How do you find two positive numbers such that the sum of the squares of the two numbers is 169 and the difference between the two numbers is 7?

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Answer 1 · 2016-12-11T06:59:58

5, 12

Let $x$ be the larger of the 2 positive numbers
Let $y$ be the smaller of the 2 positive numbers

$x^2 + y^2 = 169$

$x - y = 7$

$=> x = 7 + y$

$x^2 + y^2 = 169$

$=> (y + 7)^2 + y^2 = 169$

$=> y^2 + 14y + 49 + y^2 = 169$

$=> 2y^2 + 14y - 120 = 0$

$=> 2(y^2 + 7y - 60) = 0$

$=> y^2 + 7y - 60 = 0$

$=> (y + 12)(y - 5) = 0$

$=> y = -12, y = 5$

However, we want $y > 0$

$=> y = 5$

$x - y = 7$

$=> x - 5 = 7$

$=> x = 12$