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
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