A Generic Quadratic Equation $ax^2+bx+c=0.$
Pop Quiz: Name $3$ methods for solving quadratics.
Three Methods
- Factoring
- Completing the Square
- Quadratic Formula
An Incomplete List of Kinds of Equations which Lead to Quadratics
- Rational Equations
- Radical Equations
- "Stuffed-Quadratic" Equations
Example: A Rational Equation
Solve the rational equation. Be sure to check for extraneous solutions. If necessary, leave any radicals unsimplified. $$\displaystyle \frac{v^2}{v^2- v - 20}=\displaystyle \frac{2 v}{v^2-2 v - 15}$$
$$
\begin{array}{lll}
&\displaystyle \frac{v^2}{v^2- v - 20}=\displaystyle \frac{2 v}{v^2-2 v - 15}&\mbox{}\\
&\displaystyle \frac{v^2}{(v-5)(v+4)}=\displaystyle \frac{2 v}{(v-5)(v+3)}&\mbox{factor denominators}\\
\implies &\displaystyle \color{magenta}{(v-5)(v+4)(v-3)}\frac{v^2}{(v-5)(v+4)}=\displaystyle \color{magenta}{(v-5)(v+4)(v-3)} \frac{2 v}{(v-5)(v+3)}&\mbox{}\\
\implies &\displaystyle v^2(v+3)=2v(v+4)&\mbox{cancel denominators}\\
\implies &\displaystyle v^3+3v^2=2v^2+8v&\mbox{distribute}\\
\implies &\displaystyle v^3+3v^2 \color{magenta}{-2v^2-8v}=2v^2+8v\color{magenta}{-2v^2-8v}&\mbox{get terms to one side}\\
\implies &\displaystyle v^3+v^2-8v=0&\mbox{combine like terms}\\
\implies &\displaystyle v(v^2+v-8)=0&\mbox{factor}\\
\implies &\displaystyle v=0 \,\,\,\,\,\mbox{or}\,\,\,\, v^2+v-8=0 &\mbox{by the Zero Factor Property}\\
\implies &\displaystyle v=0 \,\,\,\,\,\mbox{or}\,\,\,\, v=\frac{-1 \pm \sqrt{33}}{2} &\mbox{using Quadratic Formula}\\
\end{array}
$$
Check Each Solution
It is clear that $v=0$ solves the equation.
Now, $\displaystyle v=\displaystyle \frac{-1+ \sqrt{33}}{2}\approx 2.3723.$ Plugging this value into both sides, of the original equation, $$ \displaystyle \frac{2.3723^2}{2.3723^2- 2.3723 - 20}\approx -0.3360990021 $$ and $$ \displaystyle \frac{2 \cdot 2.3723}{2.3723^2-2 \cdot 2.3723 - 15}\approx -0.3360961727 $$ So, this solution checks.
Now, $\displaystyle v=\displaystyle \frac{-1- \sqrt{33}}{2}\approx -3.3723.$ Plugging this value into both sides, of the original equation, $$ \displaystyle \frac{(-3.3723)^2}{(-3.3723)^2- (-3.3723) - 20}\approx -2.163991221 $$ and $$ \displaystyle \frac{2 \cdot (-3.3723)}{(-3.3723)^2-2 \cdot (-3.3723) - 15}\approx -2.163806296 $$ So, this solution also checks.
All solutions are bonafide solutions. (No extraneous solutions.)
In the homework, the correct option will look like $$v=0\,\,\,\,\,\mbox{or}\,\,\,\,\displaystyle v=\displaystyle \frac{-1+ \sqrt{33}}{2}\,\,\,\,\,\mbox{or}\,\,\,\,\displaystyle v=\displaystyle \frac{-1- \sqrt{33}}{2}$$
It is clear that $v=0$ solves the equation.
Now, $\displaystyle v=\displaystyle \frac{-1+ \sqrt{33}}{2}\approx 2.3723.$ Plugging this value into both sides, of the original equation, $$ \displaystyle \frac{2.3723^2}{2.3723^2- 2.3723 - 20}\approx -0.3360990021 $$ and $$ \displaystyle \frac{2 \cdot 2.3723}{2.3723^2-2 \cdot 2.3723 - 15}\approx -0.3360961727 $$ So, this solution checks.
Now, $\displaystyle v=\displaystyle \frac{-1- \sqrt{33}}{2}\approx -3.3723.$ Plugging this value into both sides, of the original equation, $$ \displaystyle \frac{(-3.3723)^2}{(-3.3723)^2- (-3.3723) - 20}\approx -2.163991221 $$ and $$ \displaystyle \frac{2 \cdot (-3.3723)}{(-3.3723)^2-2 \cdot (-3.3723) - 15}\approx -2.163806296 $$ So, this solution also checks.
All solutions are bonafide solutions. (No extraneous solutions.)
In the homework, the correct option will look like $$v=0\,\,\,\,\,\mbox{or}\,\,\,\,\displaystyle v=\displaystyle \frac{-1+ \sqrt{33}}{2}\,\,\,\,\,\mbox{or}\,\,\,\,\displaystyle v=\displaystyle \frac{-1- \sqrt{33}}{2}$$
Example: A Radical Equation
Solve the radical equation. Leave any radicals unsimplified. Consider only real-number solutions. $$\sqrt{m^2-5 m + 7}=2 m$$
$$
\begin{array}{lll}
&\displaystyle \sqrt{m^2-5 m + 7}=2 m&\mbox{}\\
\implies &\displaystyle \left(\sqrt{m^2-5 m + 7}\right)^2=(2 m)^2&\mbox{square both sides}\\
\implies &\displaystyle m^2-5 m + 7=4m^2&\mbox{simplify}\\
\implies &\displaystyle m^2-5 m + 7\color{magenta}{-m^2+5m-7}=4m^2\color{magenta}{-m^2+5m-7}&\mbox{get everything to one side}\\
\implies &\displaystyle 0=3m^2+5m-7&\mbox{simplify}\\
\implies &\displaystyle m=\frac{-5 \pm \sqrt{109}}{6} &\mbox{using Quadratic Formula}\\
\end{array}
$$
Check Both Solutions
Plugging $\displaystyle m=\displaystyle \frac{-5 +\sqrt{109}}{6}\approx 0.9067$ into both sides, of the original equation, $$ \displaystyle \sqrt{0.9067^2-5\cdot 0.9067 + 7}\approx 1.813451099 $$ and $$ \displaystyle 2\cdot (0.9067)\approx 1.8134 $$ So, this solution checks.
On the other hand, plugging $\displaystyle v=\displaystyle \frac{-5 - \sqrt{109}}{6}\approx -2.5734$ into both sides, of the original equation, $$ \displaystyle \sqrt{(-2.5734)^2-5\cdot (-2.5734) + 7}\approx 5.146784196 $$ and $$ \displaystyle 2\cdot (-2.5734)\approx -5.1468 $$ Since both sides are not equal (opposite signs), this solution is an extraneous (fake) solution.
The correct option in the homework will look like $v=\displaystyle \frac{-5 +\sqrt{109}}{6}.$
Plugging $\displaystyle m=\displaystyle \frac{-5 +\sqrt{109}}{6}\approx 0.9067$ into both sides, of the original equation, $$ \displaystyle \sqrt{0.9067^2-5\cdot 0.9067 + 7}\approx 1.813451099 $$ and $$ \displaystyle 2\cdot (0.9067)\approx 1.8134 $$ So, this solution checks.
On the other hand, plugging $\displaystyle v=\displaystyle \frac{-5 - \sqrt{109}}{6}\approx -2.5734$ into both sides, of the original equation, $$ \displaystyle \sqrt{(-2.5734)^2-5\cdot (-2.5734) + 7}\approx 5.146784196 $$ and $$ \displaystyle 2\cdot (-2.5734)\approx -5.1468 $$ Since both sides are not equal (opposite signs), this solution is an extraneous (fake) solution.
The correct option in the homework will look like $v=\displaystyle \frac{-5 +\sqrt{109}}{6}.$
Example: A "Stuffed Quadratic" $$z^4-17z^2+16=0$$
$$
\begin{array}{lll}
&\displaystyle z^4-17z^2+16=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{z^2})^2-17\color{magenta}{z^2}+16=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}^2-17\color{magenta}{x}+16=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{x}-1)(\color{magenta}{x}-16)=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}-1=0\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}-16=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}=1\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}=16&\mbox{}\\
\implies &\displaystyle \color{magenta}{z^2}=1\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{z^2}=16&\mbox{}\\
\implies &\displaystyle z=\pm 1\,\,\,\,\,\mbox{or}\,\,\,\,z=\pm 4&\mbox{}\\
\end{array}
$$
Recognizing and Solving "Stuffed Quadratics"
Step 1: Massage the equation into the form $$a \left( \color{magenta}{\mbox{stuff}} \right)^2+b \left( \color{magenta}{\mbox{stuff}} \right)+c=0$$
Step 2: Use quadratic techniques to get $\color{magenta}{\mbox{stuff}}$ by itself: $$\color{magenta}{\mbox{stuff}}=\mbox{number}_1 \,\,\,\,\,\,\, \mbox{ or } \,\,\,\,\,\,\, \color{magenta}{\mbox{stuff}}=\mbox{number}_2.$$ Step 3: Solve the equation(s) in Step 2.
More Examples of "Stuffed Quadratics"
$v^{2/3}-2v^{1/3}=8$
$$
\begin{array}{lll}
&\displaystyle v^{2/3}-2v^{1/3}=8&\mbox{}\\
\implies &\displaystyle v^{2/3}-2v^{1/3}-8=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{v^{1/3}})^2-2\color{magenta}{v^{1/3}}-8=0&\mbox{Step 1}\\
\implies &\displaystyle \color{magenta}{x}^2-2\color{magenta}{x}-8=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{x}-4)(\color{magenta}{x}+2)=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}-4=0\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}+2=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}=4\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}=-2&\mbox{}\\
\implies &\displaystyle \color{magenta}{v^{1/3}}=4\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{v^{1/3}}=-2&\mbox{Step 2}\\
\implies &\displaystyle v=4^3\,\,\,\,\,\mbox{or}\,\,\,\,v=(-2)^3&\mbox{}\\
\implies &\displaystyle v=64\,\,\,\,\,\mbox{or}\,\,\,\,v=-8&\mbox{}\\
\end{array}
$$
$(h+2)^2-(h+2)-20=0$
$$
\begin{array}{lll}
&\displaystyle (h+2)^2-(h+2)-20=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{h+2})^2-(\color{magenta}{h+2})-20=0&\mbox{Step 1}\\
\implies &\displaystyle \color{magenta}{x}^2-\color{magenta}{x}-20=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{x}+4)(\color{magenta}{x}-5)=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}+4=0\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}-5=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}+4=0\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}-5=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}=-4\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}=5&\mbox{}\\
\implies &\displaystyle \color{magenta}{h+2}=-4\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{h+2}=5&\mbox{Step 2}\\
\implies &\displaystyle h=-6\,\,\,\,\,\mbox{or}\,\,\,\,h=3&\mbox{}\\
\end{array}
$$
$x^{-2}+8x^{-1}+12=0$
$$
\begin{array}{lll}
&\displaystyle x^{-2}+8x^{-1}+12=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{x^{-1}})^2+8\color{magenta}{x^{-1}}+12=0&\mbox{Step 1}\\
\implies &\displaystyle \color{magenta}{u}^2+8\color{magenta}{u}+12=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{u}+2)(\color{magenta}{u}+6)=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{u}+2=0\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{u}+6=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{u}=-2\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{u}=-6&\mbox{}\\
\implies &\displaystyle \color{magenta}{x^{-1}}=-2\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x^{-1}}=-6&\mbox{Step 2}\\
\implies &\displaystyle \frac{1}{x}=-2\,\,\,\,\,\mbox{or}\,\,\,\,\frac{1}{x}=-6&\mbox{}\\
\implies &\displaystyle x=-\frac{1}{2}\,\,\,\,\,\mbox{or}\,\,\,\,x=-\frac{1}{6}&\mbox{}\\
\end{array}
$$
$u^4+u^2-72=0$
$$
\begin{array}{lll}
&\displaystyle u^4+u^2-72=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{u^2})^2+\color{magenta}{u^2}-72=0&\mbox{Step 1}\\
\implies &\displaystyle \color{magenta}{x}^2+\color{magenta}{x}-72=0&\mbox{}\\
\implies &\displaystyle (\color{magenta}{x}-8)(\color{magenta}{x}+9)=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}-8=0\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}+9=0&\mbox{}\\
\implies &\displaystyle \color{magenta}{x}=8\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}=-9&\mbox{}\\
\implies &\displaystyle \color{magenta}{u^2}=8\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{u^2}=-9&\mbox{Step 2}\\
\implies &\displaystyle u=\pm\sqrt{8}\,\,\,\,\,\mbox{or}\,\,\,\,u=\pm\sqrt{-9}&\mbox{}\\
\implies &\displaystyle u=\pm 2\sqrt{2}\,\,\,\,\,\mbox{or}\,\,\,\,u=\pm 3i &\mbox{}\\
\end{array}
$$
Bonus Example: A Radical Equation (If there's time.)
Solve the radical equation. Completely simplify any radical expressions. Consider only real-number solutions. $$ c-4\sqrt{c}+3=0$$
$$
\begin{array}{lll}
&\displaystyle c-4\sqrt{c}+3=0&\mbox{}\\
\implies &\displaystyle c+3=4\sqrt{c}&\mbox{}\\
\implies &\displaystyle (c+3)^2=(4\sqrt{c})^2&\mbox{}\\
\implies &\displaystyle c^2+6c+9=16c&\mbox{}\\
\implies &\displaystyle c^2-10c+9=0&\mbox{}\\
\implies &\displaystyle (c-1)(c-9)=0&\mbox{}\\
\implies &\displaystyle c-1=0\,\,\,\,\,\mbox{or}\,\,\,\,c-9=0&\mbox{}\\
\implies &\displaystyle c=1\,\,\,\,\,\mbox{or}\,\,\,\,c=9&\mbox{}\\
\end{array}
$$
Scenic Alternative $$ \begin{array}{lll} &\displaystyle c-4\sqrt{c}+3=0&\mbox{}\\ \implies &\displaystyle (\color{magenta}{\sqrt{c}})^2-4\color{magenta}{\sqrt{c}}+3=0&\mbox{}\\ \implies &\displaystyle \color{magenta}{x}^2-4\color{magenta}{x}+3=0&\mbox{}\\ \implies &\displaystyle (\color{magenta}{x}-1)(\color{magenta}{x}-3)=0&\mbox{}\\ \implies &\displaystyle \color{magenta}{x}-1=0\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}-3=0&\mbox{}\\ \implies &\displaystyle \color{magenta}{x}=1\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}=3&\mbox{}\\ \implies &\displaystyle \color{magenta}{\sqrt{c}}=1\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{\sqrt{c}}=3&\mbox{}\\ \implies &\displaystyle c=1^2\,\,\,\,\,\mbox{or}\,\,\,\,c=3^2&\mbox{}\\ \implies &\displaystyle c=1\,\,\,\,\,\mbox{or}\,\,\,\,c=9&\mbox{}\\ \end{array} $$
Scenic Alternative $$ \begin{array}{lll} &\displaystyle c-4\sqrt{c}+3=0&\mbox{}\\ \implies &\displaystyle (\color{magenta}{\sqrt{c}})^2-4\color{magenta}{\sqrt{c}}+3=0&\mbox{}\\ \implies &\displaystyle \color{magenta}{x}^2-4\color{magenta}{x}+3=0&\mbox{}\\ \implies &\displaystyle (\color{magenta}{x}-1)(\color{magenta}{x}-3)=0&\mbox{}\\ \implies &\displaystyle \color{magenta}{x}-1=0\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}-3=0&\mbox{}\\ \implies &\displaystyle \color{magenta}{x}=1\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{x}=3&\mbox{}\\ \implies &\displaystyle \color{magenta}{\sqrt{c}}=1\,\,\,\,\,\mbox{or}\,\,\,\,\color{magenta}{\sqrt{c}}=3&\mbox{}\\ \implies &\displaystyle c=1^2\,\,\,\,\,\mbox{or}\,\,\,\,c=3^2&\mbox{}\\ \implies &\displaystyle c=1\,\,\,\,\,\mbox{or}\,\,\,\,c=9&\mbox{}\\ \end{array} $$