# alg 150 help

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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?

• alg 150 help -

if x is the length, 2x-9 = width

x(2x-9) = 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(√409-9)^2)
= 1/8 (1225-27√409)
d = √(that)

that is an unusual answer. I suspect a typo somewhere. If so, make the adjustment and redo the calculations.

• alg 150 help -

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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