Solve the following quadratic equation by factorization:
$\frac{x\ -\ 3}{x\ +\ 3}\ -\ \frac{x\ +\ 3}{x\ -\ 3}\ =\ \frac{48}{7},\ x\ ≠\ 3,\ -3$

Given:

Given quadratic equation is $\frac{x\ -\ 3}{x\ +\ 3}\ -\ \frac{x\ +\ 3}{x\ -\ 3}\ =\ \frac{48}{7},\ x\ ≠\ 3,\ -3$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$

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

$\frac{x^2-3x-3x+9-(x^2+3x+3x+9)}{(x)^2-(3)^2}=\frac{48}{7}$

$\frac{x^2-6x+9-x^2-6x-9}{x^2-9}=\frac{48}{7}$

$\frac{-12x}{x^2-9}=\frac{48}{7}$

$7(-12x)=48(x^2-9)$   (On cross multiplication)

$-7x=4(x^2-9)$

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

$4x^2+16x-9x-36=0$    ($16-9=7$ and $16\times(-9)=4\times(-36)$)

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

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

$4x-9=0$ or $x+4=0$

$4x=9$ or $x=-4$

$x=\frac{9}{4}$ or $x=-4$

The roots of the given quadratic equation are $\frac{9}{4}$ and $-4$.

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Updated on: 10-Oct-2022

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