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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 '14 at 0:06

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Lost1, Elias, user86418, Sami Ben Romdhane
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Just like that? No "please", no showing self effort, ideas, tries...? –  DonAntonio Nov 4 '12 at 13:06

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. $


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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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