Question

The width of a rectangle is 8 less than twice its length. if the area of the rectangle is 137 cm^2, what is the length of the diagonal?

Answer 1

Length (L): L

width (w): 2L - 8

Area (A) = L x w

       137 = L(2L - 8)

       137 = 2L² - 8L

         0 = 2L² - 8L - 137

L = $[-(-8) +/- \sqrt{(-8)^{2}  - 4(2)(-137) }]/2(2)$

   =  $[8 +/- \sqrt{(64  + 1096 }]/4$        

  = $[8 +/- \sqrt{(1160 }]/4$  

  = $[8 +/- 2\sqrt{(290 }]/4$  

  = $[4 +/- \sqrt{(290 }]/2$

since length cannot be negative, disregard [4 - √290]/2

L = [4 + √290]/2  = 10.51

width (w): 2L - 8   = 2(10.51) - 8   = 21.02 - 8   = 13.02

Answer: width = 13.02 cm, length = 10.51 cm