SOLUTION: A painting canvas has a length 9 inches longer than its width. If the area is 90in to the second power, what are the lengths of the legs of the field?

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A painting canvas has a length 9 inches longer than its width. If the area is 90in to the second power, what are the lengths of the legs of the field?      Log On


   

Question 87337: A painting canvas has a length 9 inches longer than its width. If the area is 90in to the second power, what are the lengths of the legs of the field?
Answer by tutorcecilia(2152) About Me  (Show Source):
You can put this solution on YOUR website!
Use the formula for the area of a rectangle
Area = (length) ( width)
Area = 90
Length = Width + 9
Width = W
.
Plug-in the values and solve for the w-term]
90=(w+9)(w)
90 = w^2+9w [set the equation equal to zero]
90-90 = w^2+9w-90
0=w^2+9w-90 [solve for the w-term]
0=(w+15)(w-6)[factor; set each factor equal to zero and solve for w]
w+15=0
w=-15 [disregard a negative measurement]
or
w-6=0
w=6 [use this measurement because it is positive]
.
Length = w+90=6+90=96
Check by plugging all values back into the original equation:
A=lw
90=(w+9)(w)
90=(6+9)(6)
90=(15)(6)
90=90 [checks out]