Question

The son of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
​

Answer 1

Information provided with us:

• The son of a father is twice the square of the age of his son
• Eight years hence, the age of the father will be 4 years more than three times the age of the son.

We have to calculate:

• Present ages of son and his father

Assuming son's age,

• Son's age = x

Assuming father's age:

• Father's age = 2x²

After eight years:

Adding 8 to both the ages,

• Son's age = x + 8
• Father's age = 2x² + 8

4 years more than three times the age of the son:

We have,

• Son's age = x + 8

So, after 4 years,

• 3 (x + 8) + 4

Equation formed:

• 2x² + 8 = 3 (x + 8) + 4

Solving it now!!

➠ 2x² + 8 = 3 × (x + 8) + 4

➠ 2x² + 8 = 3x + 24 + 4

Adding 24 and 4,

➠ 2x² + 8 = 3x + 28

Transposing 3x + 28 to L.H.S.,

➠ 2x² + 8 - 3x - 28 = 0

➠ 2x² - 3x - 20 = 0

Forming factors,

➠ 2x² - 8x + 5x - 20 = 0

Factorising them,

➠ 2x (x - 4) + 5 (x - 4)

Grouping them,

➠ (x - 4) (2x + 5)

Hence, value of x is 4.

Present ages:

• Son = 4 years
• Father = 2x² → 2 × 4 × 4 → 2 × 16 → 32 years.

Learn More:

• https://brainly.in/question/38610284
• https://brainly.in/question/42623694