Divide 16 into 2 parts such that twice the sq | vedclass

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(A) Let the smaller part be $x$. Then the larger part is $(16 - x)$.According to the problem,twice the square of the larger part exceeds the square of

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$10, 6$

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(A) Let the smaller part be $x$. Then the larger part is $(16 - x)$.According to the problem,twice the square of the larger part exceeds the square of the smaller part by $164$:$2(16 - x)^2 - x^2 = 164$$2(256 + x^2 - 32x) - x^2 = 164$$512 + 2x^2 - 64x - x^2 = 164$$x^2 - 64x + 348 = 0$Factoring the quadratic equation:$x^2 - 6x - 58x + 348 = 0$$x(x - 6) - 58(x - 6) = 0$$(x - 6)(x - 58) = 0$So,$x = 6$ or $x = 58$.Since the sum of the two parts is $16$,$x$ cannot be $58$.Therefore,the smaller part is $6$ and the larger part is $16 - 6 = 10$.

Divide $16$ into $2$ parts such that twice the square of the larger part exceeds the square of the smaller part by $164$.

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