Question

The length of a rectangle is seven less than twice the length of it’s width if the area of the rectangle is 15 square meters, find the value of x

Answer 1

Let's denote the width of the rectangle as \( x \). According to the problem, the length \( L \) of the rectangle can be expressed as:

\[
L = 2x - 7
\]

The area \( A \) of a rectangle is calculated using the formula:

\[
A = L \times W
\]

Given that the area is 15 square meters, we can substitute the expressions for \( L \) and \( W \):

\[
15 = (2x - 7)x
\]

Now, expanding the right-hand side:

\[
15 = 2x^2 - 7x
\]

Next, rearranging this equation to set it to zero gives us:

\[
2x^2 - 7x - 15 = 0
\]

Now we can use the quadratic formula to solve for \( x \). The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation, \( a = 2 \), \( b = -7 \), and \( c = -15 \). First, we calculate the discriminant:

\[
b^2 - 4ac = (-7)^2 - 4 \times 2 \times (-15) = 49 + 120 = 169
\]

Now substituting back into the quadratic formula:

\[
x = \frac{-(-7) \pm \sqrt{169}}{2 \times 2}
\]
\[
x = \frac{7 \pm 13}{4}
\]

This gives us two possible solutions:

1.
\[
x = \frac{7 + 13}{4} = \frac{20}{4} = 5
\]
2.
\[
x = \frac{7 - 13}{4} = \frac{-6}{4} = -1.5
\]

Since \( x \) represents a width, it cannot be negative. Therefore, the only valid solution is:

\[
x = 5
\]

Thus, the width of the rectangle is \( 5 \) meters.