Skip to main content
Logo image

Mathematics in Trades and Life

Megan Ossiander-Gobeille, Mark Fitch

Section 5.2 Solving Quadratics

We know what quadratic data looks like and how quadratics are written. Next we will learn to solve expressions with quadratics.

Subsection 5.2.1 Solving Quadratics

For our purposes we will consider two methods of solving quadratics. Both of these are processes to memorize and practice.

Subsubsection 5.2.1.1 Solving Quadratics with Inversion

The first is for very simple quadratics of the form \(ax^2+c=0.\)
Example 5.2.1.
Find all solutions to \(10-21x^2=0\text{.}\)
Solution.
We can solve this by undoing each operation.
\begin{align*} 10-21x^2 \amp = 0.\\ -21x^2 \amp = -10.\\ x^2 \amp = \frac{-10}{-21}.\\ x^2 \amp = \frac{10}{21}.\\ \sqrt{x^2} \amp = \sqrt{10/21}.\\ x \amp = \pm \sqrt{10/21}.\\ x \amp \approx \pm 0.69. \end{align*}
Notice that we end up with two results. This is typical of quadratics.
Example 5.2.2.
Find all solutions to \(7(x-3)^2-4=0\text{.}\)
Solution.
We can solve this by undoing each operation.
\begin{align*} 7(x-3)^2-4 \amp = 0.\\ 7(x-3)^2 \amp = 4.\\ (x-3)^2 \amp = \frac{4}{7}.\\ \sqrt{(x-3)^2} \amp = \sqrt{\frac{4}{7}}.\\ x-3 \amp = \pm \sqrt{\frac{4}{7}}.\\ x \amp = \pm \sqrt{\frac{4}{7}}+3.\\ x \amp \approx 3.76, 2.24. \end{align*}
Checkpoint 5.2.3.
Find the solutions to \(6(x-2)^2-24=0\text{.}\)
Largest solution:
Smallest solution:
Answer 1.
\(4\)
Answer 2.
\(0\)
Solution.
\begin{equation*} \begin{aligned} 6(x-2)^2-24 \amp = 0\\ 6(x-2)^2 \amp = 24\\ (x-2)^2 \amp = 4\\ x-2 \amp = \pm 2\\ x \amp = 2 \pm 2\\ x \amp = 4, 0. \end{aligned} \end{equation*}
Example 5.2.4.
Recall the lift equation in Example 3.3.8. If \(\rho=0.002309 \; \text{slugs}/\text{ft}^3\text{,}\) \(S=174.0 \; \text{ft}^2\text{,}\) and \(C_L=0.5001\text{,}\) what velocity in miles per hour is needed to produce \(L=3500. \; \text{lbs}\text{?}\)
Solution.
We start by filling in the information we know in the equation.
\begin{align*} 3500. \amp = \frac{1}{2}(0.002309)(174.0)(0.5001)v^2.\\ 3500. \amp \approx 0.1005v^2.\\ \frac{3500.}{0.1005} \amp \approx v^2.\\ 34830 \amp \approx v^2.\\ \sqrt{34830} \amp \approx \sqrt{v^2}.\\ 186.6 \amp \approx v. \end{align*}
Note the units for velocity are feet per second. Now we need to convert units (like Example 2.3.12).
\begin{equation*} \frac{186.6 \; \text{ft}}{\text{s}} \cdot \frac{\text{mi}}{5280. \; \text{ft}} \cdot \frac{3600 \; \text{s}}{\text{hr}} \approx 273.7 \frac{\text{mi}}{\text{hr}} \end{equation*}

Subsubsection 5.2.1.2 Solving Formulas with Quadratics

Example 5.2.5.
Solve the lift equation for \(v\text{.}\)
Solution.
\begin{align*} L \amp = \frac{1}{2}\rho S C_L v^2.\\ 2L \amp = 2\frac{1}{2}\rho S C_L v^2.\\ 2L \amp = \rho S C_L v^2.\\ \frac{2L}{\rho} \amp = \frac{\rho S C_L v^2}{\rho}.\\ \frac{2L}{\rho} \amp = S C_L v^2.\\ \frac{2L}{\rho S} \amp = \frac{S C_L v^2}{S}.\\ \frac{2L}{\rho S} \amp = C_L v^2.\\ \frac{2L}{\rho S C_L} \amp = \frac{C_L v^2}{C_L}.\\ \frac{2L}{\rho S C_L} \amp = v^2.\\ \sqrt{\frac{2L}{\rho S C_L}} \amp = \sqrt{v^2}.\\ \sqrt{\frac{2L}{\rho S C_L}} \amp = v. \end{align*}
Example 5.2.6.
The load factor imposed on aircraft is given by
\begin{equation*} n=\left(\frac{v}{v_s}\right)^2 \end{equation*}
where \(v\) is the speed and \(v_s\) is the stall speed. Solve for \(v\) if \(v_s=54\text{.}\)
Solution.
\begin{align*} n \amp = \left(\frac{v}{v_s}\right)^2.\\ n \amp = \left(\frac{v}{54}\right)^2.\\ \sqrt{n} \amp = \sqrt{\left(\frac{v}{54}\right)^2}.\\ \sqrt{n} \amp = \frac{v}{54}.\\ 54\sqrt{n} \amp = 54\frac{v}{54}.\\ 54\sqrt{n} \amp = v. \end{align*}

Subsubsection 5.2.1.3 Solving Quadratics with the Formula

When the quadratic has more than a square term, e.g., \(11x^2+32x-3=0\) we cannot undo each operation. For this class we will solve all of these using the quadratic formula. For \(ax^2+bx+c=0\) the solutions are given by
\begin{equation*} x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. \end{equation*}
Example 5.2.7.
Find all solutions to \(11x^2+32x-3=0\text{.}\)
Solution.
We note that \(a=11\text{,}\) \(b=32\text{,}\) \(c=-3\text{.}\)
\begin{align*} x & = \frac{-32 \pm \sqrt{(32)^2-4(11)(-3)}}{2(11)}\\ & = \frac{-32 \pm \sqrt{1024+132}}{22}\\ & = \frac{-32 \pm \sqrt{1156}}{22}\\ & = \frac{-32 \pm 34}{22}\\ & = \frac{1}{11}, -3 \end{align*}
Example 5.2.8.
Find all solutions to \(8x^2-5x=2x^2+21\text{.}\)
Solution.
We first need to collect all terms on one side and combine them.
\begin{align*} 8x^2-5x & = 2x^2+21.\\ 6x^2-5x-21 & = 0. \end{align*}
Now we can use the formula.
\begin{align*} x & = \frac{-(-5) \pm \sqrt{(-5)^2-4(6)(-21)}}{2(6)}\\ & = \frac{5 \pm \sqrt{25+504}}{12}\\ & = \frac{5 \pm \sqrt{529}}{12}\\ & = \frac{5 \pm 23}{12}\\ & = \frac{7}{3}, -\frac{3}{2} \end{align*}
Checkpoint 5.2.9.
Find the solutions to \(8x^2-30x+25=0\text{.}\)
Largest solution:
Smallest solution:
Answer 1.
\(\frac{5}{2}\)
Answer 2.
\(\frac{5}{4}\)
Solution.
\begin{equation*} \begin{aligned} x \amp = \frac{-(-30) \pm \sqrt{(-30)^2-4(8)(25)}}{2(8)}\\ \amp = \frac{30 \pm \sqrt{900-800}}{16}\\ \amp = \frac{30 \pm \sqrt{100}}{16}\\ \amp = \frac{30 \pm 10}{16}\\ \amp = \frac{5}{2}, \frac{5}{4} \end{aligned} \end{equation*}

Exercises 5.2.2 Exercises

1. Contextless Practice.

2. Contextless Practice.

3. Contextless Practice.

4. Contextless Practice.

5. Parabolic Arc Approximation.

6. Application.

7. Application.

8. Change of Area.

9. Change of Area.

PrevTopNext