Question

The solutions of the equation $(3|x|-3{)}^2=|x|+7$ 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 $-\frac{1}{9},-2$

Let us find the domain of $\sqrt{x(x-3)}$
$x(x-3)\ge 0$
$x>3$ or $x<0$
Now, for the equation let us assume, $x>0\Rightarrow |x|=x$
The equation will be
$(3x-3{)}^2=x+7$
$9x^2+9-18x=x+7$
$9x^2-19x+2=0$
$9x^2-18x-x+2=0$
$(9x-1)(x-2)=0$
$x=\frac{1}{9},x=2$
Both the solutions will be discarded as they don't belong to the domain $x>3$ or $x<0$
So, let us now consider the equation for $x\le 0$
The equation becomes $(-3x-3{)}^2=-x+7$
$9x^2+9+18x=-x+7$
$9x^2+19x+2=0$
$(9x+1)(x+2)=0$
$x=-2,-\frac{1}{9}$
which lie in the domain $x<0$. So option C is the correct option.