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 -\(4\over 3\).
