Radical Equations
Learning Objectives
- Solve Radical Equations
- Isolate a radical term in an equation using algebra and rules for radicals and exponents
- Identify a radical equation with no solutions, or extraneous solutions
- Applications with radical equations
- Kinetic Energy
- Volume
- Free-fall
Isolate a radical term
A basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical. (The reason for using powers will become clear in a moment.) This is the same type of strategy you used to solve other, non-radical equations—rearrange the expression to isolate the variable you want to know, and then solve the resulting equation. There are two key ideas that you will be using to solve radical equations. The first is that if [latex] a=b[/latex], then [latex] {{a}^{2}}={{b}^{2}}[/latex]. (This property allows you to square both sides of an equation and remain certain that the two sides are still equal.) The second is that if the square root of any nonnegative number x is squared, then you get x: [latex] {{\left( \sqrt{x} \right)}^{2}}=x[/latex]. (This property allows you to “remove” the radicals from your equations.) Let’s start with a radical equation that you can solve in a few steps:[latex] \sqrt{x}-3=5[/latex].Example
Solve. [latex] \sqrt{x}-3=5[/latex]Answer: Add 3 to both sides to isolate the variable term on the left side of the equation.
[latex] \begin{array}{r}\sqrt{x}-3\,\,\,=\,\,\,5\\\underline{+3\,\,\,\,\,\,\,+3}\end{array}[/latex]
Collect like terms.[latex] \sqrt{x}=8[/latex]
Square both sides to remove the radical, since [latex] {{(\sqrt{x})}^{2}}=x[/latex]. Make sure to square the 8 also! Then simplify.[latex]\begin{array}{r}{{(\sqrt{x})}^{2}}={{8}^{2}}\\x=64\end{array}[/latex]
Answer
[latex-display]x=64[/latex] is the solution to [latex] \sqrt{x}-3=5[/latex-display]Example
Solve. [latex] \sqrt{x+8}=3[/latex]Answer: Notice how the radical contains a binomial: [latex]x+8[/latex]. Square both sides to remove the radical.
[latex] {{\left( \sqrt{x+8} \right)}^{2}}={{\left( 3 \right)}^{2}}[/latex]
[latex] {{\left( \sqrt{x+8} \right)}^{2}}=x+8[/latex]. Now simplify the equation and solve for x.[latex] \begin{array}{r}x+8=9\\x=1\end{array}[/latex]
Check your answer. Substituting 1 for x in the original equation yields a true statement, so the solution is correct.[latex] \begin{array}{r}\sqrt{1+8}=3\\\sqrt{9}=3\\3=3\end{array}[/latex]
Answer
[latex] x=1[/latex] is the solution to [latex] \sqrt{x+8}=3[/latex].Example
Solve. [latex] 1+\sqrt{2x+3}=6[/latex]Answer: Begin by subtracting 1 from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.
[latex] \begin{array}{r}1+\sqrt{2x+3}-1=6-1\\\sqrt{2x+3}=5\,\,\,\,\,\,\,\,\,\,\\{{\left( \sqrt{2x+3} \right)}^{2}}={{\left( 5 \right)}^{2}}\,\,\,\end{array}[/latex]
Simplify the equation and solve for x.[latex] \begin{array}{r}2x+3=25\\2x=22\\x=11\end{array}[/latex]
Check your answer. Substituting 11 for x in the original equation yields a true statement, so the solution is correct.[latex] \begin{array}{r}1+\sqrt{2(11)+3}=6\\1+\sqrt{22+3}=6\\1+\sqrt{25}=6\\1+5=6\\6=6\end{array}[/latex]
Answer
[latex] x=11[/latex] is the solution for [latex] 1+\sqrt{2x+3}=6[/latex].Solving Radical Equations
Follow the following four steps to solve radical equations.- Isolate the radical expression.
- Square both sides of the equation: If [latex]x=y[/latex] then [latex]x^{2}=y^{2}[/latex].
- Once the radical is removed, solve for the unknown.
- Check all answers.
Identify a radical equation with no solutions or extraneous solutions
Following rules is important, but so is paying attention to the math in front of you—especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.Example
Solve. [latex] \sqrt{a-5}=-2[/latex]Answer: Square both sides to remove the term [latex]a–5[/latex] from the radical.
[latex] {{\left( \sqrt{a-5} \right)}^{2}}={{(-2)}^{2}}[/latex]
Write the simplified equation, and solve for a.[latex]\begin{array}{r}a-5=4\\a=9\end{array}[/latex]
Now check the solution by substituting [latex]a=9[/latex] into the original equation. It does not check![latex] \begin{array}{r}\sqrt{9-5}=-2\\\sqrt{4}=-2\\2\ne -2\end{array}[/latex]
Answer
No solution.Example
Solve. [latex] x+4=\sqrt{x+10}[/latex]Answer: Square both sides to remove the term [latex]x+10[/latex] from the radical.
[latex] {{\left( x+4 \right)}^{2}}={{\left( \sqrt{x+10} \right)}^{2}}[/latex]
Now simplify and solve the equation. Combine like terms, and then factor.[latex] \begin{array}{r}\left( x+4 \right)\left( x+4 \right)=x+10\\{{x}^{2}}+8x+16=x+10\\{{x}^{2}}+8x-x+16-10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\{{x}^{2}}+7x+6=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\left( x+6 \right)\left( x+1 \right)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Set each factor equal to zero and solve for x.[latex] \begin{array}{c}\left( x+6 \right)=0\,\,\text{or}\,\,\left( x+1 \right)=0\\x=-6\text{ or }x=-1\end{array}[/latex]
Now check both solutions by substituting them into the original equation. Since [latex]x=−6[/latex] produces a false statement, it is an extraneous solution.[latex] \begin{array}{l}-6+4=\sqrt{-6+10}\\\,\,\,\,\,\,\,\,-2=\sqrt{4}\\\,\,\,\,\,\,\,\,-2=2\\\text{FALSE!}\\\\\\-1+4=\sqrt{-1+10}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=\sqrt{9}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=3\\\text{TRUE!}\end{array}[/latex]
Answer
[latex]x=−1[/latex] is the only solutionExample
Solve. [latex] 4+\sqrt{x+2}=x[/latex]Answer: Isolate the radical term.
[latex] \sqrt{x+2}=x-4[/latex]
Square both sides to remove the term [latex]x+2[/latex] from the radical.[latex] {{\left( \sqrt{x+2} \right)}^{2}}={{\left( x-4 \right)}^{2}}[/latex]
Now simplify and solve the equation. Combine like terms, and then factor.[latex] \begin{array}{l}x+2={{x}^{2}}-8x+16\\\,\,\,\,\,\,\,\,\,\,\,0={{x}^{2}}-8x-x+16-2\\\,\,\,\,\,\,\,\,\,\,\,0={{x}^{2}}-9x+14\\\,\,\,\,\,\,\,\,\,\,\,0=\left( x-7 \right)\left( x-2 \right)\end{array}[/latex]
Set each factor equal to zero and solve for x.[latex] \begin{array}{c}\left( x-7 \right)=0\text{ or }\left( x-2 \right)=0\\x=7\text{ or }x=2\end{array}[/latex][latex] [/latex]
Now check both solutions by substituting them into the original equation. Since [latex]x=2[/latex] produces a false statement, it is an extraneous solution.[latex] \begin{array}{r}4+\sqrt{7+2}=7\\4+\sqrt{9}=7\\4+3=7\\7=7\\\text{TRUE!}\\\\4+\sqrt{2+2}=2\\4+\sqrt{4}=2\\4+2=2\\6=2\\\text{FALSE!}\end{array}[/latex]
Answer
[latex]x=7[/latex] is the only solution.Example
Solve [latex]\sqrt{2x+3}+\sqrt{x - 2}=4[/latex].Answer: As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.
Answer
The only solution is [latex]x=3[/latex]Applications with 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 ([latex]E_{k}[/latex], 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 [latex] v=\sqrt{\frac{2{{E}_{k}}}{m}}[/latex].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.
[latex]\begin{array}{l}E_{k}=\text{unknown}\\\,\,m=1000\\\,\,\,\,v=30\end{array}[/latex]
Substitute values into the formula.[latex] 30=\sqrt{\frac{2{{E}_{k}}}{1,000}}[/latex]
Solve the radical equation for Ek.[latex] \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}[/latex]
Now check the solution by substituting it into the original equation.[latex] \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}[/latex]
Answer
The Kinetic Energy is 450,000 Joules.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: [latex]r=\sqrt[3]{\frac{3V}{2\pi }}[/latex].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 [latex]\pi =3.14[/latex].Answer: The radius of a cone given it's volume can be found with the following formula: [latex]r=\sqrt[3]{\frac{3V}{2\pi }}[/latex], [latex]r\ge 0\\[/latex] Cube both sides to eliminate the cube root.
Therefore, the volume is about 100 cubic feet.
Answer
[latex-display]V\approx{100}\frac{\text{cu}}{ft}[/latex-display]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 [latex]h=600 - 16{t}^{2}[/latex]. 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 [latex]\pm\sqrt{\frac{h-600}{-16}}[/latex]. We want time to be only positive since we are talking about a measurable quantity, so we will restrict our answers to just [latex]+\sqrt{\frac{h-600}{-16}}[/latex] 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]
Answer
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 [latex]\frac{h-600}{-16}[/latex] 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. Radical equations play a significant role in science, engineering, and even music. Sometimes you may need to use what you know about radical equations to solve for different variables in these types of problems.Licenses & Attributions
CC licensed content, Original
- Kinetic Energy - Radical Equation Application. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Volume of a Cone -Radical Equation Application. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution.
CC licensed content, Shared previously
- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay, et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at: http://cnx.org/contents/[email protected]:1/Preface.
- Ex 1: Solve a Basic Radical Equation - Square Roots. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Unit 18: Exponential and Logarithmic Functions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
- Ex 2: Solve Radical Equations - Square Roots. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Ex 4: Solve Radical Equations - Square Roots. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Ex 7: Solve Radical Equations - Two Square Roots. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.