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 $w$, $A=l\times w$. We set up quadratic equations based on the given side - length expressions and areas.

Step2: Rearrange to standard quadratic form

Transform equations like $x(x + 3)=80$ to $x^{2}+3x - 80=0$, $ax^{2}+bx + c = 0$ form.

Step3: Apply quadratic formula

Use $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ to solve for $x$.

Step4: Reject negative solutions

Since side - lengths of a rectangle cannot be negative, we reject the negative values of $x$.

Step5: Find side - lengths

Substitute the valid $x$ values back into the expressions for the side - lengths of the rectangles.

Answer:

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

• The lengths of the sides are approximately $x\approx7.57$ and $x + 3\approx10.57$.

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

• First, set up the equation $x(3x - 1)=20$.
• Expand to get $3x^{2}-x - 20=0$.
• Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ where $a = 3$, $b=-1$, $c=-20$.
• $x=\frac{1\pm\sqrt{(-1)^{2}-4\times3\times(-20)}}{2\times3}=\frac{1\pm\sqrt{1 + 240}}{6}=\frac{1\pm\sqrt{241}}{6}$.
• $x=\frac{1\pm15.52}{6}$. 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: $2x^{2}+20x-100 = 0$, divide by 2 to get $x^{2}+10x - 50=0$.
• Using the quadratic formula $x=\frac{-10\pm\sqrt{10^{2}-4\times1\times(-50)}}{2\times1}=\frac{-10\pm\sqrt{100 + 200}}{2}=\frac{-10\pm\sqrt{300}}{2}=\frac{-10\pm17.32}{2}$.
• $x=\frac{-10 + 17.32}{2}\approx3.66$ or $x=\frac{-10-17.32}{2}\approx - 13.66$. 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: $x^{2}+7x-x - 7=500$, so $x^{2}+6x-507 = 0$.
• Using the quadratic formula $x=\frac{-6\pm\sqrt{6^{2}-4\times1\times(-507)}}{2\times1}=\frac{-6\pm\sqrt{36+2028}}{2}=\frac{-6\pm\sqrt{2064}}{2}=\frac{-6\pm45.43}{2}$.
• $x=\frac{-6 + 45.43}{2}\approx19.72$ or $x=\frac{-6-45.43}{2}\approx - 25.72$. Since length cannot be negative, $x\approx19.72$, $x - 1\approx18.72$ and $x + 7\approx26.72$.