Question

Is$(\sqrt{x}-\frac{1}{\sqrt{x}})=\sqrt{-5}$ and x≠ 0, then what is the value of $\frac{3(x^2+3)-6-3x}{4x}$ ?

• A. -3
• B. -2
• C. 3
• D. 2

Answer 1

Answer: (A)

-3

Answer 2

The correct option is (A): -3
$(\sqrt{x}=\frac{1}{\sqrt{x}})^2=\sqrt-5$
$(\sqrt{x}=\frac{1}{\sqrt{x}})^2=-5$
$x-2+\frac{1}{x}=-5$
x2 - 2x + 1 = -5x
x2 + 3x + 1 = 0
$=\frac{3(3x+6)-6-3x}{4x}$
$\frac{-3+6-6-3x}{4x}$
$\frac{-12x}{4x}$
$=-3$