Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Answers
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
Step-by-step explanation: mark me brainiest
Step-by-step explanation:
Let the larger part be x. Then, the smaller part = 16−x
By hypothesis, we have:
2x^2 = (16−x)^2+164
⇒x^2+32x−420 = 0
⇒(x+42)(x−10) = 0
⇒x = −42 or x = 10
∵ x can not be a negative
⇒x=10
Hence, the required parts are 10 and (16−10) = 6