To solve this problem, we need to find the length and width of the rectangle given that its length is one unit more than its width, and the area is 30 square units. Let's break it down step-by-step:

Step 1: Define the variables

1. Width: Let's denote the width of the rectangle by $w$.
2. Length: Since the length is one unit more than the width, we denote the length by $w + 1$.

Step 2: Write the equation for the area

The area of a rectangle is calculated by multiplying its length by its width. According to the problem, the area is 30 square units:

$\text{Area} = \text{Length} \times \text{Width}$

Substitute the expressions for length and width:

$30 = (w + 1) \cdot w$

Step 3: Formulate and simplify the quadratic equation

Now, expand the equation:

$30 = w^2 + w$

Rearrange the equation to standard quadratic form:

$w^2 + w - 30 = 0$

Step 4: Solve the quadratic equation

To solve the quadratic equation, we can use the quadratic formula:

$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where:

• $a = 1$ (coefficient of $w^2$)
• $b = 1$ (coefficient of $w$)
• $c = -30$ (constant term)

Substitute these values into the quadratic formula:

$w = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1}$

Simplify the expression under the square root:

$w = \frac{-1 \pm \sqrt{1 + 120}}{2}$

$w = \frac{-1 \pm \sqrt{121}}{2}$

Calculate the square root:

$w = \frac{-1 \pm 11}{2}$

Step 5: Find the possible values for $w$

Now, solve for the two potential values of $w$:

1. $w = \frac{-1 + 11}{2} = \frac{10}{2} = 5$
2. $w = \frac{-1 - 11}{2} = \frac{-12}{2} = -6$

Since a width cannot be negative, we discard $w = -6$ and keep $w = 5$.

Step 6: Determine the length

Now that we have $w = 5$, we can find the length:

$\text{Length} = w + 1 = 5 + 1 = 6$

Step 7: Verify the solution

To ensure our solution is correct, let's check the area:

$\text{Area} = \text{Length} \times \text{Width} = 6 \times 5 = 30$

The calculated area matches the given area, so our solution is correct.

Final Answer

• Width: 5 units
• Length: 6 units

Would you like any additional details or have any questions?

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5 Related Questions to Explore

1. How would the problem change if the length was 2 units more than the width?
2. What if the area of the rectangle was 40 square units instead?
3. Can you solve the problem using a different method, such as factoring?
4. How do you graphically represent the solution to a quadratic equation?
5. What happens if the quadratic equation has no real solutions?

Tip

When solving quadratic equations, always check if factoring is easier before using the quadratic formula.