Answer :
Certainly! Let's solve the problem step-by-step:
1. Define the variables and the given conditions:
- Let [tex]\( w \)[/tex] be the width of the garden.
- The length of the garden is specified to be four feet less than twice the width, so we can express the length [tex]\( l \)[/tex] as:
[tex]\[ l = 2w - 4 \][/tex]
- The area of the rectangular garden is given as 200 square feet. The area [tex]\( A \)[/tex] of a rectangle is calculated as:
[tex]\[ A = l \times w \][/tex]
Substituting the given area, we get:
[tex]\[ lw = 200 \][/tex]
2. Substitute the expression for [tex]\( l \)[/tex] into the area equation:
[tex]\[ (2w - 4)w = 200 \][/tex]
3. Distribute and set up the equation:
[tex]\[ 2w^2 - 4w = 200 \][/tex]
4. Rearrange the equation into standard quadratic form:
[tex]\[ 2w^2 - 4w - 200 = 0 \][/tex]
5. To simplify, divide the entire equation by 2:
[tex]\[ w^2 - 2w - 100 = 0 \][/tex]
6. Solve the quadratic equation using the quadratic formula:
The quadratic formula is given by:
[tex]\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( w^2 - 2w - 100 = 0 \)[/tex], we have [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -100 \)[/tex].
Plugging these values into the quadratic formula:
[tex]\[ w = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-100)}}{2(1)} \][/tex]
[tex]\[ w = \frac{2 \pm \sqrt{4 + 400}}{2} \][/tex]
[tex]\[ w = \frac{2 \pm \sqrt{404}}{2} \][/tex]
[tex]\[ w = \frac{2 \pm 2\sqrt{101}}{2} \][/tex]
[tex]\[ w = 1 \pm \sqrt{101} \][/tex]
7. Determine the feasible value for [tex]\( w \)[/tex]:
We have two potential solutions:
[tex]\[ w = 1 + \sqrt{101} \][/tex]
[tex]\[ w = 1 - \sqrt{101} \][/tex]
Since the width must be a positive value, we discard [tex]\( w = 1 - \sqrt{101} \)[/tex]. Thus:
[tex]\[ w = 1 + \sqrt{101} \][/tex]
8. Calculate the corresponding length:
Recall that the length [tex]\( l \)[/tex] is:
[tex]\[ l = 2w - 4 \][/tex]
Substituting [tex]\( w = 1 + \sqrt{101} \)[/tex]:
[tex]\[ l = 2(1 + \sqrt{101}) - 4 \][/tex]
[tex]\[ l = 2 + 2\sqrt{101} - 4 \][/tex]
[tex]\[ l = 2\sqrt{101} - 2 \][/tex]
9. Round the final answers to the nearest tenth:
- First, approximate [tex]\( \sqrt{101} \)[/tex]:
[tex]\[ \sqrt{101} \approx 10.05 \][/tex]
- Then use it to approximate [tex]\( w \)[/tex] and [tex]\( l \)[/tex]:
[tex]\[ w \approx 1 + 10.05 = 11.05 \Rightarrow \text{Rounded } w \approx 11.1 \][/tex]
[tex]\[ l \approx 2 \times 10.05 - 2 = 20.1 - 2 = 18.1 \Rightarrow \text{Rounded } l \approx 18.1 \][/tex]
Therefore, the width of the garden is approximately [tex]\( 11.1 \)[/tex] feet, and the length of the garden is approximately [tex]\( 18.1 \)[/tex] feet.
1. Define the variables and the given conditions:
- Let [tex]\( w \)[/tex] be the width of the garden.
- The length of the garden is specified to be four feet less than twice the width, so we can express the length [tex]\( l \)[/tex] as:
[tex]\[ l = 2w - 4 \][/tex]
- The area of the rectangular garden is given as 200 square feet. The area [tex]\( A \)[/tex] of a rectangle is calculated as:
[tex]\[ A = l \times w \][/tex]
Substituting the given area, we get:
[tex]\[ lw = 200 \][/tex]
2. Substitute the expression for [tex]\( l \)[/tex] into the area equation:
[tex]\[ (2w - 4)w = 200 \][/tex]
3. Distribute and set up the equation:
[tex]\[ 2w^2 - 4w = 200 \][/tex]
4. Rearrange the equation into standard quadratic form:
[tex]\[ 2w^2 - 4w - 200 = 0 \][/tex]
5. To simplify, divide the entire equation by 2:
[tex]\[ w^2 - 2w - 100 = 0 \][/tex]
6. Solve the quadratic equation using the quadratic formula:
The quadratic formula is given by:
[tex]\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( w^2 - 2w - 100 = 0 \)[/tex], we have [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -100 \)[/tex].
Plugging these values into the quadratic formula:
[tex]\[ w = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-100)}}{2(1)} \][/tex]
[tex]\[ w = \frac{2 \pm \sqrt{4 + 400}}{2} \][/tex]
[tex]\[ w = \frac{2 \pm \sqrt{404}}{2} \][/tex]
[tex]\[ w = \frac{2 \pm 2\sqrt{101}}{2} \][/tex]
[tex]\[ w = 1 \pm \sqrt{101} \][/tex]
7. Determine the feasible value for [tex]\( w \)[/tex]:
We have two potential solutions:
[tex]\[ w = 1 + \sqrt{101} \][/tex]
[tex]\[ w = 1 - \sqrt{101} \][/tex]
Since the width must be a positive value, we discard [tex]\( w = 1 - \sqrt{101} \)[/tex]. Thus:
[tex]\[ w = 1 + \sqrt{101} \][/tex]
8. Calculate the corresponding length:
Recall that the length [tex]\( l \)[/tex] is:
[tex]\[ l = 2w - 4 \][/tex]
Substituting [tex]\( w = 1 + \sqrt{101} \)[/tex]:
[tex]\[ l = 2(1 + \sqrt{101}) - 4 \][/tex]
[tex]\[ l = 2 + 2\sqrt{101} - 4 \][/tex]
[tex]\[ l = 2\sqrt{101} - 2 \][/tex]
9. Round the final answers to the nearest tenth:
- First, approximate [tex]\( \sqrt{101} \)[/tex]:
[tex]\[ \sqrt{101} \approx 10.05 \][/tex]
- Then use it to approximate [tex]\( w \)[/tex] and [tex]\( l \)[/tex]:
[tex]\[ w \approx 1 + 10.05 = 11.05 \Rightarrow \text{Rounded } w \approx 11.1 \][/tex]
[tex]\[ l \approx 2 \times 10.05 - 2 = 20.1 - 2 = 18.1 \Rightarrow \text{Rounded } l \approx 18.1 \][/tex]
Therefore, the width of the garden is approximately [tex]\( 11.1 \)[/tex] feet, and the length of the garden is approximately [tex]\( 18.1 \)[/tex] feet.