Question

A rectangle's length is 4 feet more than its width. If the area of the rectangle is 396 square feet, what is its width, in feet?

Answer 1

ok?

Answer 2

can you tell me what the nukmbers are

Answer 3

yes

Answer 4

@Aero_Exis I don't know what is the error I think it's fine.
\[
\text{width} = x_{ft} \\
\text{length} = (x+4)_{ft} \\
\text{area} = \text{width} \cdot \text{length} =x_{ft} \cdot (x+4)_{ft} = (x^2 + 4x)_{ft^2} = 396_{ft^2}\\
(x^2 + 4x)_{ft^2} = 396_{ft^2} \\
x^2 +4x -396 = 0 \\
x_{1,2} = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-396) } }{2 \cdot 1} \\
x_{1,2} = \frac{-4 \pm \sqrt{16 +1584 } }{2 \cdot 1} = \frac{-4 \pm \sqrt{1600} }{2} = \frac{-4 \pm 40}{2} \\
x_{1} = \frac{-4 + 40}{2} = \frac{36}{2} = 18\\
x_{2} = \frac{-4 -40}{2} = \frac{-44}{2} = -22
\]
But, since width can't be negative that means
\[
x \ge 0 \implies x = 18 \\
\text{width} = x_{ft} = 18_{ft}
\]
Is there anything wrong that I miss?

Answer 5

also 18 * 22 = 396..