Certainly! Let’s solve the problem step-by-step.
1. Identify the given information and relationship:
- The length of the rectangle is 5 inches more than the width.
- Denote the width of the rectangle by [tex]\( x \)[/tex].
- Therefore, the length of the rectangle can be written as [tex]\( x + 5 \)[/tex] inches.
- The area of the rectangle is given by the equation [tex]\( x^2 + 5x = 300 \)[/tex].
2. Form the equation for the area:
- The area of a rectangle is calculated as [tex]\( \text{width} \times \text{length} \)[/tex].
- According to the given information, the area is [tex]\( 300 \)[/tex] square inches.
- Substitute the relationships into the area formula:
[tex]\[
x \times (x + 5) = 300
\][/tex]
3. Expand and rearrange the equation:
- Multiply through to get a single quadratic equation:
[tex]\[
x^2 + 5x = 300
\][/tex]
- This is already in standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex].
4. Solve the quadratic equation:
- Rewrite the equation as:
[tex]\[
x^2 + 5x - 300 = 0
\][/tex]
- Now, we need to solve this quadratic equation for [tex]\( x \)[/tex].
5. Determine the solutions:
- By solving the quadratic equation, we find two solutions. We determine that the positive solution is:
[tex]\[
x = 15
\][/tex]
6. Verify the dimensions:
- Since the width [tex]\( x \)[/tex] is 15 inches:
- The length, which is 5 inches more than the width, is:
[tex]\[
x + 5 = 15 + 5 = 20 \text{ inches}
\][/tex]
Thus, the measures of the rectangle are:
Width [tex]\( = 15 \)[/tex] inches
Length [tex]\( = 20 \)[/tex] inches
