Math, asked by saurav5030, 1 year ago

The age 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 his son. find their present ages.​

Answers

Answered by niral
21

Answer:

Step-by-step explanation:

Let the present age of son be x  years

Present age of his father be 2x²  years

Age of son after 8 years =(x+8) years

Age of his father after 8 years  years= (2x²+8) years

Given age of man will be 4 years more than 3 times his sons age.

Hence , 2x²+8=3(x+8)+4

→ 2x²+8=3x+28

→ 2x²-3x-20=0

→ 2x²-8x+5x-20=0

→ 2x(x-4)+5(x-4)=0

→ (x-4)(2x+5)=0

x-4=0  or 2x+5=0

∴ x=4    or x= -5/2

Hence x=4 since age cannot be negative

→ 2x²=2(4)²=32

Therefore present age of son is 4 years and present age of his father is 32 years.


niral: mark me as brainliest answer.
Answered by VishalSharma01
57

Answer:

Step-by-step explanation:

Solution :-

Let x be the present age of son.

And the present age of man is 2x² years.

8 years hence :-

Age of the son = x + 8 years.

Age of the man = 2x² + 8 years.

According to the Question,

(2x² + 8) = 3(x + 8) + 4

2x² - 3x - 20 = 0

⇒ 2x² - 8x + 5x - 20 = 0

⇒ 2x(x - 4) + 5(x - 4) = 0

⇒ (x - 4) (2x + 5) = 0

⇒ x - 4 = 0 or 2x + 5 = 0

x = 4, - 5/2 (As x can't be negative)

x = 4

Son's present age = x = 4 years.

Man's present age = 2x² = 2(4)² = 2(16) = 32 years.

Hence, the present age of son and man is 4 years and 32 years.

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