To solve this problem, let's define variables and set up an equation based on the information given.

1. Let the width of the poster be $w$ inches.

2. The length of the poster is stated to be "10 more inches than three times its width." So, we can express the length as $3w + 10$.

3. The area of the poster is given as 88 square inches, and the area of a rectangle is calculated by multiplying its length and width. Therefore:

   $w \times (3w + 10) = 88$

Solving the Equation

Expanding this equation, we get:

$3w^2 + 10w = 88$

Rearrange to form a standard quadratic equation:

$3w^2 + 10w - 88 = 0$

We can solve this quadratic equation using the quadratic formula:

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

where $a = 3$, $b = 10$, and $c = -88$.

Plugging in the Values

$w = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 3 \cdot (-88)}}{2 \cdot 3}$ $w = \frac{-10 \pm \sqrt{100 + 1056}}{6}$ $w = \frac{-10 \pm \sqrt{1156}}{6}$ $w = \frac{-10 \pm 34}{6}$

This gives us two potential solutions for $w$:

1. $w = \frac{-10 + 34}{6} = \frac{24}{6} = 4$
2. $w = \frac{-10 - 34}{6} = \frac{-44}{6} = -\frac{22}{3}$

Since width cannot be negative, we take $w = 4$.

Finding the Length

Now, substitute $w = 4$ into the expression for the length:

$\text{Length} = 3w + 10 = 3(4) + 10 = 12 + 10 = 22$

Solution

The dimensions of the poster are:

• Width: 4 inches
• Length: 22 inches

Would you like more details, or do you have any questions?

Related Questions

1. How would the solution change if the area were different, say 100 square inches?
2. How do we handle cases when quadratic equations yield complex roots?
3. Can we verify the solution by checking the area with the found dimensions?
4. What if the length was "10 inches less than three times the width"?
5. How can we solve similar problems with non-linear relationships between length and width?

Tip

When solving quadratic equations, always check if factorization is possible for a quicker solution, especially with simpler integer values.