SOLUTION: The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 81 "cm"^2, what is the length of the diagonal?

Question 789258: The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 81 "cm"^2, what is the length of the diagonal?
Answer by waynest(281) (Show Source): You can put this solution on YOUR website!
w = 2L - 6
A = Lw
81 = L(2L -6)
81 = 2L^2 -6L
81 - 81 = 2L^2 - 6L - 81
0 = 2L^2 -6L - 81
use quadratic formula
(-b +-sqrtb^2 - 4(a)(c))/2(a) a = 2, b = -6, c = -81
(-(-6) +-sqrt-6^2 - 4(2)9-81))/2(2)
(6 +-sqrt16 + 648)/4
(6 +-sqrt684)/4
(6 +- 6sqrt19)/4
x =(6 +- 6sqrt19)/4
x = (3 +-3sqrt19)/2
x = (3 + 3sqrt19)/2
x = 8.038
x = (3 - 3sqrt19)/2
x = -5.038
you can't have a negative value so use 8.038
c^2 = a^2 + b^2 a = 8.038, b = 2(8.038) - 6 = 10.077
c^2 = 8.038^2 + 10.077^2
c^2 = 166.155
sqrtc^2 = sqrt166.155
c = 12.890
the diagonal is 12.890 cm
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