Question

divide 16 into two parts such that twice the square of the longer part exceeds the sqare of the smaller part by 164

Answer 1

Let x and (16 - x) are two parts of 16 where (16 - x) is longer and x is smaller .
A/C to question,
2 × square of longer = square of smaller + 164
⇒ 2 × (16 - x)² = x² + 164
⇒ 2 × (256 + x² - 32x ) = x² + 164
⇒ 512 + 2x² - 64x = x² + 164
⇒ x² - 64x + 512 - 164 = 0
⇒ x² - 64x + 348 = 0
⇒x² - 58x - 6x + 348 = 0
⇒ x(x - 58) - 6(x - 58) = 0
⇒(x - 6)(x - 58) = 0
⇒ x = 6 and 58

But x ≠ 58 because x < 16
so, x = 6 and 16 - x = 10

Hence, answer is 6 and 10

Answer 2

Solution:-

given by:-

divide 16 into two parts let x is longer part and
(16-x ) is smaller part

acourdind to quetion:-

$2 \times {x}^{2} = {(16 - x)}^{2} + 164 \\ 2 {x}^{2} = 256 + {x}^{2} - 32x + 164 \\ {x}^{2} + 32x - 420 = 0\\ {x}^{2} + 42x - 10x - 420 = 0 \\ x(x + 42) - 10(x + 42) = 0\\ (x + 42)(x - 10) = 0 \\ x = - 42 \: does \: not \: exits \\ (x = 10 ) \: longer \: part \\ (16 - x) = (16 - 10) \\ 6 \: is \: smaller \: part \\ 10 \: is \: longer \: part$
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