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### Learning Outcomes

• Solve an application using a formula containing a radical expression

Radical equations in real-life applications may contain two or more variables. Solving such an equation involves isolating the unknown variable, which may be contained within a radical such as a square or cube root, while substituting known values for the remaining variables.

$S=\sqrt{\dfrac{x}{32}}\qquad \qquad T=2\sqrt{L}\qquad \qquad r=\sqrt{\dfrac{V}{h\pi}}$

Each of the above equations is an example of a radical equation. Use the examples given in the situations below to practice solving applications involving radical equations.

## Kinetic Energy

One way to measure the amount of energy that a moving object (such as a car or roller coaster) possesses is by finding its Kinetic Energy. The Kinetic Energy ($E_{k}$, measured in Joules) of an object depends on the object’s mass (m, measured in kg) and velocity (v, measured in meters per second), and can be written as $v=\sqrt{\frac{2{{E}_{k}}}{m}}$.

### Example

What is the Kinetic Energy of an object with a mass of $1,000$ kilograms that is traveling at $30$ meters per second?

Answer: Identify variables and known values.

$\begin{array}{l}E_{k}=\text{unknown}\\\,\,m=1000\\\,\,\,\,v=30\end{array}$

Substitute values into the formula.

$30=\sqrt{\frac{2{{E}_{k}}}{1,000}}$

Solve the radical equation for  Ek.

$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( 30 \right)}^{2}}={{\left( \sqrt{\frac{2{{E}_{k}}}{1,000}} \right)}^{2}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,900=\frac{2{{E}_{k}}}{1,000}\\\\900\cdot 1,000=\frac{2{{E}_{k}}}{1,000}\cdot 1,000\\\\\,\,\,\,\,\,\,\,900,000=2{{E}_{k}}\\\\\,\,\,\,\,\,\,\,\frac{900,000}{2}=\frac{2{{E}_{k}}}{2}\\\\\,\,\,\,\,\,\,\,\,\,450,000={{E}_{k}}\end{array}$

Now check the solution by substituting it into the original equation.

$\begin{array}{l}30=\sqrt{\frac{2\cdot 450,000}{1,000}}\\30=\sqrt{\frac{900,000}{1,000}}\\30=\sqrt{900}\\30=30\end{array}$

The Kinetic Energy is $450,000$ Joules.

Here is another example of finding the kinetic energy of an object in motion. https://youtu.be/NbRXMihA1fY

### Volume

Harvester ants found in the southwest of the U.S. create a vast interlocking network of tunnels for their nests. As a result of all this excavation, a very common above-ground hallmark of a harvester ant nest is a conical mound of small gravel or sand [footnote]Taber, Stephen Welton. The World of the Harvester Ants. College Station: Texas A & M University Press, $1998$.[/footnote] The radius of  a cone whose height is is equal to twice it's radius is given as: $r=\sqrt[3]{\frac{3V}{2\pi }}$.

### Example

A mound of gravel is in the shape of a cone with the height equal to twice the radius. Calculate the volume of such a mound of gravel whose radius is $3.63$ ft. Use $\pi =3.14$.

Answer: The radius of a cone given it's volume can be found with the following formula: $r=\sqrt[3]{\frac{3V}{2\pi }}$, $r\ge 0$ Cube both sides to eliminate the cube root.

$\begin{cases}3.63 & =\sqrt[3]{\frac{3V}{2\pi }} \\(3.63)^3 =\left(\sqrt[3]{\frac{3\cdot V}{2\cdot 3.14}}\right)^3\\47.83=\frac{3V}{2\pi}\\47.83\cdot2\cdot3.14=3V\\100\approx{V}\end{cases}$

Therefore, the volume is about 100 cubic feet.

[latex-display]V\approx{100}\frac{\text{cu}}{ft}[/latex-display]

Here is another example of finding Volume given the radius of a cone. https://youtu.be/gssQYeV87u8

## Free-Fall

When you drop an object from a height, the only force acting on it is gravity (and some air friction) and it is said to be in free-fall. We can use math to describe the height of an object in free fall after a given time because we know how to quantify the force of the earth pulling on us - the force of gravity. An object dropped from a height of 600 feet has a height, h, in feet after t seconds have elapsed, such that $h=600 - 16{t}^{2}$. In our next example we will find the time at which the object is at a given height by first solving for t.

### Example

Find the time is takes to reach a height of 400 feet by first finding an expression for t.

Answer: We are asked two things.  First, we will solve the height equation for t, then we will find how long it takes for the object's height above the ground to be 400 feet. Solve for t: [latex-display]\begin{array}{ccc}h=600 - 16{t}^{2}\\h-600=-16t^2\\\frac{h-600}{-16}=t^2\\\pm\sqrt{\frac{h-600}{-16}}=t\end{array}[/latex-display] At this point, let's stop and talk about whether it makes sense to include both $\pm\sqrt{\frac{h-600}{-16}}$.  We want time to be only positive since we are talking about a measurable quantity, so we will restrict our answers to just $+\sqrt{\frac{h-600}{-16}}$ We want to know at what time the height will be 400 ft., so we can substitute 400 for h. [latex-display]\begin{array}{ccc}\sqrt{\frac{400-600}{-16}}=t\\\sqrt{\frac{-200}{-16}}=t\\\sqrt{12.5}=t\\3.54=t\end{array}[/latex-display]

It takes $3.54$ seconds for the object to be at a height of $400$ ft.

### Analysis of the solution

We have made a point of restricting the radicand of radical expressions to non-negative numbers. In the previous example, we divided by a negative number, then took the square root to solve for t.  In this example is it possible to get a negative number in the radicand?  In other words, for what values for height would we have an issue where we may be taking the square root of a negative number?  We can use algebra to answer this question. Let's translate our question into an inequality.  Again, for what values of h would we get a negative quantity under the radical? The radicand is $\frac{h-600}{-16}$ so if we set up an inequality we can solve it for h: [latex-display]\begin{array}{ccc}\frac{h-600}{-16}\lt0\\-16\cdot\frac{h-600}{-16}\lt0\cdot{-16}\\h-600\gt0\\h\gt600\end{array}[/latex-display] We can interpret this as "when the height is greater than $600$ ft. the radicand will be negative and therefore not a real number." If you re-read the question, you will see that heights greater than $600$ don't even make sense, because the object starts at a height of 600 feet and is falling toward the ground, so height is decreasing. Understanding what domain our variables have is important in applications problems so we can get answers that make sense.