# One year ago , the father was 8 times as old as his son . Now his age is the square of his son's age .Find their present ages.

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by dweejareddy

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by dweejareddy

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father's age one yr ago= 8(x-1).

now his age = x^2

x^2 -1=8(x-1)

x^2=8x-7

x^2-8x+7=0

(x-7)(x-1)

x=7

Son's age now is 7

nd father's age is =7*7=49

f-1=8(s-1)⇒f=8s-7

f=s²

putting (i) here

8s-7=s²

⇒s²-8s+7=0

⇒(s-7)(s-1)

so s=7 or s=1

⇒f=49 or f=1

since f=1 and s=1 in not possible.

The correct answer is son's age=7 and father's age=49