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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Solution

Let x and (16 - x) are two parts of 16 where (16 - x) is longer and x is smaller.
According to the question,

2 $\times$ square of longer = square of smaller + 164

⇒ $2 \times (16 - x)^{2} = x^{2} + 164$

⇒ $2 \times (256 + x^{2} - 32 x) = x^{2} + 164$

⇒ $512 + 2 x^{2} - 64 x = x^{2} + 164$

⇒ $x^{2} - 64 x + 512 - 164 = 0$

⇒ $x^{2} - 64 x + 348 = 0$

⇒ $x^{2} - 58 x - 6 x + 348 = 0$

⇒ $x (x - 58) - 6 (x - 58) = 0$

⇒ $(x - 6) (x - 58) = 0$

⇒ $x = 6 a n d 58$

But $x \neq 58$ because $x < 16$

so, $x = 6$ and $16 - x = 10$

Hence, the answer is 6 and 10

Suggest Corrections

124

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