Question

What are their present ages ?​

Answer 1

Answer:

Mother : 42 years old

Son : 12 years old

Step-by-step explanation:

Six years ago, the age of mother was equal to the square of her son's age.

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

Find the present age of the mother and son.

Six years ago

• $x^2$ was the mother's age.
• $x$ was the son's age.

Three years hence

• $3(x+9)$ is the mother's age.
• $(x+9)$ is the son's age.

Difference of their age is same.

$x^2-x=2(x+9)\\x^2-3x-18=0\\x=6,-3$

∴$x=6$

Present

Mother's age : $x^2+6=42$

Son's age : $x+6=12$

Answer 2

Answer:

Son = 12 years

Mother = 42 years 

Explanation - 

Six years ago let the son age = x

According to the given condition. Mother = x² 

Present ages now:

Son = x + 6

Mother = x²  + 6

3 years later:

Son = x + 6 + 3 = x + 9

Mother =  x²  + 6 + 3 =  x²  + 9

Solving x - 

x²  + 9 = 3(x + 9)

x²  + 9 = 3x + 27

x²  - 3x - 18 = 0

Sum= -3

Product= -18

Numbers= -6,3

By factorisation method,

x² - 6x + 3x - 18 = 0

(x² - 6x) + (3x - 18) = 0

x( x - 6 ) + 3( x - 6 ) = 0

(x - 6) (x + 3) =0

x = 6 or x = -3(rejected, because age

can never be negative)

Take x = 6,

Finding their Present age -

Son = x + 6 = 6 + 6 = 12 years.

Mother =  x²  + 6 = (6)² + 6 = 42 years.

Hence : 

Son = 12 years

Mother = 42 years