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.
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