Solution of Quadratic Equation of Class 10

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.

(2) By the method of completing the square.

(3) By the quadratic formula

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