Question

For $a\le 0$, what will be the roots of the equation $x^2-2a\mid x-a\mid -3a^2=0$.

• A. $a(-1\pm \sqrt{6})$ and $a(1\pm \sqrt{2})$
• B. $a(-1+\sqrt{6})$ and $a(1-\sqrt{2})$
• C. $a(-1-\sqrt{6})$ and $a(1+\sqrt{2})$
• D. $a(-1-\sqrt{6})$ and $a(1-\sqrt{2})$

Answer 1

Answer: (B)

$a(-1+\sqrt{6})$ and $a(1-\sqrt{2})$

$x^2-2a\mid x-a\mid -3a^2=0$.

When $x<a$

$\Rightarrow x^2+2a(x-a)-3a^2=0$

$\Rightarrow x^2+2ax-5a^2=0$

$x=\frac{-2a\pm \sqrt{(2a{)}^2-4(1\left)\right(-5a^2)}}{2}$
$x=\frac{-2a\pm \sqrt{24a^2}}{2}$
$x=a(-1\pm \sqrt{6})$

But since $x<a$ and $a\le 0$

$\Rightarrow x\le 0$

So,$x=a(-1+\sqrt{6})$
When $x\ge a$

$\Rightarrow x^2-2a(x-a)-3a^2=0$

$\Rightarrow x^2-2ax-a^2=0$

$x=\frac{2a\pm \sqrt{(-2a{)}^2-4(1\left)\right(-a^2)}}{2}$
$x=\frac{-2a\pm \sqrt{8a^2}}{2}$

$x=a(-1\pm \sqrt{2})$

But since $x\ge a$ and $a\le 0$

$\Rightarrow x\ge 0$

So,$x=a(-1-\sqrt{2})$