Solve for x.

The length of a rectangle is equal to 2 times the width plus 3 feet. If the area of the rectangle is 20 square feet, what are the dimension of the rectangle?

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Mathematics

30 September, 11:33

Solve for x.

The length of a rectangle is equal to 2 times the width plus 3 feet. If the area of the rectangle is 20 square feet, what are the dimension of the rectangle?

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Answers (2)

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Amigo · Answer 1

L*w=20, so l=20/w

l=2w+3

20/w=2w+3

20=2w^2+3w

2w^2+3w-20=0

Slip and slide

w^2+3w-40

(w+8) (w-5)

(w+8/2) (w-5/2)

(w+4) (2w-5)

w = - 4 or 5/2, but since width must be positive, it is 5/2

(5/2) * l=20

l=8

Final answer: Length = 8, Width = 5/2

Vegas · Answer 2

Let width be x feet

then the length = 2x + 3 feet

then the area = x (2x + 3) = 20

2x^2 + 3x - 20 = 0

(2x - 5) (x + 4) = 0

x = 2.5 or - 4 (ignore the negative value)

so the deimnasions are width = 2.5 feet and length = 2 (2.5) + 3 = 8 feet

Solve for x.The length of a rectangle is equal to 2 times the width plus 3 feet. If the area of the rectangle is 20 square feet, what are the dimension of the rectangle?

Solve for x.

The length of a rectangle is equal to 2 times the width plus 3 feet. If the area of the rectangle is 20 square feet, what are the dimension of the rectangle?