 Quadratic Formula help to evaluate the solution of quadratic equation replacing the factorization method. A quadratic equation is of the form of ax2 + 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.  The formula to find the solution of the quadratic equation is given by:

x = -b ± √(b2-4ac)/2a

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

x1 = -b + √(b2-4ac)/2a and x2 = -b – √(b2-4ac)/2a

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

Example: Consider the quadratic equation x2-3x+2 = 0; substituting x=1 in LHS of the equation gives;

12-3+2 = 0 which is equal to RHS of the equation.

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

22-6+2 = 0 which is equal to RHS of the equation.

Here, 1 and 2 satisfy the equation x2-3x+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 x2-3x+2.

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 α is a root of the quadratic equation ax2+bx+c = 0, a ≠ 0; then, 2+bα+c = 0. We can also say that x = α is a solution of the quadratic equation or α satisfies the equation, ax2+bx+c=0.

Note:

• Roots of the quadratic equation ax2+bx+c =0 is same as zeros of the polynomial ax2+bx+c.
• By splitting the middle term, we can factorise quadratic polynomial.

Example :  x2+5x+6  can be written as x2+2x+3x+6, x2+2x+3x+6 = 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 2x2-5x+2 = 0; which is of the form of ax2+ bx +c = 0

Here, -5x can be broken down into two parts -4x – x, as the multiplication of these parts result in 4xwhich is equal to a × c = 2x2 × 2 = 4x.Thus equation becomes:

2x2 – 5x + 2 = 2x2 – 4x – x + 2

= 2x(x – 2) -1 (x – 2)

= (2x – 1)(x – 2)

Therefore, 2x2-5x+2 = 0 is same as (2x – 1) (x – 2)=0

The values of x for which 2x2-5x+2 = 0 is same as the values of x for which (2x – 1)(x – 2) = 0.

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

2x – 1 = 0 gives, 2x = 1 or  x = 1/2

x – 2 = 0 gives x = 2

Therefore, 1/2 and 2 are the roots of the equation 2x2-5x+2 = 0.

Consider the equation ax2+bx+c = 0, a ≠ 0.

Dividing the equation by a gives,

x2+ b/a x+c/a = 0

By using method of completing the square, we get

(x+b/2a)2 – (b/2a)2 + c/a = 0

(x+b/2a)2 – [(b2-4ac)/4a2]= 0

(x+b/2a)2 = (b2-4ac)/4a2

Roots of the equation are found by taking the square root of RHS. For that b2-4ac should be greater than or equal to zero.

When b2-4ac ≥ 0,

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

$x$ = $\frac {-b~±~\sqrt{b^2~-~4ac}}{2a}$ —-(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 b2-4ac < 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 ax2+bx+c = 0, which is known as quadratic formula.

### Example

Example: Find the roots of the equation  x2-5x+6 = 0 using the quadratic formula.

Solution: Comparing the equation with ax2+bx+c = 0 gives,

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

b2-4ac = (-5)2-4×1×6 = 1

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