Question

If $\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=3$, then $x=$

• A. $\sqrt{\frac{2}{3}}$
• B. $\sqrt{\frac{1}{3}}$
• C. $\sqrt{\frac{2}{5}}$
• D. $\sqrt{\frac{3}{5}}$

Answer 1

Answer: (D)

$\sqrt{\frac{3}{5}}$

After rationalizing the expression we get $\frac{{(\sqrt{1+x^2}+\sqrt{1-x^2})}^2}{2x^2}=3$
$(\sqrt{1+x^2}+\sqrt{1-x^2}{)}^2=6x^2$
$2+2\sqrt{1-x^4}=6x^2$
$1-x^4=(3x^2-1{)}^2$
$10x^4-6x^2=0$
$x$ can't be equal to zero since the denominator in the original fraction becomes zero.
So $x=\sqrt{\frac{3}{5}}$