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 = 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

Answer:

ONE PART IS X

THE OTHER IS 16-X

X²-(16-X)²=164

X²-(256-32X+X²)=164

32X-256=164

32X=420

X=13.125

THE PARTS ARE 13.125,2.875