Question

One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. Find their present ages.

Answer 1

Step-by-step explanation:

Let x be the age of the son one year ago, then the age of the man was 8x.

The present age of the son is (x+1) and that of the man is (8x+1). Then,

8x+1=(x+1) 2

8x+1=x 2+1+2x

x=0,6

As the age cannot be 0, so the value of x is 6.

So, present age of son =(x+1)=7 years

and Present age of man =(8x+1)=48 years

Answer 2

$\boxed{ \boxed{ \bf \: SOLUTION}}$

$\rm \: Let \: present \: age \: of \: his \: son = x \: year$

$\rm \: One \: year \: ago, \: his \: son's \: age = (x - 1 )\: year$

$\rm \: One \: year \: ago, \: man's \: age = 8(x - 1) \: year = (8x - 8) \: year$

$\rm \: Present \: age \: of \: man = (8x - 8 + 1) \: year = (8x - 7) \: year$

According to the question,

$\rm \: 8x - 7 = {x}^{2} \implies \: {x}^{2} - 8x + = 0$

which is the required quadratic equation.

$\rm \: Now, \: {x}^{2} - 7x - x + 7 = 0 \: \: \: \: \: \: \: \: \: [by \: factorisation]$

$\implies \: \rm \: x(x - 7) - 1(x - 7) = 0$

$\implies \: \rm \: (x - 7)(x - 1) = 0$

$\implies \: \rm \: x - 7 = 0 \: or \: x - 1 = 0$

$\implies \: \rm \: x = 7 \: or \: x = 1$

But x = 1 is not possible because if x = 1, then present age of the son and father are same.

$\rm \: so \: x = 7.$

$\rm \: Hence, \: present \: age \: of \: his \: son = 7 \: year$

$\rm \: and \: present \: age \: of \: man = 8 \times 7 - 7 = \boxed {\boxed{ \bf{ \red{49 \: year.}}}}$