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 $x \ge 3$. $(i)$The given equation i

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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 $x \ge 3$. $(i)$The given equation is $(3|x| - 3)^2 = |x| + 7$.Expanding the left side: $9|x|^2 - 18|x| + 9 = |x| + 7$.Rearranging the terms: $9|x|^2 - 19|x| + 2 = 0$.Let $t = |x|$,then $9t^2 - 19t + 2 = 0$.Factoring the quadratic: $(9t - 1)(t - 2) = 0$.Thus,$|x| = 1/9$ 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 $x = 2$,$2$ is not in the domain.For $x = -2$,$-2 \le 0$ (Valid).For $x = 1/9$,$1/9$ is not in the domain.For $x = -1/9$,$-1/9 \le 0$ (Valid).Therefore,the solutions are $-2$ and $-1/9$.

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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