Solving Quadratics

Section 5.2 Solving Quadratics

This section addresses the following topics.

Read and use mathematical models in a technical document

This section covers the following mathematical concepts.

Solve linear, rational, quadratic, and exponential equations and formulas (skill)

This section presents how to solve equations that contain quadratics. For this text we present two methods of solving quadratics. Both should be memorized and practiced.

Subsection 5.2.1 Solving Quadratics with Inversion

The first method is for very simple quadratics of the form \(ax^2+c=0.\) Notice that there is no linear term in \(x\text{.}\) This is what makes solving these simpler. The majority of quadratic models in this text can be solved with this technique.

Example 5.2.1.

Identify all solutions to \(10-21x^2=0\text{.}\) Round solutions to 2 decimal places.

We can solve this by undoing each operation.

\begin{align*} 10-21x^2 \amp = 0. & & \text{Subtract to isolate} x\\ -21x^2 \amp = -10. & & \text{Divide to isolate } x\\ x^2 \amp = \frac{-10}{-21}.\\ x^2 \amp = \frac{10}{21}. & & \text{Square root to undo square.}\\ \sqrt{x^2} \amp = \sqrt{10/21}.\\ x \amp = \pm \sqrt{10/21}.\\ x \amp \approx \pm 0.6900655593.\\ x \amp \approx \pm 0.69. \end{align*}

Notice that we end up with two results. The \(\pm\) results from squaring eliminating a negative. That is, \((2)^2=4\) and \((-2)^2=4\text{.}\) So \(\sqrt{4}\) could be either 2 or -2.

Example 5.2.2.

Identify all solutions to \(7(x-3)^2-4=0\text{.}\) Round solutions to 2 decimal places.

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 3.755928946, 2.244071054 .\\ x \amp \approx 3.76, 2.24. \end{align*}

Checkpoint 5.2.3.

Significant Digits beyond Arithmetic.

There are significant digits rules for addition/subtraction and multiplication/division. We can handle signficant digits with exponents, such as \((0.0230)^2\text{,}\) because this can be interpreted as multiplication. In the case of \((0.023\underline{0})^2 = (0.023\underline{0})(0.023\underline{0}) = 0.00052\underline{9}\text{.}\)

For square roots we rely on mathematics we do not need to present here that indicates we can calculate the square root so that the number of significant digits is maintained. For example \(\sqrt{12\underline{3}0} \approx = 35.07135583 \approx 35.1\)

Example 5.2.4.

Recall the lift equation in Model 1.3.3. 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 nautical miles per hour is needed to produce \(L=3500. \text{ lbs}\text{?}\)

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 0.100\underline{4}615883v^2.\\ \frac{3500.}{0.100\underline{4}615883} \amp \approx v^2.\\ 348\underline{3}9.18639 \amp \approx v^2.\\ \sqrt{348\underline{3}9.18639} \amp \approx \sqrt{v^2}.\\ 186.\underline{6}525821 \amp \approx v.\\ 186.7 \amp \approx v. \end{align*}

We do not consider the negative square root, because a negative velocity (flying backwards) does not work. Remember to consider reality constraints.

Note the units for velocity are feet per second. Now we need to convert units (like Example 1.1.20).

\begin{equation*} \frac{186.\underline{6}525821 \; \text{ft}}{\text{s}} \cdot \frac{\text{mi}}{6076 \; \text{ft}} \cdot \frac{3600 \; \text{s}}{\text{hr}} \approx 110.\underline{5}907333 \frac{\text{mi}}{\text{hr}} \approx 110.6 \frac{\text{mi}}{\text{hr}}\text{.} \end{equation*}

Example 5.2.5.

Consider Model 5.1.3 What velocity will produce a maximum load factor of \(n_{max}=2\) if the stall speed is \(v_s=54\text{?}\)

Note that 2 is an exact number, and 54 is a measurement.

Solution.

\begin{align*} 2 \amp = \left(\frac{v}{v_s}\right)^2.\\ 2 \amp = \left(\frac{v}{54}\right)^2.\\ \sqrt{2} \amp = \sqrt{\left(\frac{v}{54}\right)^2}.\\ \sqrt{2} \amp = \frac{v}{54}.\\ 54\sqrt{2} \amp = 54\frac{v}{54}.\\ 54\sqrt{2} \amp = v.\\ 7\underline{6}.36753235 \amp \approx v.\\ 76 \amp \approx v. \end{align*}

The aircraft would have to be travelling at 76 nm/hour to be able to experience 2 G’s in a maximum performance maneuver.

Again reality constraints eliminate the possibility of the negative square root. Negative velocity (backwards motion) does not produce flight.

Subsection 5.2.2 Solving Quadratics with the Formula

When the quadratic has more than a square term, e.g., \(11x^2+32x-3=0\) we cannot solve by undoing each operation. In this text we will solve all of these quadratics 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 5.2.6.

Identify all solutions to \(11x^2+32x-3=0\text{.}\)

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*}

Checkpoint 5.2.7.

Exercises 5.2.3 Exercises

Solving Quadratic Equations.

Use these contextless problems to practice using the quadratic formula.

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2.

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4.

Quadratic Applications.

Answer these questions which involve a quadratic model. Select rounding that is appropriate or as directed.

Mathematics in Trades and Life