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 age.

Answer 1

Answer: 49 years and 7 years.

Step-by-step explanation:

Let the man's age be 'x'

Let the son's age be '8x'   (According to the question)

Man's age 1 year ago = 8x+1 ------(1)

Son's age 1 year ago = x+1 ---------(2)

According to the question,

8x+1 = (x+1)²

8x+1 = x²+1²+2×1×x

8x+1 = x²+1+2x

0= x²+2x-8x

x²-6x = 0

Now,

Taking 'x' as common,

x(x-6) = 0

x-6 = 0

x= 6

Putting the value of 'x' in equation (1) and (2):

8x+1 = 49

x+1 = 7

The age of man is 49 years and the age of son is 7 years.

Answer 2

$\huge \boxed{ \bf \: Solution}$

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

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

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

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

$\rm \: According \: to \: the \: question,$

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

$\rm \: which \: is \: the \: required \: quadratic \: equation.$

$\rm \: Now,$

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

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

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

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

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

$\rm \: But \: x =1 \: is \: not \: possible \: because \: if \: x=1, \: then \: present \\\rm \: age \: of \: the \: son \: and \: father \: are \: same. \: So, \: x=7.$

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

$\rm \: and \: present \: age \: of \: man = 8 \times \: 7 -7=49 \: year.$