Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.
Quadratic Equations
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Question

Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.

KnowledgeBoat · Accepted Answer

Let present age of daughter be x years Two years before daughter's age = (x - 2) years Man's age two years before = 3(x - 2)² So, present age of man = 3(x - 2)² + 2 Three year's later Daughter's age = (x + 3) years Man's age = (3(x - 2)² + 2 + 3) years According to question, Since, age cannot be in fraction so x ≠ ∴ x = 5, 3(x - 2)² + 2 = 3(5 - 2)² + 2 = 3(3)² + 2 = 29 The present age of daughter is 5 years and of man is 29 years.

Mathematics

Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.

Quadratic Equations

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Mathematics

Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.

Quadratic Equations

Related Questions

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If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14cm2. However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Determine the sides of two squares.

If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14cm². However, if twice the area of the larger square is added to three times the area of the smaller square, the result is 203 cm². Determine the sides of two squares.