Question

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$\huge{question\: iS \: for\: Experts}$✔✔

The age of man is twice the square of age of his son . Eight year hence, the age of the man will be 4 years more than three times the age of his son. Find the present ages.✔✔

Solution please✌

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Answer 1

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Answer 2

$\underline{\underline{\bold{Question:}}}$

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

$\underline{\bold{Solution:}}$

$\bold{\textsf{Let the son's age be x years .}}\\\\\therefore{\bold{Father's\:age =Twice\:the\:square\:of\:son's\:age. }}$

                                                        $\bold{=2x^{2}\:years. }$

8 years hence,

• The son's age = ( x+8 ) years.
• Father's age = (2x²+8) years.

According to question,

$\implies{\bold{(x+8)3+4=2x^{2} +8}}\\\\\\\implies{\bold{3x+24+4=2x^{2}+8 }}\\\\\\\implies{\bold{2x^{2}-3x-20=0 }}\\\\\\\implies{\bold{2x^{2}-8x+5x-20=0 }}\\\\\\\implies{\bold{2x(x-4)+5(x-4)=0}}\\\\\\\implies{\bold{(x-4)(2x+5)=0}}\\\\\\\implies{\bold{x=4,-\dfrac{5}{2}.}}\\\\\\\boxed{\bold{The\:possible\:value\:of\:x=4.}}$

$\boxed{\boxed{\bold{The\:son's\:age=4\:years\:and\:Father's\:age=32\:years.}}}$