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Question

The sum of the squares of two positive integer is 208. If the square of larger number is 18 times the smaller number, find the numbers.


Solution

Let's assume the two parts to be $$x$$ and $$y$$,  $$y$$ being the larger of the two numbers.
Then, from the question
$$x^{2} + y^{2} = 208$$ .......(i) and 
$$ y^{2} = 18x$$ .......(ii) 
From (i), we get $$ y^{2} = 208 - x^{2} $$
Now, putting this in (ii), we have
$$ 208 - x^{2} = 18x$$ 
$$x^{2} + 18x - 208 = 0$$
$$x^{2} + 26x - 8x - 208 = 0$$
$$x\left(x + 26 \right) - 8\left(x + 26 \right) = 0$$
As $$x$$ can't be a negative integer, so $$ x= 8$$ is valid solution.
Using $$ x= 8$$ in (ii), we get $$y^{2} = 18 \times 8 = 144$$
Thus, $$y = 12$$ only as $$y$$ is also a positive integer
Therefore, the two numbers are 8 and 12.

Mathematics

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