Question

Find the roots of the following quadratic equations by using the quadratic formula
$\frac{x-3}{x+3}-\frac{x+3}{x-3}=6\frac{6}{7},x\ne -3,3$

• A. $4,\frac{9}{4}$
• B. $-4,\frac{4}{9}$
• C. $-4,-\frac{4}{9}$
• D. $-4,\frac{9}{4}$

Answer 1

The correct option is D $-4,\frac{9}{4}$

Given, $\frac{x-3}{x+3}-\frac{x+3}{x-3}=6\frac{6}{7}$

$\Rightarrow (x-3{)}^2-(x+3{)}^2=\frac{48}{7}(x+3)(x-3)$

$\Rightarrow 7\left[(x^2-6x+9-x^2-6x-9)\right]=48(x^2-9)$

$\Rightarrow 7(-12x)=48x^2-432$

$\Rightarrow 48x^2+84x-432=0$

$\Rightarrow 4x^2+7x-36=0$

$\Rightarrow 4x^2+16x-9x-36=0$

$\Rightarrow 4x(x+4)-9x(x+4)=0$

$\Rightarrow (x+4)(4x-9)=0$

$\Rightarrow x+4=0$

$\Rightarrow x=-4$

$\Rightarrow 4x-9=0$

$\Rightarrow x=\frac{9}{4}=2\frac{1}{4}$