Question

Solve the following quadratic equation by factorization.
$\frac{5+x}{5-x}-\frac{5-x}{5+x}=3\frac{3}{4};x\ne 5,-5$

Answer 1

$\frac{5+x}{5-x}-\frac{5-x}{5+x}=3\frac{3}{4};x\ne 5,-5$

$\frac{(5+x)(5+x)-(5-x)(5-x)}{(5-x)(5+x}=\frac{15}{4}$

$\frac{(5+x{)}^2-(5-x{)}^2}{(5^2-x^2)}=\frac{15}{4}$

$\frac{(25+10x+x^2)-(25-10x+x^2)}{(25-x^2)}=\frac{15}{4}$

$\frac{(25+10x+x^2-25+10x-x^2)}{(25-x^2)}=\frac{15}{4}$

$\frac{\left(20x\right)}{(25-x^2)}=\frac{15}{4}$

$(20x\times 4)=15\times (25-x^2)$

$\left(80x\right)=375-15x^2$

$15x^2+80x-375=0$

$3x^2+16x-75=0$

$3x^2-9x+25x-75=0$

$3x(x-3)+25(x-3)=0$

$(x-3)(3x+25)=0$

Therefore,

$x=3,\frac{-25}{3}$