To solve this problem, let's first set up an equation based on the given information.
We know:
- The length [tex]\( L \)[/tex] of the rectangle is [tex]\( 5 \)[/tex] yards more than twice the width [tex]\( x \)[/tex].
- The area [tex]\( A \)[/tex] of the rectangle is [tex]\( 348 \)[/tex] square yards.
This gives us the following relationships:
1. [tex]\( L = 2x + 5 \)[/tex]
2. [tex]\( A = L \cdot x \)[/tex]
3. Plugging in the given area, we get [tex]\( 348 = (2x + 5) \cdot x \)[/tex]
Let's solve for [tex]\( x \)[/tex]:
1. Start with the equation [tex]\( 348 = (2x + 5) \cdot x \)[/tex].
2. Distribute [tex]\( x \)[/tex] on the right-hand side: [tex]\( 348 = 2x^2 + 5x \)[/tex].
3. Rewrite the equation in standard quadratic form: [tex]\( 2x^2 + 5x - 348 = 0 \)[/tex].
Next, we solve the quadratic equation [tex]\( 2x^2 + 5x - 348 = 0 \)[/tex].
The solutions to this quadratic equation are:
[tex]\[ x_1 = -\frac{29}{2} \][/tex]
[tex]\[ x_2 = 12 \][/tex]
Next, we need to find the corresponding lengths:
For [tex]\( x = -\frac{29}{2} \)[/tex]:
[tex]\[ L = 2x + 5 = 2\left(-\frac{29}{2}\right) + 5 = -29 + 5 = -24 \][/tex]
For [tex]\( x = 12 \)[/tex]:
[tex]\[ L = 2x + 5 = 2 \cdot 12 + 5 = 24 + 5 = 29 \][/tex]
Thus, the width and corresponding length for the rectangle based on the solutions are:
- Width [tex]\( x = -\frac{29}{2} \)[/tex] and length [tex]\( L = -24 \)[/tex]
- Width [tex]\( x = 12 \)[/tex] and length [tex]\( L = 29 \)[/tex]
Only the positive dimensions make physical sense in the context of this problem. Therefore, the valid solution is:
- Width [tex]\( 12 \)[/tex] yards
- Length [tex]\( 29 \)[/tex] yards
