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

Concept: For \[ \sqrt{u+iv}=a+ib \] we use \[ (a+ib)^2=u+iv \]

Step 1: Square both sides.
\[ (a+ib)^2=-4x+2i\sqrt{x^4+2x^2+9} \] Expand \[ a^2-b^2+2abi=-4x+2i\sqrt{x^4+2x^2+9} \] Comparing real and imaginary parts \[ a^2-b^2=-4x \] \[ ab=\sqrt{x^4+2x^2+9} \]

Step 2: Find \(a^2+b^2\).
Identity: \[ (a^2+b^2)^2=(a^2-b^2)^2+4a^2b^2 \] Substitute \[ =(16x^2)+4(x^4+2x^2+9) \] \[ =4(x^2+3)^2 \] Thus \[ a^2+b^2=x^2+3 \]

Step 3: Required value.
\[ a^2+b^2-6 \] \[ =(x^2+3)-6 \] \[ =2x^2 \] Thus \[ \boxed{2x^2} \]