Question

using the quadratic formula to solve problems
find the lengths of the sides of each rectangle to the nearest hundredth.
check :
x(x + 3)=80 a≈7.57(10.57)
x² + 3x = 80 a≈80.01
x² + 3x + - 80 = 0
x = \frac{-3\pm\sqrt{3² - 4(1)(-80)}}{2}
x = \frac{-3\pm\sqrt{9 + 320}}{2}=\frac{-3\pm\sqrt{329}}{2}
x≈7.57 or x≈ - 10.57
x + 3≈10.57 a negative length doesnt make sense.

Explanation:

Step1: Set up area - equations

For a rectangle with length \(l\) and width \(w\), \(A=l\times w\). So we set up equations based on the given side - expressions and areas.

Step2: Rearrange to quadratic form

Transform the area - equations into the standard quadratic form \(ax^{2}+bx + c = 0\).

Step3: Apply quadratic formula

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) to solve for \(x\).

Step4: Reject negative solutions

Since the side - lengths of a rectangle cannot be negative, we reject the negative values of \(x\) obtained from the quadratic formula.

Answer:

1. For the rectangle with sides \(x\) and \(x + 3\) and area \(A = 80\):

• The valid value of \(x\approx7.57\), and the other - side \(x + 3\approx10.57\)

1. For the rectangle with sides \(x\) and \(3x - 1\) and area \(A = 20\):

• First, set up the equation \(x(3x - 1)=20\).
• Expand it to get \(3x^{2}-x - 20 = 0\).
• Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 3\), \(b=-1\), and \(c=-20\).
• \(x=\frac{1\pm\sqrt{(-1)^{2}-4\times3\times(-20)}}{2\times3}=\frac{1\pm\sqrt{1 + 240}}{6}=\frac{1\pm\sqrt{241}}{6}\).
• \(x=\frac{1\pm15.52}{6}\). We get \(x=\frac{1 + 15.52}{6}\approx2.75\) or \(x=\frac{1-15.52}{6}\approx - 2.42\). Since length cannot be negative, \(x\approx2.75\) and \(3x - 1\approx3\times2.75-1=7.25\)

1. For the rectangle with sides \(2x\) and \(x + 10\) and area \(A = 100\):

• Set up the equation \(2x(x + 10)=100\).
• Expand to \(2x^{2}+20x-100 = 0\), then divide by 2 to get \(x^{2}+10x - 50=0\).
• Using the quadratic formula \(x=\frac{-10\pm\sqrt{10^{2}-4\times1\times(-50)}}{2\times1}=\frac{-10\pm\sqrt{100 + 200}}{2}=\frac{-10\pm\sqrt{300}}{2}=\frac{-10\pm17.32}{2}\).
• We get \(x=\frac{-10 + 17.32}{2}=3.66\) or \(x=\frac{-10-17.32}{2}=-13.66\). Since length cannot be negative, \(x\approx3.66\), \(2x\approx7.32\) and \(x + 10\approx13.66\)

1. For the rectangle with sides \(x - 1\) and \(x + 7\) and area \(A = 500\):

• Set up the equation \((x - 1)(x + 7)=500\).
• Expand to \(x^{2}+7x-x - 7 = 500\), or \(x^{2}+6x-507 = 0\).
• Using the quadratic formula \(x=\frac{-6\pm\sqrt{6^{2}-4\times1\times(-507)}}{2\times1}=\frac{-6\pm\sqrt{36+2028}}{2}=\frac{-6\pm\sqrt{2064}}{2}=\frac{-6\pm45.43}{2}\).
• We get \(x=\frac{-6 + 45.43}{2}=19.72\) or \(x=\frac{-6-45.43}{2}=-25.72\). Since length cannot be negative, \(x\approx19.72\), \(x - 1\approx18.72\) and \(x + 7\approx26.72\)