The solutions of the quadratic equation (3|x| | vedclass

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(D) The domain of definition of the function $y = \sqrt{x(x - 3)}$ is $x(x - 3) \ge 0$,which implies $x \le 0$ or $x \ge 3$ ... $(i)$.Given the equati

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$-1/9, -2$

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(D) The domain of definition of the function $y = \sqrt{x(x - 3)}$ is $x(x - 3) \ge 0$,which implies $x \le 0$ or $x \ge 3$ ... $(i)$.Given the equation $(3|x| - 3)^2 = |x| + 7$,let $t = |x|$. Then $(3t - 3)^2 = t + 7$.Expanding the left side: $9t^2 - 18t + 9 = t + 7$.Rearranging the terms: $9t^2 - 19t + 2 = 0$.Factoring the quadratic: $(9t - 1)(t - 2) = 0$.Thus,$t = 1/9$ or $t = 2$.Since $t = |x|$,we have $|x| = 1/9$ or $|x| = 2$.This gives $x = \pm 1/9$ or $x = \pm 2$.Checking these values against the domain condition $(i)$ ($x \le 0$ or $x \ge 3$):For $x = 1/9$,$1/9 
ot\le 0$ and $1/9 
ot\ge 3$ (Reject).For $x = -1/9$,$-1/9 \le 0$ (Accept).For $x = 2$,$2 
ot\le 0$ and $2 
ot\ge 3$ (Reject).For $x = -2$,$-2 \le 0$ (Accept).Therefore,the required solutions are $-1/9$ and $-2$.

The solutions of the quadratic equation $(3|x| - 3)^2 = |x| + 7$ which belong to the domain of definition of the function $y = \sqrt{x(x - 3)}$ are given by

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