Question: The width of a rectangle is a meter greater than the length. If the area of the rectangle is 72 square meters, find the dimensions of the rectangle.
1. **State the problem:** We need to find the length and width of a rectangle where the width is 1 meter greater than the length and the area is 72 square meters.
2. **Define variables:** Let the length be $x$ meters.
3. **Express width:** The width is $x + 1$ meters.
4. **Write the area formula:** Area of a rectangle is given by $$\text{Area} = \text{length} \times \text{width}$$
5. **Set up the equation:** Using the given area,
$$x(x + 1) = 72$$
6. **Expand the equation:**
$$x^2 + x = 72$$
7. **Rewrite as a quadratic equation:**
$$x^2 + x - 72 = 0$$
8. **Solve the quadratic equation using the quadratic formula:**
The quadratic formula is
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a=1$, $b=1$, and $c=-72$.
9. **Calculate the discriminant:**
$$\Delta = b^2 - 4ac = 1^2 - 4(1)(-72) = 1 + 288 = 289$$
10. **Find the square root of the discriminant:**
$$\sqrt{289} = 17$$
11. **Calculate the two possible solutions:**
$$x = \frac{-1 \pm 17}{2}$$
12. **Evaluate each solution:**
- $$x = \frac{-1 + 17}{2} = \frac{16}{2} = 8$$
- $$x = \frac{-1 - 17}{2} = \frac{-18}{2} = -9$$
13. **Interpret the solutions:** Length cannot be negative, so $x = 8$ meters.
14. **Find the width:**
$$x + 1 = 8 + 1 = 9$$ meters.
15. **Final answer:** The length is 8 meters and the width is 9 meters.