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 ag**

**o**{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

father is 49

*- Algebra House*