Question

The solution of the equation ${(3\left|x\right|-3)}^2=\left|x\right|+7$ which belongs to the domain of $\sqrt{x(x-3)}$ are given by ?

• A. $\pm \frac{1}{9},\pm 2$
• B. $\frac{1}{9},-2$
• C. $-\frac{1}{9},-2$
• D. $-\frac{1}{9},2$

Answer 1

The correct option is C $-\frac{1}{9},-2$

As quantity inside square root can't be negative,

$x(x-3)\ge 0$

$\Rightarrow x\in (-\infty ,0]\cup [3,\infty )$

Now let $|x|=y$, $(3y-3{)}^2=y+7$

$\Rightarrow 9(y-1{)}^2=y+7$

$\Rightarrow 9y^2+9-18y-y-7=0$

$\Rightarrow 9y^2-19y+2=0$

$\Rightarrow 9y^2-18y-y+2=0$

$\Rightarrow 9y(y-2)-1(y-2)=0$

$\Rightarrow (9y-1)(y-2)=0$

$\Rightarrow y=\frac{1}{9}$ or $y=2$

$\Rightarrow |x|=\frac{1}{9}$ or $|x|=2$

$\Rightarrow x=\pm \frac{1}{9}$ or $x=\pm 2$

But $x\ne \frac{1}{9},2$

$\Rightarrow x=-\frac{1}{9},-2$

$\Rightarrow$ (C).