Question

Two years ago my age was 4.5 times the age of son. 6 years ago, my age was twice the square of the age of my son. What is the present age of me and my son.​

Answer 1

Answer:

Let 2 years ago, son's age be x.

So my age(2 years ago) =4.5 x

So, 6 years ago,

My age=4.5x-4

Son's age=x-4

B/C,

4.5x-4=2((x-4)^2)

or, 4x^2–41x+72=0

Solving the quadratic, we get-

x=8 or x=2.5(not possible)

So my present age of=4.5x+2=38

My son's present age=x+2=10

Answer 2

Answer:

Step-by-step explanation:

let son's age 2yrs ago be x yrs.

so my age is(2 yrs ago) =4.5x

so 6 yrs ago

my age=4.5x-4

so son's age will be= x-4

we get the equations 4$x^{2}$-41x+72=0

on solving the quadratic eq we get

x=8 or x=2.5

x=2.5 is not possible so my present age will be 4.5x +2=38

and son's present age will be x+2=8+2=10 yrs

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