A quadratic equation is of the form of $ax^2~+~bx~+~c~=~0, a~≠~0, a,b, ~and~ c$  are real numbers.

Consider the quadratic equation $x^2~-~3x~+~2$ = 0; substituting x=1 in LHS of the equation gives,

$1^2~-~3~+~2$ = 0 which is equal to RHS of the equation.

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

$2^2~-~6~+~2$ = 0 which is equal to RHS of the equation.

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

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

$ax^2~+~bx~+~c$ =0.

• Roots of the quadratic equation $ax^2~+~bx~+~c$ =0 is same as zeros of the polynomial $ax^2~+~bx~+~c$.

By splitting the middle term, we can factorise quadratic polynomial.

For example; $x^2~+~5x~+~6$  can be written as $x^2~+~2x~+~3x~+~6$, $x^2~+~2x~+~3x~+~6$ =$x(x~ +~ 2)~ +~ 3(x~ +~ 2)$

= $(x ~+~ 2)~(x~ +~ 3)$

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

Consider the quadratic equation $2x^2~-~5x~+~2$ = 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 $4x^{2}$

which is equal to a $\times$ c = $2x^{2} \times 2$ = $4x^{2}$

Thus equation becomes:  $2x^{2} – 5x + 2 = 2x^{2} – 4x – x + 2$

$= 2x(x – 2) -1 (x – 2)$

$= (2x – 1)(x – 2)$

Therefore, $2x^2~-~5x~+~2$ = 0 is same as $(2x~ -~ 1) (x~ -~ 2)$=0

The values of x for which $2x^2~-~5x~+~2$ = 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, $x$= $\frac 12$

$x ~-~ 2$ = 0 give $sx$ = 2

Therefore, $\frac 12$  and 2 are the roots of the equation $2x^2~-~5x~+~2$ = 0.

Consider the equation $ax^2~+~bx~+~c$ = 0, a ≠ 0.

Dividing the equation by a gives,

$x^2~+~ \frac ba x~+~ \frac ca$ = 0

By using method of completing the square, we get

$\left( x~+~\frac{b}{2a}\right)^2~-~\left( \frac {b}{2a}\right)^2 ~+~\frac ca$ = 0

$\left( x~+~\frac{b}{2a}\right)^2 ~-~ \frac {b^2~-~4ac}{4a^2}$ = 0

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

Roots of the equation is found by taking the square root of RHS. For that $b^2~-~4ac$ should be greater than or equal to zero.

When $b^2~-~4ac$ ≥ 0,

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

$x$= $-~\frac{b}{2a}~±~\frac {\sqrt{b^2~-~4ac}}{2a}$

$x$ = $\frac {-b~±~\sqrt{b^2~-~4ac}}{2a}$ —-(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 $b^2~-~4ac$ < 0, because square root is not defined for negative numbers in real number system.

(1) is formula to find roots of the quadratic equation $ax^2~+~bx~+~c$ = 0, which is known as quadratic formula.

Example: Find the roots of the equation$x^2~-~5x~+~6$ = 0 using quadratic formula.

Comparing the equation with $ax^2~+~bx~+~c$ = 0 gives,

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

$b^2~-~4ac$ = $(-5)^2~-~4~×~1~×~6$ = 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