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Algebra Difficulty 1.0 Junior Prove it Canada

IMG0 If $x=2$ , what is the value of $\dfrac{x^4 + 3x^2}{x^2}$ ? In the diagram, $\triangle ABC$ is right-angled at $B$ . Also, $AB=10$ , $BC=t-1$ , and $AC=t+1$ . What is the value of $t$ ?IMG3 Suppose that $\dfrac{2}{y} + \dfrac{3}{2y} = 14$ . Determine the value of $y$ .

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Solution

For every $x \neq 0$ , we note that $\text{x 4}$ + 3x^2}{x^2} = x^2 +
3 $. Therefore, when$ x = 2 $, we have$ $\text{x 4}$ + 3x^2}{x^2} = x^2 + 3 = 2^2 + 3
= 7 $. Alternatively, when$ x = 2 $, we have$ $\text{x 4}$ + 3x^2}{x^2} = $\text{2 4}$ + 3
$\cdot$ 2^2}{2^2} = $\dfrac{28}{4}$ = 7 $. By the Pythagorean Theorem in$ $\triangle$ ABC $, we have$ $\text{AC 2 = AB 2 + BC 2 (t+1) 2 = 10 2 + (t-1) 2 t 2 + 2t + 1 = 100 + t 2 - 2t + 1 4t = 100}$ $and so$ t = 25. Alternatively, we could remember the Pythagorean triple

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Figure for this problem 5 $-$ 12 $-$ 13 and scale this triple by a factor of

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Figure for this problem 2 to obtain the Pythagorean triple

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Figure for this problem 10 $-$ 24 $-$ 26, noting that the difference between

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Figure for this problem t+1 $and$ t-1 $is$ 2 $as is the difference between$ 26 $and$ 24 $, which gives$ t + 1 = 26 $and so$ t = 25 $. Since$ $\dfrac{2}{y} + \dfrac{3}{2y}$ =
14 $, then$ $\dfrac{4}{2y} + \dfrac{3}{2y}=14$ $or$ $\dfrac{7}{2y}$ = 14$.

Therefore, $2y = \dfrac{7}{14} = \dfrac{1}{2}$ and so $y = \dfrac{1}{4}$ .

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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.