Answer
$x=\pm\sqrt{7}$
Work Step by Step
$\sqrt{11-x^{2}}-\dfrac{2}{\sqrt{11-x^{2}}}=1$
Multiply the whole equation by $\sqrt{11-x^{2}}$:
$\sqrt{11-x^{2}}\Big(\sqrt{11-x^{2}}-\dfrac{2}{\sqrt{11-x^{2}}}=1\Big)$
$(\sqrt{11-x^{2}})^{2}-2=\sqrt{11-x^{2}}$
$11-x^{2}-2=\sqrt{11-x^{2}}$
Simplify the left side:
$9-x^{2}=\sqrt{11-x^{2}}$
Square both sides:
$(9-x^{2})^{2}=(\sqrt{11-x^{2}})^{2}$
$81-18x^{2}+x^{4}=11-x^{2}$
Take all terms to the left side:
$x^{4}-18x^{2}+x^{2}+81-11=0$
$x^{4}-17x^{2}+70=0$
Solve this equation by factoring:
$(x^{2}-7)(x^{2}-10)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x^{2}-7=0$
$x^{2}=7$
$x=\pm\sqrt{7}$
$x^{2}-10=0$
$x^{2}=10$
$x=\pm\sqrt{10}$
Only $x=\sqrt{7}$ and $x=-\sqrt{7}$ make the initial equation true. The final answer is:
$x=\pm\sqrt{7}$
