Monte has a small rectangular herb garden that has a length 3 yd longer than twice its width. If the area of the garden is 50 yd^2, what is the length of the garden, to the nearest tenth of a yard?
y = 3 + 2 x
x y = 50
x (3 + 2x) = 50
2 x^2 + 3 x - 50 = 0
x = [-3 +/- sqrt(9+400)]/4
x = 4.306
y = 3 +2(4.306) = 11.6 yd
11.6
Let's assume the width of the garden is "x" yards.
According to the given information, the length of the garden is 3 yards longer than twice its width, which can be written as:
Length = 2x + 3
The area of a rectangle is given by the formula:
Area = Length * Width
Substituting the values in the equation:
50 = (2x + 3) * x
Expanding and rearranging the equation:
50 = 2x^2 + 3x
Now, let's solve this quadratic equation to find the value of x.
First, move all the terms to one side of the equation to make it equal to zero:
2x^2 + 3x - 50 = 0
We can try factoring or using the quadratic formula to solve this equation. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 2, b = 3, c = -50
x = (-(3) ± √((3)^2 - 4(2)(-50))) / (2(2))
Simplifying:
x = (-3 ± √(9 + 400)) / 4
x = (-3 ± √(409)) / 4
x ≈ (-3 ± 20.2) / 4
Now, we have two possible values for x:
x ≈ (-3 + 20.2) / 4 ≈ 4.55
x ≈ (-3 - 20.2) / 4 ≈ -5.8
Since width cannot be negative, we discard the negative value of x.
Therefore, the width of the garden is approximately 4.55 yards.
To find the length, we can substitute this value back into our earlier equation:
Length = 2x + 3
Length ≈ 2(4.55) + 3 ≈ 9.1 + 3 ≈ 12.1
Therefore, the length of the garden, to the nearest tenth of a yard, is approximately 12.1 yards.
To find the length of the garden, we can use the given information about its area and dimensions. Let's break down the problem step by step.
Step 1: Assign variables
Let's assign variables to the unknowns in the problem. We can let "x" represent the width of the garden and "y" represent the length of the garden.
Step 2: Translate the given information into equations
From the problem statement, we can determine two pieces of information:
- The length of the garden is 3 yards longer than twice its width: y = 2x + 3.
- The area of the garden is 50 square yards: xy = 50.
Step 3: Solve the equations
Now we have a system of two equations with two variables. We can solve this system simultaneously to find the values of x and y.
From equation 1, substitute the value of y in equation 2:
x(2x + 3) = 50.
Expand the equation:
2x^2 + 3x = 50.
Rearrange the equation to form a quadratic equation:
2x^2 + 3x - 50 = 0.
Step 4: Solve the quadratic equation
To solve the quadratic equation, we can factor it or use the quadratic formula. In this case, let's use the quadratic formula to find the values of x:
x = (-b ± √(b^2 - 4ac)) / (2a),
Where a = 2, b = 3, and c = -50.
Substituting the values into the formula,
x = (-(3) ± √((3)^2 - 4(2)(-50))) / (2(2)),
Simplifying,
x = (-3 ± √(9 + 400)) / 4,
x = (-3 ± √(409)) / 4.
Calculating the approximate values using a calculator,
x ≈ (-3 + √409) / 4 ≈ 3.258,
x ≈ (-3 - √409) / 4 ≈ -6.258.
Since dimensions cannot be negative, we discard the negative value. Thus, the width (x) is approximately 3.258 yards.
Step 5: Calculate the length
Using the equation y = 2x + 3 and substituting the value of x,
y ≈ 2(3.258) + 3 ≈ 9.516.
Therefore, the length of the garden, to the nearest tenth of a yard, is approximately 9.5 yards.