**Quadratic Formula** help to evaluate the solution of quadratic equation replacing the factorization method. A quadratic equation is of the form of ax^{2} + bx + c = 0, where a, b and c are real numbers, also called “**numeric coefficients”**. Here x is an unknown variable, for which we need to find the solution.Â The formula to find the solution of the quadratic equation is given by:

**x = -bÂ Â±Â âˆš(b ^{2}-4ac)/2a**

The sign of plus (+) and minus (-) represents that there are two solutions for quadratic equation, such as;

**x _{1} = -b + âˆš(b^{2}-4ac)/2a and x_{2} = -b – âˆš(b^{2}-4ac)/2a**

Both the solutions evaluated above are called the roots of the quadratic equation.

**Also, read:**

Example: Consider the quadratic equation x^{2}-3x+2Â = 0; substituting x=1 in LHS of the equation gives;

1^{2}-3+2Â = 0 which is equal to RHS of the equation.

Similarly, substituting x = 2 in LHS of the equation also gives,

2^{2}-6+2Â = 0 which is equal to RHS of the equation.

Here, 1 and 2 satisfy the equation x^{2}-3x+2=0. Therefore, those are known as the solution of the quadratic equation or roots of the equation. This also means that numbers 1 and 2 are the zeros of the polynomial x^{2}-3x+2.Â

## Roots of Quadratic Equation

We know that a second-degree polynomial will have at most two zeros. Therefore a quadratic equation will have at most two roots.

In general, if Î± is a root of the quadratic equationÂ ax^{2}+bx+cÂ = 0, a â‰ 0; then, aÎ±^{2}+bÎ±+cÂ = 0. We can also say that x = Î± is a solution of the quadratic equation or Î± satisfies the equation,Â ax^{2}+bx+c=0.

**Note:**

- Roots of the quadratic equation ax
^{2}+bx+cÂ =0 is same as zeros of the polynomial ax^{2}+bx+c. - By splitting the middle term, we can factorise quadratic polynomial.

Example :Â Â x^{2}+5x+6Â can be written as x^{2}+2x+3x+6, x^{2}+2x+3x+6Â =Â x(x + 2) + 3(x + 2)= (x+2)(x + 3)

### How to find the roots of quadratic equation?

Roots of the quadratic equation can be found by factorising the polynomial and equate it with zero.

Consider the quadratic equation 2x^{2}-5x+2Â = 0; which is of the form of ax^{2}+ bx +c = 0

Here, -5x can be broken down into two parts -4x – x, as the multiplication of these parts result in 4x^{2Â }which is equal to a Ã—Â c = 2x^{2} Ã— 2Â = 4x^{2Â }.Thus equation becomes:

2x^{2}Â – 5x + 2 = 2x^{2} – 4x – x + 2

= 2x(x – 2) -1 (x – 2)

= (2x – 1)(x – 2)

Therefore, 2x^{2}-5x+2Â = 0 is same as (2x – 1) (x – 2)=0

The values of x for which 2x^{2}-5x+2Â = 0 is same as the values of x for which (2x – 1)(x – 2)Â = 0.

IfÂ (2x-1)(x – 2)Â = 0Â ; then, either (2x – 1)Â = 0 or (x – 2)Â = 0

2x – 1Â = 0 gives, 2x = 1 orÂ Â x = 1/2

x – 2Â = 0 gives xÂ = 2

Therefore, 1/2Â and 2 are the roots of the equation 2x^{2}-5x+2Â = 0.

## Quadratic Formula Derivation

Consider the equation ax^{2}+bx+cÂ = 0, a â‰ 0.

Dividing the equation by a gives,

Â x^{2}+ b/a x+c/aÂ = 0

By using method of completing the square, we get

(x+b/2a)^{2} – (b/2a)^{2} + c/a = 0

(x+b/2a)^{2} – [(b^{2}-4ac)/4a^{2}]= 0

(x+b/2a)^{2} = (b^{2}-4ac)/4a^{2}

Roots of the equation are found by taking the square root of RHS. For that b^{2}-4acÂ should be greater than or equal to zero.

When b^{2}-4ac â‰¥ 0,

\( \left( x~+~\frac{b}{2a}\right) \) =Â Â± \( \frac {\sqrt{b^2~-~4ac}}{2a}\)

\( x \)Â = \(\frac {-b~Â±~\sqrt{b^2~-~4ac}}{2a}\) —-(1)

Therefore roots of the equation are, \(\frac {-b~+~\sqrt{b^2~-~4ac}}{2a}\) and \(\frac {-b~-~\sqrt{b^2~-~4ac}}{2a}\)

The equation will not have real roots if b^{2}-4ac < 0, because square root is not defined for negative numbers in real number system.

Equation (1) is a formula to find roots of the quadratic equation ax^{2}+bx+cÂ = 0, which is known as quadratic formula.

### Example

**Example: Find the roots of the equationÂ x ^{2}-5x+6Â = 0 using the quadratic formula.**

Solution: Comparing the equation with ax^{2}+bx+cÂ = 0 gives,

a = 1, b = -5 and c = 6

b^{2}-4ac = (-5)^{2}-4Ã—1Ã—6Â = 1

Roots of the equations are \(\frac {-b~+~\sqrt{b^2~-~4ac}}{2a}\) Â = \( \frac {5~+~1}{2} \) = \( \frac {6}{2}\) = 3 and \(\frac {-b~-~\sqrt{b^2~-~4ac}}{2a}\) Â = \( \frac {5~-~1}{2} \) = \( \frac {4}{2}\) = 2

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