SOLUTION: Solving a word problem using a quadratic equation with rational roots. if the area of a rectangle is 42ft^2 and the length is 5 feet more than twice the width what are the length

Question 886019: Solving a word problem using a quadratic equation with rational roots. if the area of a rectangle is 42ft^2 and the length is 5 feet more than twice the width
what are the length and width
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x = length
y = width
area = x * y
you get:
x * y = 42
since the length is equal to 5 more than 2 times the width, this means that:
x = 2y + 5
substitute for x in the equation of x * y = 42 to get:
(2y + 5) * y = 42
simplify this to get:
2y^2 + 5y = 42
subtract 42 from both sides of this equation to get:
2y^2 + 5y - 42 = 0
factor this equation to get:
(2y-7) * (y+6) = 0
solve for y to get:
y = 7/2 or y = -6
since y can't be negative, then y has to be equal to 7/2.
since x = 2y + 5, then replace y with 7/2 to get:
x = 2 * 7/2 + 5 which makes x = 12
your dimensions are:
x = 12
y = 7/2
multiply x * y and you get 12 * 7/2 = 6 * 7 = 42
everything checks out so those are your answers.

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