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 = (-3 ± √(3² - 4(1)(-80)))/2
x = (-3 ± √(9 + 320))/2 = (-3 ± √329)/2
x ≈ 7.57 or x ≈ - 10.57
x + 3 ≈ 10.57 a negative length doesnt make sense.

Explanation:

Step1: Set up area - equation

For a rectangle with length $l$ and width $w$, $A=l\times w$. We set up the quadratic equation based on the given side - expressions and area.

Step2: Rearrange to standard form

Transform the equation into the standard quadratic form $ax^{2}+bx + c = 0$.

Step3: Apply quadratic formula

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

Step4: Select positive root

Since side - lengths cannot be negative, we choose the positive value of $x$ to find the side - lengths of the rectangle.

Answer:

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

• The sides are approximately $7.57$ and $10.57$.

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 $a = 3$, $b=-1$, $c = - 20$.
• $x=\frac{1\pm\sqrt{(-1)^{2}-4\times3\times(-20)}}{2\times3}=\frac{1\pm\sqrt{1 + 240}}{6}=\frac{1\pm\sqrt{241}}{6}$.
• $x\approx\frac{1\pm15.52}{6}$. We take the positive root $x\approx\frac{1 + 15.52}{6}\approx2.75$. Then $3x-1\approx3\times2.75-1=7.25$. The sides are approximately $2.75$ and $7.25$.

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 simplify to $x^{2}+10x - 50=0$.
• Using the quadratic formula with $a = 1$, $b = 10$, $c=-50$.
• $x=\frac{-10\pm\sqrt{10^{2}-4\times1\times(-50)}}{2\times1}=\frac{-10\pm\sqrt{100 + 200}}{2}=\frac{-10\pm\sqrt{300}}{2}=\frac{-10\pm17.32}{2}$. We take the positive root $x\approx3.66$. Then $2x\approx7.32$ and $x + 10\approx13.66$. The sides are approximately $7.32$ and $13.66$.

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 $x^{2}+7x-x-7 = 500$, then $x^{2}+6x-507 = 0$.
• Using the quadratic formula with $a = 1$, $b = 6$, $c=-507$.
• $x=\frac{-6\pm\sqrt{6^{2}-4\times1\times(-507)}}{2\times1}=\frac{-6\pm\sqrt{36+2028}}{2}=\frac{-6\pm\sqrt{2064}}{2}=\frac{-6\pm45.43}{2}$. We take the positive root $x\approx19.72$. Then $x - 1\approx18.72$ and $x + 7\approx26.72$. The sides are approximately $18.72$ and $26.72$.