Question

Six year before, the age of mother was equal to the square of her son's

age. Three year hence, her age will be thrice the age of her son then.

Find the present ages of the mother and son???

Answer 1

Let age of son is x and age of mother is y

a/c to question,
six years before , the age of mother was equal to the square of her son's age.

so, y - 6 = (x - 6)²

or, y - 6 = x² - 12x + 36

or, y = x² - 12x + 42 ......(1)

again, after three years ,her age will be thrice the age of her son.

so, y + 3 = 3(x + 3)

or, y + 3 = 3x + 9

or, y = 3x + 6 .......(2)

put equation (2) in equation (1),

3x + 6 = x² - 12x + 42

x² - 12x + 42 - 3x - 6 = 0

x² - 15x + 36 = 0

x² - 12x - 3x + 36 = 0

x(x - 12) - 3(x - 12) = 0

(x - 3)(x - 12) = 0

x = 3, 12

y = 3x + 6 = 3 × 3 + 6 = 15
or 3 × 12 + 6 = 36 + 42

but if we put y = 15 in y = x² - 12x + 42
15 = x² - 42x + 42
x² - 42x + 27 = 0
Discriminant = (42)² - 4 × 27 < 0
so, x ≠ 3

hence, x = 12 and y = 42

so, son's age = 12 years and mother's age = 42 years

Answer 2

Answer:

Son = 12 years old

Mother = 42 years old

Step-by-step explanation:

Define x:

Let the son be x years old six year ago

His mother will be x²

Find their age now:

son = x + 6

mother = x² + 6

Find their age 3 year later:

son = x + 6 + 3 = x + 9

mother = x² + 6 + 3 = x² + 9

Solve x:

Mother's age will be thrice the age of her son

x² + 9 = 3(x + 9)

x² + 9 = 3x + 27

x² - 3x - 18 = 0

(x - 6)(x + 3)  = 0

x = 6 or x = -3 (rejected, since age cannot be negative)

Find their present age:

son = x + 6 = 6 + 6 = 12 years old

mother = x² + 6 = 6² + 6 = 42 years old

Answer: The son is 12 years old and the mother is 42 years old.