Name: Transformation of Roots of a Quadratic Equation
Uploaded: 2019-07-19
Duration: 11 min 45 s
Description: Here an equation and its roots are given, and we have to find new quadratic equations with new roots, which are simple multiples or factors of the original r...

Question

Solve

a b x^{2} - (a^{2} - b^{2}) x - a b = 0

by Sridhara Acharya's method.

Accepted Answer

Given that, $a b x^{2} - (a^{2} - b^{2}) x - a b = 0$

To find : the solution of given quadratic equation.

Now,

Comparing the equation $a b x^{2} - (a^{2} - b^{2}) x - a b = 0$ with $a x^{2} + b x + c = 0$ , we get

$a = a b, b = - (a^{2} - b^{2})$ and $c = - a b$

By Sridhara Achryya's method,

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$= \frac{- (- (a^{2} - b^{2})) \pm \sqrt{{(- (a^{2} - b^{2}))}^{2} - 4 \times a b \times (- a b)}}{2 \times a b}$

$= \frac{(a^{2} - b^{2})) \pm \sqrt{a^{4} + b^{4} - 2 a^{2} b^{2} + 4 a^{2} b^{2}}}{2 a b}$

$= \frac{(a^{2} - b^{2})) \pm \sqrt{a^{4} + b^{4} + 2 a^{2} b^{2}}}{2 a b}$

$= \frac{(a^{2} - b^{2})) \pm \sqrt{{(a^{2} + b^{2})}^{2}}}{2 a b} = \frac{(a^{2} - b^{2})) \pm \sqrt{{(a^{2} + b^{2})}^{2}}}{2 a b}$

$= \frac{(a^{2} - b^{2})) \pm (a^{2} + b^{2})}{2 a b}$

$∴ x = \frac{(a^{2} - b^{2})) + (a^{2} + b^{2})}{2 a b} = \frac{2 a^{2}}{2 a b} = \frac{a}{b}$ and

$x = \frac{(a^{2} - b^{2}) - (a^{2} + b^{2})}{2 a b} = - \frac{2 b^{2}}{2 a b} = - \frac{b}{a}$

Hence, The required solution is $x = \frac{a}{b}, - \frac{b}{a}$ .

Solve $a b x^{2} - (a^{2} - b^{2}) x - a b = 0$ by Sridhara Acharya's method.

Important Questions on Quadratic Equation in One Variable

Learn and Prepare for any exam you want !

Solve abx2-(a2-b2)x-ab=0 by Sridhara Acharya's method.

Important Questions on Quadratic Equation in One Variable

Learn and Prepare for any exam you want !

Solve $a b x^{2} - (a^{2} - b^{2}) x - a b = 0$ by Sridhara Acharya's method.