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 integerTherefore, the two numbers are 8 and 12.Mathematics

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