Divide 16 into two parts such that twice the | vedclass

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(D) Let the two parts be $x$ and $16 - x$. Assume $x$ is the larger part.According to the problem,$2x^2 - (16 - x)^2 = 164$.Expanding the equation: $2

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$10, 6$

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(D) Let the two parts be $x$ and $16 - x$. Assume $x$ is the larger part.According to the problem,$2x^2 - (16 - x)^2 = 164$.Expanding the equation: $2x^2 - (256 - 32x + x^2) = 164$.$2x^2 - 256 + 32x - x^2 = 164$.$x^2 + 32x - 420 = 0$.Factoring the quadratic equation: $x^2 + 42x - 10x - 420 = 0$.$x(x + 42) - 10(x + 42) = 0$.$(x - 10)(x + 42) = 0$.Since the parts must be positive,$x = 10$.The other part is $16 - 10 = 6$.Thus,the two parts are $10$ and $6$.

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

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