Question

The length of a rectangular poster is 7 more inches than three times its width. The area of the poster is 26 square inches. Solve for the dimensions (length and width) of the poster.
Answer
Attempt 1 out of 2

The dimensions are
inches by
inches.
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Answer 1

Let the width of the poster be \( w \) inches. According to the problem, the length \( l \) is given by the equation:

\[
l = 3w + 7
\]

The area \( A \) of the rectangle can be expressed as:

\[
A = l \times w
\]

We know the area is 26 square inches, so we can set up the equation:

\[
(3w + 7) \times w = 26
\]

Expanding this, we get:

\[
3w^2 + 7w = 26
\]

Rearranging it into a standard quadratic equation form gives:

\[
3w^2 + 7w - 26 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula where \( a = 3 \), \( b = 7 \), and \( c = -26 \):

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

Calculating the discriminant:

\[
b^2 - 4ac = 7^2 - 4 \times 3 \times (-26)
\]
\[
= 49 + 312
\]
\[
= 361
\]

Now substituting back into the quadratic formula:

\[
w = \frac{-7 \pm \sqrt{361}}{2 \times 3}
\]
\[
= \frac{-7 \pm 19}{6}
\]

Calculating the two possible values for \( w \):

1. \( w = \frac{12}{6} = 2 \)
2. \( w = \frac{-26}{6} \) (this value is negative and not applicable)

Thus, the width \( w \) is:

\[
w = 2 \text{ inches}
\]

Now we can find the length \( l \):

\[
l = 3w + 7 = 3(2) + 7 = 6 + 7 = 13 \text{ inches}
\]

So, the dimensions of the poster are:

\[
\text{Width} = 2 \text{ inches}, \quad \text{Length} = 13 \text{ inches}
\]

Final answer:

The dimensions are \( 13 \) inches by \( 2 \) inches.