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Two positive numbers differ by $5$ and the square of their sum is $169$. Find the numbers.

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closed as off-topic by tomasz, Lost1, Elias, user86418, Sami Ben Romdhane Feb 28 at 0:06

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Just like that? No "please", no showing self effort, ideas, tries...? –  DonAntonio Nov 4 '12 at 13:06
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3 Answers 3

You can solve this without knowing anything about quadratics.

You know that $x-y=5$, and the other condition implies that $x+y = 13$ or $x+y= -13$. Just consider the two cases separately and solve the linear equations simultaneously.

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$\left\{ \begin{array}{l} x>0 \\ y>0 \\ |x-y|=5 \\ (x+y)^2=169 \therefore x+y=\pm \sqrt{169}=\pm13 \end{array} \right. $

Solve.

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Here is my thought process when thinking through this problem: I know that $\sqrt{169} = 13$ since we are only working in the positive reals. Then, since we are given that the difference between two numbers (say, $x$ and $y$) is $5$, I started thinking of small integers that add up to $13$ and whose difference is $5$. From this point, it should be trivial to note that only $4$ and $9$ work.

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