Here you will learn how to find solution of quadratic equation of class 10 and different methods used to solve quadratic equation.
Let’s begin –
Solution of Quadratic Equation Class 10
(a) The general form of quadratic equation is \(ax^2 + bx + c\) = 0, a \(\ne\) 0.
The roots or solution of quadratic equation can be found in following manner.
\(a(x^2 + {b\over a}x + {c\over a})\) = 0 \(\implies\) \((x + {b\over 2a})^2\) + \(c\over a\) – \(b^2\over 4a^2\) = 0
\((x + {b\over 2a})^2\) = \(b^2\over 4a^2\) – \(c\over a\)
\(\implies\) x = \(-b \pm \sqrt{b^2 – 4ac}\over 2a\)
This \(-b \pm \sqrt{b^2 – 4ac}\over 2a\) expression can be directly used to find the two roots of a quadratic equation. This formula is known as sridharacharya formula.
(b) This expression \(b^2 – 4a\) = D is called the discriminant of the quadratic equation.
Example : Find the roots of the quadratic equation \(x^2 + 3x + 2\) = 0. Also find sum of roots and product of roots.
Solution : We have, \(x^2 + 3x + 2\) = 0
By using the formula described above,
x = \(-3 \pm \sqrt{9 – 8}\over 2\) = \(-3 \pm 1\over 2\)
\(\implies\) x = -1 and -2
Let \(\alpha\) = -1 and \(\beta\) = -2
Sum of Roots = \(\alpha + \beta\) = -3
Product of Roots = \(\alpha \beta\) = 2
Different Methods for Solving Quadratic Equation
Following three methods are used to solve quadratic equation. These methods are discussed in detail in next sections.
(1) By the method of factorisation.