Question

One year ago a man was 8 times as old as his daughter. Now his age is equal to the square of his daughter's age. The present age of daughter

Answer 1

Answer:

let the present age of daughter be x

then  father's present age is x²-1

ONE YEAR AGO

daughter's age = x-1

father's age = x²-1

ACCORDING TO QUESTION

8(x-1)=x²-1

8(x-1)=(x+1)(x-1)

8=x+1

x=7

Step-by-step explanation:

Answer 2

Step-by-step explanation:

man present age =(daughter age) ^2

let daughter age =x

man age =x^2

one year ago

man =8(daughter)

x^2-1=8(x-1)

x^2-1=8x-8

x^2-8x-1+8=0

x^2-8x+7=0

x^2-7x-x+7=0

x(x-7)-1(x-7)=0

(x-7)(x-1)=0

x=1,7

let daughter present age =x=7yrs

man age =x^2 =(7)^2=49yrs