Posted by **eddie** on Sunday, February 10, 2013 at 6:51pm.

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 -
**Steve**, Sunday, February 10, 2013 at 7:07pm
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.

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