Question

the different between the squares of two numbers is 120 the square of the smaller number is twice of greater number.find the numbers ?
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Answer 1

$let \: the \: two \: numbers \: be \: x \: and \: y \\ given \\ \\ {y}^{2} - {x}^{2} = 120 \\ {x}^{2} = 2y \\ {y}^{2} - 2y = 120 \\ {y}^{2} - 2y - 120 = 0 \\ {y}^{2} - 12y + 10y - 120 =0 \\ y(y - 12) + 10(y - 12) = 0 \\ (y + 10)(y - 12) = 0 \\ since \: numbers \: cannot \: be \: negative \\ y = 12 \\ {x}^{2} = 24 \\ x = 2 \sqrt{6} \\ \\ {12}^{2} - {(2 \sqrt{6}) }^{2} = 120$

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