Solve Quadratic Equation Using Quadratic Formula

Here you will learn how to solve quadratic equation using quadratic formula or sridharacharya formula with examples.

Let’s begin – 

Solve Quadratic Equation Using Quadratic Formula (Sridharacharya Formula)

For the equation $ax^2 + bx + c$ = 0, if $b^2 – 4ac$ $\ge$ 0, then

x = ($-b + \sqrt{b^2 -4ac}\over 2a$)  and  x = ($-b – \sqrt{b^2 -4ac}\over 2a$)

This formula was given by indian mathematician sridharacharya. Therefore , it is also called sridharacharya formula.

If $b^2 – 4ac$ < 0, i.e. negative, then $\sqrt{b^2 -4ac}$ is not real and therefore, the equation does not have any real roots.

Therefore, if $b^2 – 4ac$ $\ge$ 0, then the quadratic equation $ax^2 + bx + c$ = 0 has two roots $\alpha$ and $\beta$, given by 

$\alpha$ = ($-b + \sqrt{b^2 -4ac}\over 2a$)  and  $\beta$ = ($-b – \sqrt{b^2 -4ac}\over 2a$)

Example : Solve the following quadratic equation using quadratic formula.

(i) $6x^2 + 7x – 10$ = 0

(ii) $15x^2 – 28$ = x

Solution : 

(i) We have, $6x^2 + 7x – 10$ = 0

Here, a = 6, b = 7, c = -10

$\therefore$    D = $b^2 – 4ac$ = $(7)^2 – 4\times 6 \times (-10)$

D = 49 + 240 = 289 > 0

So, the given equation has real roots, given by

x =  $-b + \sqrt{b^2 -4ac}\over 2a$ = $-7 + \sqrt{289}\over 12$ = $10\over 12$ = $5\over 6$

or,   x = $-b – \sqrt{b^2 -4ac}\over 2a$ = $-7 – \sqrt{289}\over 12$ = $24\over 12$ = -2

Therefore, the roots of the given equations are $5\over 6$ and -2.

(ii) We have, $15x^2 – x – 28$ = 0

Here, a = 15, b = -1, c = -28

$\therefore$    D = $b^2 – 4ac$ = $(-1)^2 – 4\times 15 \times (-28)$

D = 1 + 1680 = 1681 > 0

So, the given equation has real roots, given by

x =  $-b + \sqrt{b^2 -4ac}\over 2a$ = $-(-1) + \sqrt{1681}\over 30$ = $42\over 30$ = $7\over 5$

or,   x = $-b – \sqrt{b^2 -4ac}\over 2a$ = $-(-1) – \sqrt{1681}\over 30$ = $40\over 30$ = -$4\over 3$

Therefore, the roots of the given equations are $7\over 5$ and -$4\over 3$.