Question

Find two parts of 20 such that the square
of the greater part exceeds twice the square
of the smaller part by 16.​

Answer 1

Answer:

Step-by-step explanation:

Let the smaller part be x

Hence the larger part is (20 – x)

Given,

2(20 – x)^2 = x^2 + 16

⇒ 2(400 – 40x + x^2) =  x^2 + 16

⇒ 800 – 80x + 2x^2 –  x^2 – 16 = 0

⇒ x^2– 80x + 784 = 0

Solve this to get the value of x

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Answer 2

Answer:

The two parts are 12 and 8

Step-by-step explanation:

(12)^2= 144 ------(1)

(8)^2= 64 --------(2)

64*2= 128

144-128 = 16

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