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?

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) (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]

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