Question

Find the real values of 'x' such that $x^2$ + 4 |x| - 4 = 0.

• A. $-2\pm \sqrt{8},2\pm \sqrt{8}$
• B. $\pm (\sqrt{8}-2)$
• C. $-2\pm \sqrt{8}$
• D. $\sqrt{8}+2$

Answer 1

The correct option is B $\pm (\sqrt{8}-2)$

$x^2$ + 4 |x| - 4 = 0
If x > 0, $x^2$ + 4x - 4 = 0
$x=\frac{-4\pm 4\sqrt{2}}{2}=-2\pm 2\sqrt{2}$
But as x > 0 so x = -2 + 2 $\sqrt{2}$
If x < 0 $\Rightarrow x^2$ - 4x - 4 = 0
$x=\frac{-4\pm 4\sqrt{2}}{2}$
$\Rightarrow$ x = 2 - 2$\sqrt{2}$ (As x < 0)
x = $\pm$ ($\sqrt{8}$ - 2)