Age problem

One year ago, the father was 8 times as old as his son. Now, his age is the square of his son's age.  Find their present age.

x = son's age now
x² = father's age now   {father is now the square of his son's age} 

x - 1 = son 1 year ago {subtracted 1 from present age}
x² - 1 = father's age 1 year ago {subtracted 1 from present age} 

x² - 1 = 8(x - 1)   {one year ago, father was equal to eight times son}
x² - 1 = 8x - 8   {used distributive property}
x² - 8x + 7 = 0   {subtracted 8x and added 8 to each side}
(x - 7)(x - 1) = 0   {factored into two binomials}
x - 7 = 0 or x - 1 = 0   {set each factor equal to 0}
x = 7 or x = 1   {solved each equation for x}
x = 7   {x cannot be 1, because the father would have been 0, one year ago}
x² = 49   {substituted 7, in for x, into x²} 

son is 7
father is 49

- Algebra House