Certainly! Let's derive the equation together step-by-step.
1. Given Information:
- Area of the rectangle, [tex]\( A \)[/tex] = 91 square inches.
- The length of the rectangle is 1 less than twice its width.
2. Define Variables:
- Let [tex]\( w \)[/tex] be the width of the rectangle.
- The length [tex]\( l \)[/tex] can be expressed as [tex]\( 2w - 1 \)[/tex].
3. Area of the Rectangle:
- The area [tex]\( A \)[/tex] of a rectangle is given by the product of its length and width.
- So, [tex]\( A = w \times l \)[/tex].
4. Substitute Known Values:
- Using the given information, [tex]\( 91 = w \times (2w - 1) \)[/tex].
- This simplifies to [tex]\( 91 = w(2w - 1) \)[/tex].
5. Set Up the Equation:
- Distribute [tex]\( w \)[/tex] through the parentheses: [tex]\( 91 = 2w^2 - w \)[/tex].
- Rearrange to form a standard quadratic equation: [tex]\( 2w^2 - w - 91 = 0 \)[/tex].
Therefore, the equation that can be used to find the width [tex]\( w \)[/tex] of the rectangle is [tex]\( 2w^2 - w - 91 = 0 \)[/tex].
Now, let's match this with the given options:
1. First option:
[tex]\( 2w - 1 \)[/tex]
2. Second option (to be completed):
[tex]\( w^2 + \square w + \square = 0 \)[/tex]
To match [tex]\( 2w^2 - w - 91 = 0 \)[/tex]:
- [tex]\( 2w^2 \rightarrow 2w^2 \)[/tex]
- [tex]\( -w \rightarrow -1w \)[/tex]
- [tex]\( -91 \rightarrow -91 \)[/tex]
So, we can write [tex]\( 2w - 1 \)[/tex] in front and complete the second part as follows:
[tex]\( w^2 + 2 \ (2w^2-1w) - 91 = 0 \)[/tex]
In conclusion, we have the equation:
[tex]\( 2w - 1 \)[/tex]
[tex]\( w^2 + 91 \ \ (-1) w - 91 = 0\)[/tex]
So when rearranged, the match fits:
1. First select [tex]\(2 w-1\)[/tex],
2. Second option \( 2w^2 -1 w -91 = 0\ or
w^2 + 2w (times 2*w) -1w - 91)$
