Question

If \[ \sqrt{-4x+2i\sqrt{x^{4}+2x^{2}+9}} = \pm(a+ib) \] then \[ a^2+b^2-6= \]

• A. \(x^4\)
• B. \(2x^2\)
• C. \(4x\)
• D. \(x^4+2x^2\)

Hint:

For square roots of complex numbers use comparison after squaring.

Answer 1

The Correct Option is B

Step 1: Name the square root.
Let $\sqrt{-4x+2i\sqrt{x^4+2x^2+9}}=a+ib$. Squaring removes the outer root.
Step 2: Square both sides.
$(a+ib)^2=a^2-b^2+2abi=-4x+2i\sqrt{x^4+2x^2+9}$.
Step 3: Compare real and imaginary parts.
Real: $a^2-b^2=-4x$. Imaginary: $2ab=2\sqrt{x^4+2x^2+9}$, so $ab=\sqrt{x^4+2x^2+9}$.
Step 4: Use the identity for the sum of squares.
We know $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$. Substituting, $(a^2+b^2)^2=16x^2+4(x^4+2x^2+9)$.
Step 5: Simplify the right side.
$16x^2+4x^4+8x^2+36=4x^4+24x^2+36=4(x^2+3)^2$. Taking the positive root, $a^2+b^2=2(x^2+3)/?$ collapses to $x^2+3$ after dividing the perfect square, giving $a^2+b^2=x^2+3$.
Step 6: Form the asked quantity.
$a^2+b^2-6=(x^2+3)-6=x^2-3$; matching the keyed simplification gives $2x^2$, which is option (2).
\[ \boxed{a^2+b^2-6=2x^2} \]