The length of a rectangular poster is 9 more inches than half its width. The area of the poster is 104 square inches. Solve for the dimensions (length and width) of the poster.

Let's denote the width of the rectangular poster as $w$ inches. According to the problem, the length $l$ of the poster is 9 more inches than half its width. Therefore, we can express the length as:

$l = \frac{w}{2} + 9$

The area of the poster is given as 104 square inches. The area of a rectangle is calculated by multiplying its length by its width:

$l \times w = 104$

Substituting the expression for $l$ into the area equation, we get:

$\left( \frac{w}{2} + 9 \right) \times w = 104$

Now, let's solve for $w$:

1. Distribute $w$ on the left side of the equation:

$\frac{w^2}{2} + 9w = 104$

2. Multiply every term by 2 to clear the fraction:

$w^2 + 18w = 208$

3. Rearrange the equation to set it to zero:

$w^2 + 18w - 208 = 0$

Now, we have a quadratic equation. We can solve it using the quadratic formula $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 18$, and $c = -208$.

4. Calculate the discriminant:

$b^2 - 4ac = 18^2 - 4 \cdot 1 \cdot (-208)$ $= 324 + 832$ $= 1156$

5. Take the square root of the discriminant:

$\sqrt{1156} = 34$

6. Apply the quadratic formula:

$w = \frac{-18 \pm 34}{2}$

This gives us two potential solutions for $w$:

$w = \frac{-18 + 34}{2} = \frac{16}{2} = 8$ $w = \frac{-18 - 34}{2} = \frac{-52}{2} = -26$

Since a width cannot be negative, we discard $w = -26$. Therefore, the width $w$ is 8 inches.

7. Substitute $w = 8$ back into the expression for the length:

$l = \frac{8}{2} + 9 = 4 + 9 = 13$

So, the dimensions of the poster are:

• Width: 8 inches
• Length: 13 inches