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

Let the width be = x
Then the Length would be four more that is = x+4
Right?

So
We are given that area = length * width = 396

Thus (x+4)*x=396

Then you'd get a quadratic equation.
Then find the value of x. :)

Understood? :)

Answer 2

no i tried inputting the numbers and can't get 396

Answer 3

?

Answer 4

I'll continue from where @Agamani stopped
By the quadratic formula
\[
ax^2 + bx + c = 0 \\
x_{1,2} = \frac{-b \pm \sqrt{ b^2 - 4ac }}{2a} \\
\]
so
\[
(x+4)x = 396 \\
x^2 + 4x = 396 \\
x^2 + 4x - 396 = 0\\
x_{1,2} = \frac{-4 \pm \sqrt{ 4^2 - 4 \cdot 1 \cdot (-396) }}{2 \cdot 1} \\
x_{1,2} = \frac{-4 \pm \sqrt{ 16 + 4 \cdot 396 }}{2} \\
x_{1,2} = \frac{-4 \pm \sqrt{ 16 + 1584 }}{2} \\
x_{1,2} = \frac{-4 \pm \sqrt{ 1600 }}{2} \\
x_{1,2} = \frac{-4 \pm 40}{2} \\
x_{1} = \frac{-4 + 40}{2} = \frac{36}{2} = 18\\
x_{2} = \frac{-4 - 40}{2} = -\frac{44}{2} = -22\]
Since the width of a rectangle cannot be negative then the only result to make sense is x1.
\[
x = 18 \\
width = x = 18_{feet} \\
length = x + 4 = 18 + 4 = 22_{feet}
\]