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

Let’s begin –

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

(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 -$$4\over 3$$.