Question

Two years ago my age was 9/2 times my son's age.six years ago my age was twice the square of my sons age . Find my sons present age

Answer 1

Answer: So, the present age of his son is 6 years.

Step-by-step explanation:

Since we have given that

Let the son's age  two years ago be x.

Let his father's age two years ago be $\frac{9}{2}x$

Six years ago,

Son's age becomes (x-6)

Father's age becomes $\frac{9}{2}x-6$

According to question, his age was twice the square of his son's age.

$\frac{9}{2}x-6=(x-6)^2\\\\\frac{9x-12}{6}=x^2-12x+36\\\\9x-12=6x^2-72x+216\\\\6x^2-72x-9x+216+12=0\\\\6x^2-81x+228=0\\\\2x^2-27x+76=0\\\\2x^2-8x-19x+76=0\\\\2x(x-4)-19(x-4)=0\\\\(2x-19)(x-4)=0\\\\x=\frac{19}{2},4$

Age can't be in fraction, so x = 4

Two years ago, his son's age is 4 years.

So, the present age of his son is 4+2=6 years.

Answer 2

Step-by-step explanation:

Answer: So, the present age of his son is 6 years.

Step-by-step explanation:

Since we have given that

Let the son's age two years ago be x.

Let his father's age two years ago be \frac{9}{2}x

2

9

x

Six years ago,

Son's age becomes (x-6)

Father's age becomes \frac{9}{2}x-6

2

9

x−6

According to question, his age was twice the square of his son's age.

\begin{gathered}\frac{9}{2}x-6=(x-6)^2\\\\\frac{9x-12}{6}=x^2-12x+36\\\\9x-12=6x^2-72x+216\\\\6x^2-72x-9x+216+12=0\\\\6x^2-81x+228=0\\\\2x^2-27x+76=0\\\\2x^2-8x-19x+76=0\\\\2x(x-4)-19(x-4)=0\\\\(2x-19)(x-4)=0\\\\x=\frac{19}{2},4\end{gathered}

2

9

x−6=(x−6)

2

6

9x−12

=x

2

−12x+36

9x−12=6x

2

−72x+216

6x

2

−72x−9x+216+12=0

6x

2

−81x+228=0

2x

2

−27x+76=0

2x

2

−8x−19x+76=0

2x(x−4)−19(x−4)=0

(2x−19)(x−4)=0

x=

2

19

,4

Age can't be in fraction, so x = 4

Two years ago, his son's age is 4 years.

So, the present age of his son is 4+2=6 years.