Question

Find the number of real roots of $\sqrt{\frac{2x^2+1}{x^2-1}}+6\sqrt{\frac{x^2-1}{2x^2+1}}=5$

Answer 1

$\sqrt{\frac{2x^2+1}{x^2-1}}+6\sqrt{\frac{x^2-1}{2x^2+1}}=5$
$Let$
$\frac{2x^2+1}{x^2-1}=t$
$\sqrt{t}+\frac{6}{\sqrt{t}}=5$
$squaringbothsides$
${(\sqrt{t}+\frac{6}{\sqrt{t}})}^2={\left(5\right)}^2$
$\Rightarrow t+\frac{36}{t}+12=25[\because {(a+b)}^2=a^2+b^2+2ab]$
$\Rightarrow \frac{t^2+36+12t}{t}=25$
$\Rightarrow t^2+12t+36=25t$
$\Rightarrow t^2+12t+36-25t=0$
$\Rightarrow t^2-13t+36=0$
$\Rightarrow (t-9)(t-4)=0$
$t=9,4$
$\frac{2x^2+1}{x^2-1}=9$
$\Rightarrow 2x^2+1=9x^2-9$
$\Rightarrow 7x^2=10$
$\Rightarrow x^2=\frac{10}{7}$
$x=\pm \sqrt{\frac{10}{7}}$
$\frac{2x^2+1}{x^2-1}=4$
$\Rightarrow 2x^2+1=4x^2-4$
$\Rightarrow 2x^2=5$
$\Rightarrow x^2=\frac{5}{2}$
$x=\pm \sqrt{\frac{5}{2}}$
$\text{Hence the given equation has 4 real roots}$