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Algebra Difficulty 1.0 Junior Prove it Canada

IMG0 What is the value of $\dfrac{3^2-2^3}{2^{3}-3^{2}}$ ? What is the value of $\sqrt{\sqrt{81}+\sqrt{9}-\sqrt{64}}$ ? Determine all real numbers $x$ for which $\dfrac{1}{\sqrt{x^2 + 7}} = \dfrac{1}{4}$ .

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Solution

Evaluating, $\dfrac{3^2 - 2^3}{2^3 - 3^2} = \dfrac{9 - 8}{8-9} = \dfrac{1}{-1} = -1$ .

Alternatively, since $2^3 - 3^2 = -(3^2 - 2^3)$ , then $\dfrac{3^2 - 2^3}{2^3 - 3^2} = -1$ .
Evaluating, $\sqrt{\phantom{\left(\right.}\hspace{-2mm}\sqrt{81}+\sqrt{9}-\sqrt{64}} = \sqrt{9 + 3 - 8} = \sqrt{4} = 2$ .
Since $\dfrac{1}{\sqrt{x^2 + 7}} = \dfrac{1}{4}$ , then $\sqrt{x^2 + 7} = 4$ .

This means that $x^2 + 7 = 4^2 = 16$ and so $x^2 = 9$ . Since $x^2 = 9$ , then $x = \pm 3$ .

We can check by substitution that both of these values are
solutions.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.