving. Also, the value of the discriminant can be understood how the schedule looks like something (see below).

Discriminant and roots equation

The value of the discriminant shows the number of square equation:
- if \\ (D \\) is positive - the equation will have two roots;
- if \\ (D \\) is zero - only one root;
- If \\ (D \\) is negative - no roots.

This is not necessary to learn, it is easy to come to this conclusion, just knowing that from the discriminant (that is, \\ (\\ sqrt (D) \\) is included in the formula for calculating the roots of the equation: \\ (x_ (1) \u003d \\) \\ (\\ .

If discriminant is positive

In this case, the root of it is some positive number, and therefore \\ (x_ (1) \\) and \\ (x_ (2) \\) will be different by value, because in the first formula \\ (\\ sqrt (d) \\) adds , and in the second - subtracted. And we have two different roots.

Example : Find the roots of the equation \\ (x ^ 2 + 2x-3 \u003d 0 \\)
Decision :

Answer : \\ (x_ (1) \u003d 1 \\); \\ (x_ (2) \u003d - 3 \\)

If the discriminant is zero

And how many roots will it be if the discriminant is zero? Let's talk.

The root formulas look like this: \\ (x_ (1) \u003d \\) \\ (\\ FRAC (-b + \\ sqrt (d)) (2a) \\) and \\ (x_ (2) \u003d \\) \\ (\\ FRAC (-b- \\ SQRT (D)) (2A) \\). And if the discriminant is zero, then the root of it is also zero. Then it turns out:

\\ (X_ (1) \u003d \\) \\ (\\ FRAC (-B + \\ SQRT (D)) (2A) \\) \\ (\u003d \\) \\ (\\ FRAC (-B + \\ SQRT (0)) (2A) \\) \\ (\u003d \\) \\ (\\ FRAC (-b + 0) (2a) \\) \\ (\u003d \\) \\ (\\ FRAC (-B) (2A) \\)

\\ (X_ (2) \u003d \\) \\ (\\ FRAC (-B- \\ SQRT (D)) (2A) \\) \\ (\u003d \\) \\ (\\ FRAC (-B- \\ SQRT (0)) (2A) \\) \\ (\u003d \\) \\ (\\ FRAC (-B-0) (2A) \\) \\ (\u003d \\) \\ (\\ FRAC (-B) (2A) \\)

That is, the values \u200b\u200bof the roots of the equation will coincide, because the addition or subtraction of zero does not change anything.

Example : Find the roots of the equation \\ (x ^ 2-4x + 4 \u003d 0 \\)
Decision :

Answer : \\ (x \u003d 2 \\)

Square equation is an equation that looks like aX 2 + DX + C \u003d 0. In it, the value a, B. and from any numbers but Not equally zero.

All square equations are divided into several species, namely:

Equations in which only one root.
-Evaluation with two different roots.
-Evaluation in which there are no roots at all.

This distinguishes linear equations in which the root is always united, from square. In order to understand how much the number of roots in the expression and need Discriminant square equation.

Let's say our equation AX 2 + DX + C \u003d 0. So Discriminant square equation -

D \u003d b 2 - 4 AC

And it needs to be remembered forever. With this equation, we determine the number of roots in the square equation. And we do this as follows:

When D is less than zero, there are no roots in the equation.
- When D is zero, there is only one root.
- When D is larger, respectively, in the two root equation.
Remember that the discriminant shows how many roots in the equation, without changing signs.

Consider for clarity:

It is necessary to find out what the number of roots in this square equation.

1) x 2 - 8x + 12 \u003d 0
2) 5x 2 + 3x + 7 \u003d 0
3) x 2 -6x + 9 \u003d 0

Enter the values \u200b\u200bin the first equation, we find the discriminant.
a \u003d 1, b \u003d -8, c \u003d 12
D \u003d (-8) 2 - 4 * 1 * 12 \u003d 64 - 48 \u003d 16
Discriminant with a plus sign, which means two roots in this equality.

Do the same with the second equation
a \u003d 1, b \u003d 3, c \u003d 7
D \u003d 3 2 - 4 * 5 * 7 \u003d 9 - 140 \u003d - 131
The value is minus, which means no roots in this equality.

The following equation is decomposable by analogy.
a \u003d 1, b \u003d -6, c \u003d 9
D \u003d (-6) 2 - 4 * 1 * 9 \u003d 36 - 36 \u003d 0
As a result, we have one root in the equation.

It is important that in each equation we discharged the coefficients. Of course, this is not a lot of a long process, but it helped us not to get confused and prevented the appearance of errors. If you often solve such equations, then the calculations can be made mentally and in advance to know how many roots in the equation.

Consider another example:

1) x 2 - 2x - 3 \u003d 0
2) 15 - 2x - x 2 \u003d 0
3) x 2 + 12x + 36 \u003d 0

Unlock first
a \u003d 1, b \u003d -2, c \u003d -3
D \u003d (- 2) 2 - 4 * 1 * (-3) \u003d 16, which is more zero, then two roots, bring them
x 1 \u003d 2+? 16/2 * 1 \u003d 3, x 2 \u003d 2-? 16/2 * 1 \u003d -1.

We declare second
a \u003d -1, b \u003d -2, c \u003d 15
D \u003d (-2) 2 - 4 * 4 * (-1) * 15 \u003d 64, which is more zero and also has two roots. Let's bring them:
x 1 \u003d 2+? 64/2 * (-1) \u003d -5, x 2 \u003d 2-? 64/2 * (- 1) \u003d 3.

Unlock the third
a \u003d 1, b \u003d 12, c \u003d 36
D \u003d 12 2 - 4 * 1 * 36 \u003d 0, which is zero and has one root
x \u003d -12 +? 0/2 * 1 \u003d -6.
It is not difficult to solve these equations.

If we are given an incomplete square equation. Such as

1x 2 + 9x \u003d 0
2x 2 - 16 \u003d 0

These equations differ from those that were higher, as it is not complete, there is no third value in it. But despite this it is easier than a complete square equation and does not need to search for a discriminant.

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