Question

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

The roots are$3$, $\frac{-25}{3}$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

$\Rightarrow$ $\frac{(5{)}^2+10x+x^2-((5{)}^2-10x+x^2)}{(5{)}^2-(x{)}^2}=\frac{15}{4}$

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

$\Rightarrow$ $\frac{20x}{25-x^2}=\frac{15}{4}$

$\Rightarrow$ $\frac{4x}{25-x^2}=\frac{3}{4}$

$\Rightarrow$ $16x=75-3x^2$

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

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

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

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

$\Rightarrow$ $3x+25=0$ and $x-3=0$

$\Rightarrow$ $x=\frac{-25}{3}$ and $x=3$

$\therefore$ We can see that roots given in question are correct.