Find two numbers, one of which is double the other, if the square of their sum exceeds the sum of their squares by 100.
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Originally Posted by magentarita Find two numbers, one of which is double the other, if the square of their sum exceeds the sum of their squares by 100. So: $\displaystyle (2x + x)^2 - ( (2x)^2 + (x)^2 ) = 100$ And then it's a simple solve.
Originally Posted by janvdl So: $\displaystyle (2x + x)^2 - ( (2x)^2 + (x)^2 ) = 100$ And then it's a simple solve. Yes, I can take it from here.
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