Question

Question 43
A square is altered so that one dimension is increased by 5 meters and the other dimension is increased by 3 meters. If the area of the resulting rectangle is 99 square meters, find the area of the original square. Just type the number.
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1 point
Question 44
The length of a rectangle is seven less than twice the length of its width. If the area of the rectangle is 15 square meters, find the value of x. Just type the number.

Answer 1

**Question 43:**

Let the side length of the original square be \( s \). When one side is increased by 5 meters and the other by 3 meters, the dimensions become \( s + 5 \) and \( s + 3 \).

Thus, the area of the rectangle is given by:

\[
(s + 5)(s + 3) = 99
\]

Expanding the left side:

\[
s^2 + 3s + 5s + 15 = 99
\]
\[
s^2 + 8s + 15 = 99
\]

Now, rearranging the equation:

\[
s^2 + 8s + 15 - 99 = 0
\]
\[
s^2 + 8s - 84 = 0
\]

Next, we can solve this quadratic equation using the quadratic formula:

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

where \( a = 1, b = 8, c = -84 \):

\[
s = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-84)}}{2 \cdot 1}
\]
\[
s = \frac{-8 \pm \sqrt{64 + 336}}{2}
\]
\[
s = \frac{-8 \pm \sqrt{400}}{2}
\]
\[
s = \frac{-8 \pm 20}{2}
\]

This gives us the two potential solutions:

\[
s = \frac{12}{2} = 6 \quad \text{and} \quad s = \frac{-28}{2} = -14
\]

Since a side length can't be negative, we take \( s = 6 \) meters. Therefore, the area of the original square is:

\[
s^2 = 6^2 = 36
\]

The area of the original square is **36**.

---

**Question 44:**

Let the width of the rectangle be \( x \). Then, the length is given as \( 2x - 7 \). The area of the rectangle can be expressed as:

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

Expanding gives us:

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

Now, using the quadratic formula:

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

where \( a = 2, b = -7, c = -15 \):

\[
x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot (-15)}}{2 \cdot 2}
\]
\[
x = \frac{7 \pm \sqrt{49 + 120}}{4}
\]
\[
x = \frac{7 \pm \sqrt{169}}{4}
\]
\[
x = \frac{7 \pm 13}{4}
\]

This gives us:

\[
x = \frac{20}{4} = 5 \quad \text{and} \quad x = \frac{-6}{4} = -1.5
\]

Since width cannot be negative, we take \( x = 5 \) meters.

Thus, the value of \( x \) is **5**.