Question

The width of a rectangle is the length minus 6 units. The area of the rectangle is 7 square units. What is the length, in units, of the rectangle?

Answer 1

To find the length of the rectangle, we can start by letting the length of the rectangle be represented by \( L \). According to the problem, the width \( W \) of the rectangle can be expressed as:

\[
W = L - 6
\]

The area \( A \) of a rectangle is given by the formula:

\[
A = L \times W
\]

We know the area is 7 square units, so we can substitute the expression for width into the area formula:

\[
L \times (L - 6) = 7
\]

Expanding this equation, we get:

\[
L^2 - 6L = 7
\]

Next, we rearrange the equation to set it to zero:

\[
L^2 - 6L - 7 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula, which is given by:

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

In this case, \( a = 1 \), \( b = -6 \), and \( c = -7 \). Plugging these values into the formula gives:

\[
L = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}
\]

Calculating the discriminant:

\[
(-6)^2 = 36
\]
\[
-4 \cdot 1 \cdot (-7) = 28
\]
\[
b^2 - 4ac = 36 + 28 = 64
\]

Now substituting back into the formula:

\[
L = \frac{6 \pm \sqrt{64}}{2}
\]
\[
\sqrt{64} = 8
\]
\[
L = \frac{6 \pm 8}{2}
\]

This will give us two possible solutions:

1. \( L = \frac{6 + 8}{2} = \frac{14}{2} = 7 \)
2. \( L = \frac{6 - 8}{2} = \frac{-2}{2} = -1 \)

Since the length cannot be negative, we take the positive value:

\[
L = 7
\]

Thus, the length of the rectangle is:

\[
\boxed{7} \text{ units}
\]

Finally, let's verify this solution by calculating the width:

\[
W = L - 6 = 7 - 6 = 1
\]

Now checking the area:

\[
A = L \times W = 7 \times 1 = 7 \text{ square units}
\]

This confirms that the solution is correct. The length of the rectangle is \( 7 \) units.