Question

the age of a man is twice the square of the age of his son. eight years hence, the age of the man will be 4 years more than three times the age of his son. find their present age and age after 4years and 8years​

Answer 1

Answer:

this is ok na ................................

Answer 2

Solution :

Let the age of son's be r years & the age of man's be 2r² years respectively.

A/q

$\underbrace{\bf{After\:8\:years\::}}}}$

The age of son's will be (r+8) years.

The age of man's will be (2r²+8) years.

So;

$\longrightarrow\sf{2r^{2} +8=3(r+8)+4}\\\\\longrightarrow\sf{2r^{2} +8=3r+24+4}\\\\\longrightarrow\sf{2r^{2} +8=3r+28}\\\\\longrightarrow\sf{2r^{2} -3r=28-8}\\\\\longrightarrow\sf{2r^{2} -3r=20}\\\\\longrightarrow\sf{2r^{2} -3r-20=0}\\\\\longrightarrow\sf{2r^{2} -8r+5r-20=0}\\\\\longrightarrow\sf{2r(r-4)+5(r-4)=0}\\\\\longrightarrow\sf{(r-4)(2r+5)=0}\\\\\longrightarrow\sf{r-4=0\:\:\:Or\:\:\:2r+5=0}\\\\\longrightarrow\sf{r=4\:\:\:Or\:\:\:2r=-5}$

$\longrightarrow\bf{r=4\:\:\:Or\:\:\:r\neq -5/2}$

We know that negative value isn't acceptable.

$\boxed{\sf{The\:present\:age\:of\:son's\:(r)=\boxed{\bf{4\:years}}}}}\\\boxed{\sf{The\:present\:age\:of\:father's\:(2r^{2} )=2\times 16=\boxed{\bf{32\:years}}}}}$