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 -\(4\over 3\).

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