A rectangular pool measures 5 yd by 6 yd. A concrete deck of uniform width is constructed around the pool. The deck and pool together cover an area of 72 yds². How wide is the deck?
Common sense would say that if you add 3 yds onto the pool dimensions, that would create a width of 8 and length of 9, which would make the area 72 yds². Here is the algebraic proof:
width of pool = 5 yds
length of pool = 6 yds
x = width of deck
x + 5 = width of pool + width of deck
x + 6 = length of pool + width of deck
(x + 5)(x + 6) = 72 {multiplied width x length and set equal to area of 72}
x² + 11x + 30 = 72 {used foil method / distributive property}
x² + 11x - 42 = 0 {subtracted 72 from each side}
(x + 14)(x - 3) = 0 {factored into two binomials}
x + 14 = 0 or x - 3 = 0 {set each factor equal to zero}
x = -14 or x = 3 {solved each equation for x}
width of deck cannot be -14
width of deck = 3 yds
Ask the House
width of pool = 5 yds
length of pool = 6 yds
x = width of deck
x + 5 = width of pool + width of deck
x + 6 = length of pool + width of deck
(x + 5)(x + 6) = 72 {multiplied width x length and set equal to area of 72}
x² + 11x + 30 = 72 {used foil method / distributive property}
x² + 11x - 42 = 0 {subtracted 72 from each side}
(x + 14)(x - 3) = 0 {factored into two binomials}
x + 14 = 0 or x - 3 = 0 {set each factor equal to zero}
x = -14 or x = 3 {solved each equation for x}
width of deck cannot be -14
width of deck = 3 yds
Ask the House
