Radical Arithmetic: Now that we know how to work with radicals ($+$,$-$,$\times$,$\div$), we reap the reward of getting to solve radical equations.
A Prototype Example: $\sqrt{r-3}=5$
More Examples
$3=\sqrt{6 p+2}-9$
$$
\begin{array}{lll}
&\displaystyle 3=\sqrt{6 p+2}-9&\mbox{}\\
\implies &\displaystyle 12=\sqrt{6 p+2}&\mbox{}\\
\implies &\displaystyle 12^2=\left(\sqrt{6 p+2}\right)^2&\mbox{}\\
\implies &\displaystyle 144=6 p+2&\mbox{}\\
\implies &\displaystyle 142=6 p&\mbox{}\\
\implies &\displaystyle \frac{142}{6}= p&\mbox{}\\
\implies &\displaystyle p=\frac{71}{3}&\mbox{}\\
\end{array}
$$
Check
$$ \begin{array}{lll} \displaystyle \sqrt{6 p+2}-9&\displaystyle=\sqrt{6 \cdot \frac{71}{3}+2}-9 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{2 \cdot 71+2}-9 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{142+2}-9 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{144}-9 &\mbox{}\\ \displaystyle &\displaystyle=12-9 &\mbox{}\\ \displaystyle &\displaystyle=3 \, \color{blue}{\checkmark}&\mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle \sqrt{6 p+2}-9&\displaystyle=\sqrt{6 \cdot \frac{71}{3}+2}-9 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{2 \cdot 71+2}-9 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{142+2}-9 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{144}-9 &\mbox{}\\ \displaystyle &\displaystyle=12-9 &\mbox{}\\ \displaystyle &\displaystyle=3 \, \color{blue}{\checkmark}&\mbox{}\\ \end{array} $$
$\sqrt{z-2}+9=7$
$$
\begin{array}{lll}
&\displaystyle \sqrt{z-2}+9=7&\mbox{}\\
\implies &\displaystyle \sqrt{z-2}=-2&\mbox{}\\
\implies &\displaystyle \left(\sqrt{z-2}\right)^2=(-2)^2&\mbox{}\\
\implies &\displaystyle z-2=4&\mbox{}\\
\implies &\displaystyle z=6&\mbox{}\\
\end{array}
$$
Check
$$ \begin{array}{lll} \displaystyle \sqrt{z-2}+9&\displaystyle=\sqrt{6-2}+9 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{4}+9 &\mbox{}\\ \displaystyle &\displaystyle=2+9 &\mbox{}\\ \displaystyle &\displaystyle=11 &\mbox{}\\ \displaystyle &\displaystyle\neq 3 \, \color{red}{\times} &\mbox{}\\ \end{array} $$ The solution $z=6$ is called an extraneous solution.
$$ \begin{array}{lll} \displaystyle \sqrt{z-2}+9&\displaystyle=\sqrt{6-2}+9 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{4}+9 &\mbox{}\\ \displaystyle &\displaystyle=2+9 &\mbox{}\\ \displaystyle &\displaystyle=11 &\mbox{}\\ \displaystyle &\displaystyle\neq 3 \, \color{red}{\times} &\mbox{}\\ \end{array} $$ The solution $z=6$ is called an extraneous solution.
$\sqrt[3]{6 q-2}-4=3$
$$
\begin{array}{lll}
&\displaystyle \sqrt[3]{6 q-2}-4=3&\mbox{}\\
\implies &\displaystyle \sqrt[3]{6 q-2}=7&\mbox{}\\
\implies &\displaystyle \left(\sqrt[3]{6 q-2}\right)^3=7^3&\mbox{}\\
\implies &\displaystyle 6q-2=343&\mbox{}\\
\implies &\displaystyle 6q=345&\mbox{}\\
\implies &\displaystyle q=\frac{345}{6}&\mbox{}\\
\implies &\displaystyle q=\frac{115}{2}&\mbox{}\\
\end{array}
$$
Check
$$ \begin{array}{lll} \displaystyle \sqrt[3]{6 q-2}-4&\displaystyle=\sqrt[3]{6 \cdot \frac{115}{2} -2}-4 &\mbox{}\\ \displaystyle &\displaystyle= \sqrt[3]{3\cdot 115 -2}-4&\mbox{}\\ \displaystyle &\displaystyle= \sqrt[3]{345 -2}-4&\mbox{}\\ \displaystyle &\displaystyle= \sqrt[3]{343}-4&\mbox{}\\ \displaystyle &\displaystyle= 7-4&\mbox{}\\ \displaystyle &\displaystyle= 3 \,\color{blue}{\checkmark}&\mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle \sqrt[3]{6 q-2}-4&\displaystyle=\sqrt[3]{6 \cdot \frac{115}{2} -2}-4 &\mbox{}\\ \displaystyle &\displaystyle= \sqrt[3]{3\cdot 115 -2}-4&\mbox{}\\ \displaystyle &\displaystyle= \sqrt[3]{345 -2}-4&\mbox{}\\ \displaystyle &\displaystyle= \sqrt[3]{343}-4&\mbox{}\\ \displaystyle &\displaystyle= 7-4&\mbox{}\\ \displaystyle &\displaystyle= 3 \,\color{blue}{\checkmark}&\mbox{}\\ \end{array} $$
Process for Solving Radical Equations
1) Isolate radical on one side of the equation.
2) Raise both sides to a power equal to the index.
3) Simplify.
4) If no more radicals remain, solve the equation. Otherwise go back to step 1).
5) Check for extraneous (fake) solutions.
Even More Examples
$3=\sqrt[5]{5 y+4}-2$
$$
\begin{array}{lll}
&\displaystyle 3=\sqrt[5]{5 y+4}-2&\mbox{}\\
\implies &\displaystyle 5=\sqrt[5]{5 y+4}&\mbox{isolate radical}\\
\implies &\displaystyle 5^5=\left(\sqrt[5]{5 y+4}\right)^5&\mbox{raise both sides to a power equal to the index}\\
\implies &\displaystyle 3125=5 y+4&\mbox{simplify, no more radicals remain}\\
\implies &\displaystyle 3121=5 y&\mbox{}\\
\implies &\displaystyle y=\frac{3121}{5}&\mbox{}\\
\end{array}
$$
Check
$$ \begin{array}{lll} \displaystyle \sqrt[5]{5 y+4}-2&\displaystyle=\sqrt[5]{5 \cdot \frac{3121}{5}+4}-2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt[5]{3121+4}-2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt[5]{3125}-2 &\mbox{}\\ \displaystyle &\displaystyle=5-2 &\mbox{}\\ \displaystyle &\displaystyle=3 \,\color{blue}{\checkmark}&\mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle \sqrt[5]{5 y+4}-2&\displaystyle=\sqrt[5]{5 \cdot \frac{3121}{5}+4}-2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt[5]{3121+4}-2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt[5]{3125}-2 &\mbox{}\\ \displaystyle &\displaystyle=5-2 &\mbox{}\\ \displaystyle &\displaystyle=3 \,\color{blue}{\checkmark}&\mbox{}\\ \end{array} $$
$\sqrt{ x^2+x+7}-x=9$
$$
\begin{array}{lll}
&\displaystyle \sqrt{ x^2+x+7}-x=9&\mbox{}\\
\implies &\displaystyle \sqrt{ x^2+x+7}=x+9&\mbox{isolate radical}\\
\implies &\displaystyle \left(\sqrt{ x^2+x+7}\right)^2=(x+9)^2&\mbox{raise both sides to a power equal to the index}\\
\implies &\displaystyle x^2+x+7=x^2+18x+81&\mbox{simplify, no radicals remain}\\
\implies &\displaystyle x+7=18x+81&\mbox{}\\
\implies &\displaystyle x-74=18x&\mbox{}\\
\implies &\displaystyle -74=17x&\mbox{}\\
\implies &\displaystyle x=-\frac{74}{17}&\mbox{}\\
\end{array}
$$
Check
$$ \begin{array}{lll} \displaystyle \sqrt{ x^2+x+7}-x&\displaystyle=\sqrt{ \left(-\frac{74}{17}\right)^2+\left(-\frac{74}{17}\right)+7}-\left(-\frac{74}{17}\right) &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{ \frac{5476}{289}-\frac{74}{17}+7}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{5476}{289}-\frac{74}{17}\cdot \color{magenta}{\frac{17}{17}}+7\cdot \color{magenta}{\frac{17^2}{17^2}}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{5476}{289}-\frac{1258}{289}+\frac{2023}{289}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{5476-1258+2023}{289}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{6241}{289}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt{6241}}{\sqrt{289}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\frac{79}{17}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\frac{153}{17} &\mbox{}\\ \displaystyle &\displaystyle=9\,\color{blue}{\checkmark} &\mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle \sqrt{ x^2+x+7}-x&\displaystyle=\sqrt{ \left(-\frac{74}{17}\right)^2+\left(-\frac{74}{17}\right)+7}-\left(-\frac{74}{17}\right) &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{ \frac{5476}{289}-\frac{74}{17}+7}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{5476}{289}-\frac{74}{17}\cdot \color{magenta}{\frac{17}{17}}+7\cdot \color{magenta}{\frac{17^2}{17^2}}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{5476}{289}-\frac{1258}{289}+\frac{2023}{289}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{5476-1258+2023}{289}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{6241}{289}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sqrt{6241}}{\sqrt{289}}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\frac{79}{17}+\frac{74}{17} &\mbox{}\\ \displaystyle &\displaystyle=\frac{153}{17} &\mbox{}\\ \displaystyle &\displaystyle=9\,\color{blue}{\checkmark} &\mbox{}\\ \end{array} $$
$\sqrt{ 11 t + 2}= \sqrt{ 3 t + 7}$
$$
\begin{array}{lll}
&\displaystyle \sqrt{ 11 t + 2}= \sqrt{ 3 t + 7} &\mbox{}\\
\implies &\displaystyle \left(\sqrt{ 11 t + 2}\right)^2= \left(\sqrt{ 3 t + 7}\right)^2&\mbox{}\\
\implies &\displaystyle 11t+2= 3t+7&\mbox{}\\
\implies &\displaystyle 11t= 3t+5&\mbox{}\\
\implies &\displaystyle 8t=5&\mbox{}\\
\implies &\displaystyle t=\frac{5}{8}&\mbox{}\\
\end{array}
$$
Check
Left-hand Side: $$ \begin{array}{lll} \displaystyle \sqrt{ 11 t + 2} &\displaystyle=\sqrt{11 \cdot \frac{5}{8} + 2} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{55}{8} + \frac{16}{8}} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{71}{8}} &\mbox{}\\ \end{array} $$ Right-hand Side: $$ \begin{array}{lll} \displaystyle \sqrt{ 3 t + 7} &\displaystyle=\sqrt{ 3 \cdot \frac{5}{8} + 7} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{15}{8} + \frac{56}{8}} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{71}{8}} &\mbox{}\\ \end{array} $$ Since the left-hand side and right-hand side agree, the solution checks.
Left-hand Side: $$ \begin{array}{lll} \displaystyle \sqrt{ 11 t + 2} &\displaystyle=\sqrt{11 \cdot \frac{5}{8} + 2} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{55}{8} + \frac{16}{8}} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{71}{8}} &\mbox{}\\ \end{array} $$ Right-hand Side: $$ \begin{array}{lll} \displaystyle \sqrt{ 3 t + 7} &\displaystyle=\sqrt{ 3 \cdot \frac{5}{8} + 7} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{15}{8} + \frac{56}{8}} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{71}{8}} &\mbox{}\\ \end{array} $$ Since the left-hand side and right-hand side agree, the solution checks.
$\sqrt{z}+1=\sqrt{z+12}$
$$
\begin{array}{lll}
&\displaystyle \sqrt{z}+1=\sqrt{z+12} &\mbox{radical already isolated on RHS}\\
\implies &\displaystyle \left(\sqrt{z}+1\right)^2=\left(\sqrt{z+12}\right)^2&\mbox{raise both sides to a power equal to the index}\\
\implies &\displaystyle z+2\sqrt{z}+1=z+12&\mbox{simplify, radicals remain, so back to step 1}\\
\implies &\displaystyle 2\sqrt{z}=11&\mbox{}\\
\implies &\displaystyle \sqrt{z}=\frac{11}{2}&\mbox{isolate radical}\\
\implies &\displaystyle z=\left(\frac{11}{2}\right)^2&\mbox{raise both sides to a power equal to the index}\\
\implies &\displaystyle z=\frac{121}{4}&\mbox{}\\
\end{array}
$$
Check
LHS: $$ \begin{array}{lll} \displaystyle \sqrt{z}+1&\displaystyle=\sqrt{\frac{121}{4}}+1 &\mbox{}\\ \displaystyle &\displaystyle=\frac{11}{2}+\frac{2}{2} &\mbox{}\\ \displaystyle &\displaystyle=\frac{13}{2}&\mbox{}\\ \end{array} $$ RHS: $$ \begin{array}{lll} \displaystyle \sqrt{z+12}&\displaystyle=\sqrt{\frac{121}{4}+12} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{121}{4}+\frac{48}{4}} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{169}{4}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{13}{2} &\mbox{}\\ \end{array} $$ Since the left-hand side and right-hand side agree, the solution checks.
LHS: $$ \begin{array}{lll} \displaystyle \sqrt{z}+1&\displaystyle=\sqrt{\frac{121}{4}}+1 &\mbox{}\\ \displaystyle &\displaystyle=\frac{11}{2}+\frac{2}{2} &\mbox{}\\ \displaystyle &\displaystyle=\frac{13}{2}&\mbox{}\\ \end{array} $$ RHS: $$ \begin{array}{lll} \displaystyle \sqrt{z+12}&\displaystyle=\sqrt{\frac{121}{4}+12} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{121}{4}+\frac{48}{4}} &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{\frac{169}{4}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{13}{2} &\mbox{}\\ \end{array} $$ Since the left-hand side and right-hand side agree, the solution checks.
Bonus Example
$\sqrt{r+4}-r+2=0$
$$
\begin{array}{lll}
&\displaystyle \sqrt{r+4}-r+2=0&\mbox{}\\
\implies &\displaystyle \sqrt{r+4}=r-2&\mbox{isolate radical}\\
\implies &\displaystyle \left(\sqrt{r+4}\right)^2=(r-2)^2&\mbox{raise both sides to a power equal to the index}\\
\implies &\displaystyle r+4=r^2-4r+4&\mbox{simplify, no radicals remain}\\
\implies &\displaystyle 0=r^2-5r&\mbox{}\\
\implies &\displaystyle (r-5)r=0&\mbox{}\\
\implies &\displaystyle r-5=0 \mbox{ or } r=0&\mbox{}\\
\implies &\displaystyle r=5 \mbox{ or } r=0&\mbox{}\\
\end{array}
$$
Check
$r=0$ $$ \begin{array}{lll} \displaystyle \sqrt{r+4}-r+2&\displaystyle=\sqrt{0+4}-0+2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{4}+2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{4}+2 &\mbox{}\\ \displaystyle &\displaystyle=2+2 &\mbox{}\\ \displaystyle &\displaystyle=4 &\mbox{}\\ \displaystyle &\displaystyle\neq 0 \, \color{red}{\times} &\mbox{}\\ \end{array} $$ $r=0$ is a fake (extraneous) solution.
$r=5$ $$ \begin{array}{lll} \displaystyle \sqrt{r+4}-r+2&\displaystyle=\sqrt{5+4}-5+2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{9}-3 &\mbox{}\\ \displaystyle &\displaystyle=3-3 &\mbox{}\\ \displaystyle &\displaystyle= 0 \, \color{blue}{\checkmark} &\mbox{}\\ \end{array} $$ $r=5$ is a genuine, bonafide solution.
$r=0$ $$ \begin{array}{lll} \displaystyle \sqrt{r+4}-r+2&\displaystyle=\sqrt{0+4}-0+2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{4}+2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{4}+2 &\mbox{}\\ \displaystyle &\displaystyle=2+2 &\mbox{}\\ \displaystyle &\displaystyle=4 &\mbox{}\\ \displaystyle &\displaystyle\neq 0 \, \color{red}{\times} &\mbox{}\\ \end{array} $$ $r=0$ is a fake (extraneous) solution.
$r=5$ $$ \begin{array}{lll} \displaystyle \sqrt{r+4}-r+2&\displaystyle=\sqrt{5+4}-5+2 &\mbox{}\\ \displaystyle &\displaystyle=\sqrt{9}-3 &\mbox{}\\ \displaystyle &\displaystyle=3-3 &\mbox{}\\ \displaystyle &\displaystyle= 0 \, \color{blue}{\checkmark} &\mbox{}\\ \end{array} $$ $r=5$ is a genuine, bonafide solution.
Application
When a driver applies the brakes in a moving car, the distance $d$ a car skids (the length of the skid marks in feet) and the speed $S$ the car (in miles per hour) was going when the brakes were applied are related by the formula $S = \sqrt{30fd}$ where $f$ accounts for the friction between the tires and road surface. (For example, dry roads have higher friction than wet roads.)
(a) Suppose that on a certain road, $f=0.8.$ What does the formula simplify to?
For $f=0.8,$ the formula simplifies to
$$
S = \sqrt{30fd}= \sqrt{30\cdot 0.8\cdot d}=\sqrt{24d}
$$
(b) Suppose on a wet road an accident investigator determined a friction coefficient of $f=0.8.$ The length the skid marks of the first vehicle in the accident was $175$ feet. Use the formula to find the speed of the vehicle before the brakes were applied. Round your answer to the nearest mile per hour.
The problem tells us that $d=175.$ So,
$$
S =\sqrt{24d}=\sqrt{24\cdot 175}\approx 64.80740698
$$
We conclude that the car was travelling about $65$ miles per hour before slamming on the brakes.
(c) The other vehicle was known to be travelling $72$ miles per hour. Use the formula to find the length of the skid marks of the other vehicle.
The above gives $S=72.$ We use the above to find the length $d$ of the skid marks.
$$
\begin{array}{lll}
&\displaystyle S =\sqrt{24d}&\mbox{}\\
\implies &\displaystyle 72 =\sqrt{24d}&\mbox{}\\
\implies &\displaystyle 72^2 =\left(\sqrt{24d}\right)^2&\mbox{}\\
\implies &\displaystyle 5184 =24d&\mbox{}\\
\implies &\displaystyle d=\frac{5184}{24}&\mbox{}\\
\implies &\displaystyle d=216&\mbox{}\\
\end{array}
$$
We expect that the car skidded about $216$ feet before coming to a complete stop.
Check $$ \begin{array}{lll} \displaystyle \sqrt{24d}&\displaystyle \sqrt{24\cdot 216}= &\mbox{}\\ \displaystyle &\displaystyle=72 \, \color{blue}{\checkmark} &\mbox{}\\ \end{array} $$
Check $$ \begin{array}{lll} \displaystyle \sqrt{24d}&\displaystyle \sqrt{24\cdot 216}= &\mbox{}\\ \displaystyle &\displaystyle=72 \, \color{blue}{\checkmark} &\mbox{}\\ \end{array} $$