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

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

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Answer 2

Answer:

Let One part be x.

So, the other part is (20−x)

According to question.

3x

2

=(20−x)+10

3x

2

=30−x

3x

2

+x−30=0

3x

2

+10x−9x−30=0

x(3x+10)−3(3x+10)=0

(3x+10)(x−3)=0

x=3 , x=

3

−10

Hence, the two parts =3,17