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Let present age of the son be x years ∴ Present age of the father = x 2 years one year ago, Age of the son = (x – 1) years and Age of the father = (x 2 – 1) years According to the given condition, x 2 – 1 = 8(x – 1) ⇒ x 2 – 1 – 8x + 8 = 0 x 2 – 8x + 7 = 0 ⇒ (x – 1) (x– 7) = 0 ⇒ x – 1 = 0, x – 7 = 0 i.e., x = 1, x = 7 When x = 1, age of son = age of father, which is impossible ∴ x = 1 is rejected. ∴ x = 7 Thus present age of son = 7 years and present age of father = 49 years.

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Quadratic Equations

Long Answer Type

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183. 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.

Let present age of the son be x years
∴ Present age of the father = x² years
one year ago,
Age of the son = (x – 1) years
and Age of the father = (x² – 1) years
According to the given condition,
x² – 1 = 8(x – 1)
⇒ x² – 1 – 8x + 8 = 0
x² – 8x + 7 = 0
⇒ (x – 1) (x– 7) = 0
⇒ x – 1 = 0, x – 7 = 0
i.e., x = 1, x = 7
When x = 1, age of son = age of father, which is impossible
∴ x = 1 is rejected.
∴ x = 7
Thus present age of son = 7 years
and present age of father = 49 years.

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