Question

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

Answer 1

Answer:

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 = x^2 + 164

⇒ 2 * (256 + x^2 - 32x ) = x^2 + 164

⇒ 512 + 2x^2 - 64x = x^2 + 164

⇒ x^2 - 64x + 512 - 164 = 0

⇒ x^2 - 64x + 348 = 0

⇒x^2 - 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

Let the two required numbers be x and y such that x > y.

Then,

$\sf\green{x + y = 16 \: and \: {2x}^{2} - {y}^{2} = 164}$

$\sf\red{ {2x}^{2} - {(16 - x) }^{2} = 164}$

$\sf\red{{x}^{2} + 32x - 420 = 0}$

$\sf\red{(x -1 0)(x + 42) = 0}$

$\sf\red{x = 10 \: or \: x = - 42}$

But x ≠ -42 ,so x = 10

Hence, the required two numbers are 10 & 6.