Question

Two years ago my age was 4 1/2(it is in mixed fraction )times the age of my son . Six year ago my age was twice the square of the age of my son . What is the present age of my son​

Answer 1

Answer:

The present age of son is 10 years

Step-by-step explanation:

Given :

• Two years ago, my age was $4 \dfrac{1}{2}$  times the age of my son.

• Six years ago, my age was twice the square of the age of my son.

To find :

the present age of son

Solution :

Let the present age of son be 'a' years

and my present age be 'b' years

Two years ago,

my age = (b - 2) years

son's age = (a - 2) years

As given,

b - 2 = 4¹/₂ (a - 2)

b - 2 = 9/2 (a - 2)

2(b - 2) = 9(a - 2)

2b - 4 = 9a - 18

 9a - 2b = 18 - 4

 9a - 2b = 14 --[1]

Six years ago,

my age = (b - 6) years

son's age = (a - 6) years

As given,

b - 6 = 2(a - 6)²

b - 6 = 2(a² + 6² - 2[a][6])

b - 6 = 2(a² + 36 - 12a)

b - 6 = 2a² + 72 - 24a

b = 2a² + 72 - 24a + 6

b = 2a² - 24a + 78

Substitute the value of b in equation [1],

9a - 2b = 14

9a - 2(2a² - 24a + 78) = 14

9a - 4a² + 48a - 156 - 14 = 0

 -4a² + 57a - 170 = 0

 4a² - 57a + 170 = 0

 

Solving the quadratic equation by quadratic formula,

 $\boxed{\tt x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}}$

4a² - 57a + 170 = 0

 $\sf a=\dfrac{-(-57) \pm \sqrt{(-57)^2-4(4)(170)}}{2(4)} \\\\ \sf a=\dfrac{57 \pm \sqrt{3249-2720}}{8} \\\\ \sf a=\dfrac{57 \pm \sqrt{529}}{8}\\\\ \sf a=\dfrac{57 \pm 23}{8}\\\\ \sf a=\dfrac{57+23}{8} ; a=\dfrac{57-23}{8} \\\\ \sf a=\dfrac{80}{8}; \dfrac{34}{8} \\\\ \sf a=10;4.25$

The age of son can be either 10 years or 4.25 years.

But if we take the second condition where son's age is (a - 6) years ; if we substitute a = 4.25 , we get negative value which is not acceptable.

Hence, the present age of son is 10 years.