Question 847422: 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
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! x = son's age now
x² = father's age now
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
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