Question

Two years ago my age was
4 × 1 ÷ 2 times the age of son . 6 years ago , my age was twice the square of the age of my son . what is present age of me and my son .

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

The present age of my son is 10 years and my present age is 38 years old

Solution:

Let the age of son 2 years ago be "x"

Given that,

Two years ago my age was  4 x 1 ÷ 2 times the age of son

Which means,

$\text{ My age two years ago } = 4\frac{1}{2} \times x\\\\\text{ My age two years ago } = 4.5x$

6 years ago , my age was twice the square of the age of my son

My age 6 years ago = 4.5x - 4

Son age 6 years ago = x - 4

By given condition,

6 years ago , my age was twice the square of the age of my son

$4.5x - 4 = 2(x-4)^2\\\\4.5x - 4 = 2(x^2 - 8x + 16)\\\\4.5x - 4 = 2x^2 - 16x + 32\\\\2x^2 - 16x - 4.5x + 32 + 4 = 0\\\\2x^2 - 20.5x + 36 = 0$

Solve by quadratic formula,

$x = 8\ and\ x = \frac{9}{4}$

Discard the fraction

Take x = 8

Therefore,

age of son 2 years ago = "x"

Present age of son = x + 2 = 8 + 2 = 10

My age two years ago = 4.5x

My Present age = 4.5x + 2 = 4.5(8) + 2 = 38

Thus the present age of my son is 10 years and my present age is 38 years old

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