Question

The age of a father is twice the square of the age of his son. Eight years hence, te age of the father will be 4 more than 3 times the age of the son. Find their present ages.​

Answer 1

Step-by-step explanation:

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Answer 2

Given :-

• The age of the father is twice the square of the age of his son.

• Eight year hence,the age of the father will be 4 years more than 3 times the age of his son.

To find :-

• Theri present ages

Solution :-

Let the present age of the son be x year

and the present age of father be 2x² years.

$\bf\dag\:{\underline{Eight\:Years\:hence}}:-$

• Age of son = (x + 8) years

• Age of father = (2x² + 8) years

According to the question,

(2x² + 8) = 3(x + 8) + 4

⤇ 2x² + 8 = 3x + 24 + 4

⤇ 2x² + 8 = 3x + 28

⤇ 2x² = 3x + 28 - 8

⤇ 2x² = 3x + 20

⤇ 2x² - 3x - 20 = 0

⤇ 2x² - 8x + 5x - 20 = 0

⤇ 2x(x - 4) + 5(x - 4) = 0

⤇ (x - 4) (2x + 5) = 0

Now,

x - 4 = 0

⤇ x = 0 + 4

⤇ x = 4

Then,

2x + 5 = 0

⤇ 2x = 0 - 5

⤇ 2x = -5

⤇ x = $\rm\dfrac{-5}{2}$

We know that,age cannot be negative so,here

x = 4

Now,

⤇ 2x²

⤇ 2 × (4)²

⤇ 2 × 16

⤇ 32

Hence,the present age of son will be 4 years and present age of father will be 32 years.