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

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Answer 2

Answer:

12,8

Step-by-step explanation:

Let square of smaller part be x

and square of greater part = 20-x

According to que.,

(20-x)² = 2x² + 16

400-40x+x² = 2x² + 16

2x²-x²+40x+16-400 = 0

x²+40x-384 = 0

Now solving equation by factorization method,

x² + 40x - 384 = 0

x² +48x - 8x - 384 =0

x(x+48) - 8(x+48) =0

(x-8) (x+48) = 0

x=8 & x= -48

Hence -48 is negative, so it is not possible

Thus smaller part is x=8

and greater part is 20-x = 20-8

= 12

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