Question

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

• A. $\pm \frac{1}{\sqrt{2}}\{\sqrt{(x^2-x+1)}+i\sqrt{(x^2-x+1)}$
• B. $\pm \frac{1}{\sqrt{2}}\{\sqrt{(x^2+x+1)}-i\sqrt{(x^2-x+1)}$
• C. $\pm \frac{1}{\sqrt{2}}\{\sqrt{(x^2+x+1)}+i\sqrt{(x^2-x+1)}$
• D. $\pm \frac{1}{\sqrt{2}}\{\sqrt{(x^2-x+1)}+i\sqrt{(x^2+x+1)}$

Answer 1

The correct option is C $\pm \frac{1}{\sqrt{2}}\{\sqrt{(x^2+x+1)}+i\sqrt{(x^2-x+1)}$

We have:
$x+i\sqrt{x^4+x^2+1}=x+i\sqrt{{\left[\right(x^2+1)}^2-x^2]}=x+i\sqrt{(x^2-x+1)(x^2+x+1)}=\frac{1}{2}\left[\right(x^2-x+1).i^2+2i\sqrt{(x^2-x+1)}.\sqrt{(x^2+x+1)}+(x^2+x+1\left)\right]={[\pm \frac{i\sqrt{(x^2-x+1)}+\sqrt{(x^2+x+1)}}{\sqrt{2}}]}^2.So,squareroot:\pm \frac{i\sqrt{(x^2-x+1)}+\sqrt{(x^2+x+1)}}{\sqrt{2}}.$
Hence, (C) is correct.