Question

Find root of $\sqrt{\frac{x}{1-x}+\frac{1-x}{x}}=\frac{13}{6}$ where $(x=\frac{4}{13})$

Answer 1

$\sqrt{\frac{x}{1-x}+\frac{1-x}{x}}=\frac{13}{6}$

$\sqrt{\frac{x^2+(1-x{)}^2}{(1-x)x}}=\frac{13}{6}$

$\Rightarrow \sqrt{\frac{x^2+1+x^2-2x}{x(1-x)}}=\frac{13}{6}$

squaring both sides

$\frac{x^2+1+x^2-2x}{x-x^2}=\frac{13}{6}$

$\Rightarrow 6(2x^2-2x+1)=13(x-x^2)$

$12x^2-12x+6=13x-13x^2$

$12x^2+13x^2-12x-13x+6=0$

$25x^2-25x+6=0$

$25x^2-15x-10x+6=0$

$5x(5x-3)-2(5x+3)=0$

$(5x-2)(5x-3)$

$x=\frac{2}{5}$ or $\frac{3}{5}$