#### Question

One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

#### Solution

Let the present age of the son be x years.

∴ Present age of man = x^{2} years

One year ago,

Son’s age = (x – 1) years

Man’s age = (x^{2} – 1) years

It is given that one year ago; a man was 8 times as old as his son.

∴ (x^{2} – 1) = 8(x – 1)

x^{2} – 8x – 1 + 8 = 0

x^{2} – 8x + 7 = 0

(x – 7) (x – 1) = 0

x = 7, 1

If x = 1, then x^{2} = 1, which is not possible as father’s age cannot be equal to son’s age.

So, x = 7.

Present age of son = x years = 7 years

Present age of man = x^{2} years = 49 years

Is there an error in this question or solution?

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Solution One Year Ago, a Man Was 8 Times as Old as His Son. Now His Age is Equal to the Square of His Son’S Age. Find Their Present Ages. Concept: Quadratic Equations.

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