Question

. Two years ago, a man's age was three times the square of his son's age. In three years' time
his age will be four times his son's age. Find their present ages.​

Answer 1

let age of man be x years

age of his son be y years

so,

two years ago,

x-2=3(y-2)^2 -----(i)

three years after,

x+3=4(y+3)

X+3=4y+12

x=4y+9 -----(ii)

now, put the value of x from eq.ii in the eq.i.

x-2=3(y-2)^2

4y+9-2=3(y^2+4-4y)

4y+7=3y^2+12-12y

4y+12y+7-12=3y^2

16y-5=3y^2.

3y^2-16y+5=0

3y^2-15y-1y+5=0

3y(y-5)-1(y-5)=0

(3y-1)(y-5)=0

from it we get,

y=1/3 ( not possible)

and y=5

so, sons age is 5 years

and for man age put value of y in eq.ii

x=4y+9

x=(4x5)+9

x=29 years.

I hope this is helpful to you ☺️