Question

a landscaper wants to create a 12-foot-long diagonal path through a rectangular garden. the width of the garden is x feet and the length of the garden is 4 more than the width. he uses the pythagorean theorem to write an equation to determine the width of the garden. 1. \\((x)^2 + (x + 4)^2 = (12)^2\\) 2. \\(x^2 + x^2 + 8x + 16 = 144\\) 3. \\(2x^2 + 8x - 128 = 0\\) what are the approximate dimensions of the garden? \\(\circ\\) 6.2 ft by 2.2 ft \\(\circ\\) 6.2 ft by 10.2 ft \\(\circ\\) 10.2 ft by 2.2 ft

Explanation:

Step1: Simplify the quadratic equation

We have the quadratic equation \(2x^{2}+8x - 128=0\). Divide the entire equation by 2 to simplify it:
\(\frac{2x^{2}+8x - 128}{2}=\frac{0}{2}\)
\(x^{2}+4x - 64 = 0\)

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For the equation \(x^{2}+4x - 64=0\), we have \(a = 1\), \(b = 4\) and \(c=- 64\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(4)^{2}-4\times1\times(-64)=16 + 256=272\)
Then, find the square root of the discriminant \(\sqrt{\Delta}=\sqrt{272}\approx16.49\)
Now, substitute into the quadratic formula:
\(x=\frac{-4\pm16.49}{2}\)
We have two solutions:
\(x_{1}=\frac{-4 + 16.49}{2}=\frac{12.49}{2}\approx6.2\)
\(x_{2}=\frac{-4-16.49}{2}=\frac{-20.49}{2}\approx - 10.2\) (We discard the negative solution since width cannot be negative)

Step3: Find the length

The length of the garden is \(x + 4\). Substitute \(x\approx6.2\) into \(x + 4\):
\(x+4\approx6.2 + 4=10.2\)

Answer:

6.2 ft by 10.2 ft