Question

Find the square root of $\sqrt{2x}+\sqrt{-x^4-1}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let $\sqrt{\sqrt{2}x+\sqrt{-x^4-1}}=a+ib$
Square on both side
$\sqrt{2}x+i\sqrt{x^4+1}=(a^2-b^2)+2iab$
Compare real and imaginary part on both sides
$a^2-b^2=\sqrt{2}x....\left(1\right)$
$2ab=\sqrt{x^4+1}$
$4a^2{b}^2=x^4+1$
$a^2+b^2=\sqrt{(a^2-b^2{)}^2+4a^2{b}^2}$
So, $a^2+b^2=\sqrt{2x^2+x^4+1}$
$a^2+b^2=x^2+1.....\left(2\right)$
$a^2-b^2=\sqrt{2}x$
$2a^2=\sqrt{2}x+x^2+1$
$a=\pm \sqrt{\frac{\sqrt{2}x+x^2+1}{2}}$
$2b^2=x^2+1-\sqrt{2x}$
$b=\pm \sqrt{\frac{x^2-\sqrt{2}x+1}{2}}$
$a+ib=\pm (\sqrt{\frac{\sqrt{2}x+x^2+1}{2}}+i\sqrt{\frac{x^2-\sqrt{2}x+1}{2}})$
$=\pm \frac{1}{\sqrt{2}}[\sqrt{\sqrt{2}x+x^2+1}+i\sqrt{x^2-\sqrt{2}x+1}]$