alg 150 help
posted by eddie .
the width of a rectangle is 9 less than twice the length. if the area os the rectangle is 41cm^2 what is the length of the diagonal?

if x is the length, 2x9 = width
x(2x9) = 41
2x^2  9x  41 = 0
x = 1/4 (9±√409)
since we want a positive length,
length = 1/4 (9+√409)
width = 1/2 (9+√409)9 = (√409  9)/2
diagonal d^2 = 1/16 ((9+√409)^2 + 4(√4099)^2)
= 1/8 (122527√409)
d = √(that)
that is an unusual answer. I suspect a typo somewhere. If so, make the adjustment and redo the calculations. 
The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 55 cm22, what is the length of the diagonal?
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