Question

Two years ago my age was 4and1/2 times the
age of my son. Six years ago, my age
was twice the square of the age of
my son what is the present age of
my
son?​

Answer 1

Answer:

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.

$\frac{9}{2}x - 6 = (x - 6) {}^{2}$

$\frac{9x - 12}{6} = x {}^{2} - 12x + 36$

$9x - 12 = 6x {}^{2} - 72x + 216$

$6x {}^{2} - 81x + 228 = 0$

$2x {}^{2} - 27x + 76 = 0$

$2x {}^{2} - 8x - 19x + 76 = 0$

$2x(x - 4) - 19(x - 4) = 0$

$(2x - 19)(x - 4) = 0$

$x = \frac{19}{2}4$

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.