Asked by Brian
length of a rectangle is 5y more than twice its width and area of the rectangle is 42 yards ^2, find the dimensions
Answers
Answered by
Reiny
width --- x
length ---- 2x + 5 (assuming your 5y is a typo)
x(2x+5) = 42
2x^2 + 5x - 42 = 0
(x+6)(2x-7) = 0
x = -6 or x = 7/2 , but the width cannot be negative, so
the width is 7/2 or 3.5
the length is 12
check:
3.5(12) = 42
length ---- 2x + 5 (assuming your 5y is a typo)
x(2x+5) = 42
2x^2 + 5x - 42 = 0
(x+6)(2x-7) = 0
x = -6 or x = 7/2 , but the width cannot be negative, so
the width is 7/2 or 3.5
the length is 12
check:
3.5(12) = 42
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