Question

The solutions 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:

Answer 1

We need to proceed by finding the domain of $\sqrt {x(x - 3)}$ and then consider $\left| x \right|$ equal to another variable, say $t$ and find the solution of the given equation in terms of $t$ and then convert it into the form of $x$ .

Complete answer:
Let’s start with finding the domain of $\sqrt {x(x - 3)}$
We know that any equation under square-root should always be greater than or equal to zero.
Therefore, we can write: $x(x - 3) \geqslant 0$
The domain is: $x \in ( - \infty ,0) \cup (3,\infty )$
Now let’s consider $\left| x \right| = t$
Now solving the given equation, we get:
${(3t - 3)^2} = t + 7$
$t + 7 = 9{t^2} + 9 - 18t$
$9{t^2} + 2 - 19t = 0$
To solve this quadratic equation, we will use splitting the middle terms and solve.
$9{t^2} - 18t - t + 2 = 0$
$9t(t - 2) - 1(t - 2) = 0$
$(9t - 1)(t - 2) = 0$
Therefore, $t = 2,\dfrac{1}{9}$
Since $\left| x \right| = t$ ,
$\left| x \right| = 2,\dfrac{1}{9}$
$x = \pm 2,\dfrac{{ \pm 1}}{9}$
Now using the domain $x \in ( - \infty ,0) \cup (3,\infty )$ , we know that $x = - 2,\dfrac{{ - 1}}{9}$

Therefore, the correct option is C

Note: We can also solve this question by directly considering $\left| x \right|$ which would be slightly confusing. We can also check our answer by substituting all four values in $\sqrt {x(x - 3)}$ and eliminate wrong values. One should remember the meaning of modulus, a modulus is nothing but absolute value, always greater than or equal to zero.