Square equations - preparation for the exam in mathematics

Square equation - View equation $AX ^ {2} + BX + C = 0$where $A \ NEQ 0.$

Numbers $a, b, c$referred to as the coefficients of the square equation.

The square equation may have two valid roots, one valid root or none.

The number of roots of the square equation depends on the sign of the expression, which is called discriminant.

Discriminant square equation: $D = b ^ {2} -4ac$.

If a $D.$> 0, the square equation has two roots: $X_ {1} = \ FRAC {-B + \ SQRT {D}} {2a}$ и $X_ {2} = \ FRAC {-b- \ sqrt {d}} {2a}$.

If a $D.$= 0, the square equation has the only root $X = - \ FRAC {B} {2a}$.

If a $D.$<0, the square equation has no valid roots.

We write several square equations and check how many roots they have.

one) $3x ^ {2} -4x-9 = 0$

In this equation $a = 3.$, $B = -4.$, $C = -9.$.

Discriminant equations equations $\ left (-4 \ right) ^ {2} -4 \ CDOT 3 \ CDOT \ LEFT (-9 \ RIGHT) = 16 + 108$> 0. The equation has two roots.

2) $x ^ {2} + 4x + 4 = 0$

In this equation $a = 1, \; b = 4, \; c = 4$.

Discriminant equations equations $4 ^ {2} -4 \ CDOT 1 \ CDOT 4 = 0$. The equation has the only root.

Note that in the left part of the equation $x ^ {2} + 4x + 4 = 0$There is an expression that is called a full square. Indeed, $x ^ {2} + 4x + 4 = \ left (x + 2 \ right) ^ {2}$. We applied the formula of abbreviated multiplication.

The equation $\ left (X + 2 \ RIGHT) ^ {2} = 0$It has the only root $x = -2.$.

3) $3x ^ {2} -4x + 9 = 0$.

In this equation $a = 3, \; b = -4, \; c = 9$.

Discriminant equations equations $\ Left (-4 \ Right) ^ {2} -4 \ CDOT 3 \ CDOT 9 = 16-108$<0. No roots.

4) solving equation $2x ^ {2} -3x-20 = 0$.

Discriminant equations equations $\ Left (-3 \ Right) ^ {2} -4 \ CDOT 2 \ CDOT \ LEFT (-20 \ Right) = 9 + 160 = 169$> 0.

The equation has two roots.

Root equations

$X_ {1} = \ FRAC {-b + \ sqrt {d}} {2a} = \ FRAC {3 + 13} {4} = 4$

$X_ {2} = \ FRAC {-b- \ sqrt {d}} {2a} = \ FRAC {3-13} {4} = - 2.5$

Vieta theorem

Useful theorem for solving square equations - Vieta theorem.

If a $x_ {1}$ и $x_ {2}$- Roots of the equation $AX ^ {2} + BX + C = 0$T. $x_ {1} + x_ {2} = - \ FRAC {B} {a}$, $x_ {1} x_ {2} = \ FRAC {C} {a}$.

For example, in our equation $2x ^ {2} -3x-20 = 0$The amount of roots is equal $4-2.5 = 1.5 = - \ FRAC {-3} {2}$, and the product of the roots is equal $4 \ CDOT \ LEFT (-2.5 \ RIGHT) = - 10 = \ FRAC {-20} {2}$.

The square equation can be solved in several ways. It is possible to calculate the discriminant, or use the Vieta theorem, and sometimes you can simply guess one of the roots. Or both roots.

Incomplete square equations

A square equation in which one of the coefficients B or C (or both of them) is zero, called incomplete. In such cases, it is not necessary to seek discriminant. It is easier to solve.

1) Consider the equation $2x ^ {2} = 0$.

In this equation $B = 0.$ и $C = 0.$. Obviously $x = 0.$- the only root of the equation.

2) Consider the equation $x ^ {2} -4 = 0$. Here $B = 0.$and other coefficients zero are not equal.

The easiest way to decompose the left part of the factory equations by the formula of the square difference is. We get:

$\ Left (X-2 \ Right) \ Left (X + 2 \ RIGHT) = 0$

The product of two multipliers is zero if and only if at least one of them is zero.

It means $x = 2.$or $x = -2.$.

3) Here is a similar equation: $x ^ {2} -5 = 0$.

Insofar as $5 = \ left (\ sqrt {5} \ right) ^ {2}$The equation can be written in the form:

$\ left (x- \ sqrt {5} \ right) \ left (x + \ sqrt {5} \ right) = 0$

From here $X = \ SQRT {5}$or $X = - \ SQRT {5}$.

4) Let now $B.$not zero and $C = 0.$.

Consider the equation $3x ^ {2} + 5x = 0$.

Its left part can be decomposed on multipliers, Introducing $X.$for brackets. We get:

$X \ Left (3X + 5 \ RIGHT) = 0$.

The product of two multipliers is zero if and only if at least one of them is zero.

It means $x = 0.$or $X = - \ FRAC {5} {3}$ .

Decomposition of a square three-melan

$AX ^ {2} + BX + C = A \ LEFT (X - X_ {1} \ Right) \ Left (X-X_ {2} \ Right)$.

Here $x_ {1}$ и $x_ {2}$- roots of the square equation $AX ^ {2} + BX + C = 0$.

Remember this formula. It is necessary for solving quadratic and fractional rational inequalities.

For example, our equation $2x ^ {2} -3x-20 = 0$.

His roots $X_ {1} = 4$,$X_ {2} = - 2.5$.

$2x ^ {2} -3x-20 = 2 \ left (X-4 \ Right) \ Left (x + 2.5 \ RIGHT)$.

Useful lifehaki to solve square equations.

1) It is much easier to solve a square equation if the coefficient A, which is multiplied by xx, is positive. It seems that this is a trifle, right? But how many mistakes on the exam arises due to the fact that the high school student ignores this "trifle."

For example, equation $-15x ^ {2} + 11x-2 = 0$.

It is much easier to multiply it to 1 so that the coefficient A becomes positive. We get: $15x ^ {2} -11x + 2 = 0$.

Discriminant of this equation is equal $11 ^ {2} -4 \ Cdot 15 \ Cdot 2 = 121-120 = 1$.

Root equations $X_ {1} = \ FRAC {1} {3}, \; x_ {2} = 0.4$.

2) Before deciding the square equation, look at it carefully. Maybe you can cut both parts of some parts to some not equal zero number?

Here, for example, the equation $17x ^ {2} + 34x-51 = 0$.

You can immediately count discriminant and roots. And it can be noted that all the coefficients $A, B.$ и $C.$They are divided into 17. Objects both parts of the equation to 17, we get:

$x ^ {2} + 2x-3 = 0$.

Here you can not count the discriminant, but immediately guess the first root: $X_ {1} = 1$. And the second root $x_ {2} = - 3$Easy is located on the Vieta Theorem.

3) Working with fractional coefficients is uncomfortable. For example, equation $0.01x ^ {2} + 0.05x-0.06 = 0$.

You have already guessed what to do. Multiply both parts of the equation for 100! We get:

$X ^ {2} + 5x-6 = 0$.

The roots of this equation are equal to 1 and -6.