Question

1 year ago a man was 8 times as old as his son, now his age is equal to the square of his son age find their ages. explain ? { need it urgently }

Answer 1

given,
age of son=x
age of man=x²
one year back,
age of son=x-1
age of man=x²-1
but given age of man=8(x-1)
x²-1=8(x-1)
x²-1=8x-8
x²-8x-1+8=0
x²-8x+7=0
x²-7x-x+7=0
x(x-7)-1(x-7)=0
(x-7)(x-1)=0
the age of the son=7

we cannot take son's age =0

Answer 2

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AGE PROBLEMS QUESTION SOLUTION !

Let us assume that one years ago son's age be x years.

Then ,man's age one year ago = 8x years

Present age of son be = ( x + 1 ) years

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

Therefore, 8x + 1 = ( x + 1 )²

x² - 6x = 0

x ( x - 6 ) = 0

x = 0 or x = 6

x = 6

Present age of son be = ( x + 1 ) ⇒ 7 years.

Present age of man be = ( 8x + 1 ) ⇒ 49 years.

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