Learning Objectives
- Solve a radical equation, identify extraneous solution
- Solve an equation with ratinal exponents
Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as
[latex]\begin{array}{l}\sqrt{3x+18}\hfill&=x\hfill \\ \sqrt{x+3}\hfill&=x - 3\hfill \\ \sqrt{x+5}-\sqrt{x - 3}\hfill&=2\hfill \end{array}[/latex]
Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions.
A General Note: Radical Equations
An equation containing terms with a variable in the radicand is called a radical equation.
How To: Given a radical equation, solve it.
- Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.
- If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an nth root radical, raise both sides to the nth power. Doing so eliminates the radical symbol.
- Solve the remaining equation.
- If a radical term still remains, repeat steps 1–2.
- Confirm solutions by substituting them into the original equation.
Example: Solving an Equation with One Radical
Solve [latex]\sqrt{15 - 2x}=x[/latex].
Solution
The radical is already isolated on the left side of the equal side, so proceed to square both sides.
[latex]\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ {\left(\sqrt{15 - 2x}\right)}^{2}\hfill&={\left(x\right)}^{2}\hfill \\ 15 - 2x\hfill&={x}^{2}\hfill \end{array}[/latex]
We see that the remaining equation is a quadratic. Set it equal to zero and solve.
[latex]\begin{array}{l}0\hfill&={x}^{2}+2x - 15\hfill \\ \hfill&=\left(x+5\right)\left(x - 3\right)\hfill \\ x\hfill&=-5\hfill \\ x\hfill&=3\hfill \end{array}[/latex]
The proposed solutions are [latex]x=-5[/latex] and [latex]x=3[/latex]. Let us check each solution back in the original equation. First, check [latex]x=-5[/latex].
[latex]\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ \sqrt{15 - 2\left(-5\right)}\hfill&=-5\hfill \\ \sqrt{25}\hfill&=-5\hfill \\ 5\hfill&\ne -5\hfill \end{array}[/latex]
This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.
Check [latex]x=3[/latex].
[latex]\begin{array}{l}\sqrt{15 - 2x}\hfill&=x\hfill \\ \sqrt{15 - 2\left(3\right)}\hfill&=3\hfill \\ \sqrt{9}\hfill&=3\hfill \\ 3\hfill&=3\hfill \end{array}[/latex]
The solution is [latex]x=3[/latex].
Try It
Solve the radical equation: [latex]\sqrt{x+3}=3x - 1[/latex]
Solution
[latex]x=1[/latex]; extraneous solution [latex]x=-\frac{2}{9}[/latex]
Example: Solving a Radical Equation Containing Two Radicals
Solve [latex]\sqrt{2x+3}+\sqrt{x - 2}=4[/latex].
Solution
As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.
[latex]\begin{array}{ll}\sqrt{2x+3}+\sqrt{x - 2}\hfill& =4\hfill & \hfill \\ \sqrt{2x+3}\hfill& =4-\sqrt{x - 2}\hfill & \text{Subtract }\sqrt{x - 2}\text{ from both sides}.\hfill \\ {\left(\sqrt{2x+3}\right)}^{2}\hfill& ={\left(4-\sqrt{x - 2}\right)}^{2}\hfill & \text{Square both sides}.\hfill \end{array}[/latex]
Use the perfect square formula to expand the right side: [latex]{\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}[/latex].
[latex]\begin{array}{ll}2x+3\hfill& ={\left(4\right)}^{2}-2\left(4\right)\sqrt{x - 2}+{\left(\sqrt{x - 2}\right)}^{2}\hfill & \hfill \\ 2x+3\hfill& =16 - 8\sqrt{x - 2}+\left(x - 2\right)\hfill & \hfill \\ 2x+3\hfill& =14+x - 8\sqrt{x - 2}\hfill & \text{Combine like terms}.\hfill \\ x - 11\hfill& =-8\sqrt{x - 2}\hfill & \text{Isolate the second radical}.\hfill \\ {\left(x - 11\right)}^{2}\hfill& ={\left(-8\sqrt{x - 2}\right)}^{2}\hfill & \text{Square both sides}.\hfill \\ {x}^{2}-22x+121\hfill& =64\left(x - 2\right)\hfill & \hfill \end{array}[/latex]
Now that both radicals have been eliminated, set the quadratic equal to zero and solve.
[latex]\begin{array}{ll}{x}^{2}-22x+121=64x - 128\hfill & \hfill \\ {x}^{2}-86x+249=0\hfill & \hfill \\ \left(x - 3\right)\left(x - 83\right)=0\hfill & \text{Factor and solve}.\hfill \\ x=3\hfill & \hfill \\ x=83\hfill & \hfill \end{array}[/latex]
The proposed solutions are [latex]x=3[/latex] and [latex]x=83[/latex]. Check each solution in the original equation.
[latex]\begin{array}{l}\sqrt{2x+3}+\sqrt{x - 2}\hfill& =4\hfill \\ \sqrt{2x+3}\hfill& =4-\sqrt{x - 2}\hfill \\ \sqrt{2\left(3\right)+3}\hfill& =4-\sqrt{\left(3\right)-2}\hfill \\ \sqrt{9}\hfill& =4-\sqrt{1}\hfill \\ 3\hfill& =3\hfill \end{array}[/latex]
One solution is [latex]x=3[/latex].
Check [latex]x=83[/latex].
[latex]\begin{array}{l}\sqrt{2x+3}+\sqrt{x - 2}\hfill&=4\hfill \\ \sqrt{2x+3}\hfill&=4-\sqrt{x - 2}\hfill \\ \sqrt{2\left(83\right)+3}\hfill&=4-\sqrt{\left(83 - 2\right)}\hfill \\ \sqrt{169}\hfill&=4-\sqrt{81}\hfill \\ 13\hfill&\ne -5\hfill \end{array}[/latex]
The only solution is [latex]x=3[/latex]. We see that [latex]x=83[/latex] is an extraneous solution.
Try It
Solve the equation with two radicals: [latex]\sqrt{3x+7}+\sqrt{x+2}=1[/latex].
Solution
[latex]x=-2[/latex]; extraneous solution [latex]x=-1[/latex]
Solve Equations With Rational Exponents
Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\frac{1}{2}}[/latex] is another way of writing [latex]\sqrt{16}[/latex]; [latex]{8}^{\frac{1}{3}}[/latex] is another way of writing [latex]\text{ }\sqrt[3]{8}[/latex]. The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.
We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\frac{2}{3}\left(\frac{3}{2}\right)=1[/latex], [latex]3\left(\frac{1}{3}\right)=1[/latex], and so on.
A General Note: Rational Exponents
A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:
[latex]{a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}={\left(\sqrt[n]{a}\right)}^{m}[/latex]
Example: Evaluating a Number Raised to a Rational Exponent
Evaluate [latex]{8}^{\frac{2}{3}}[/latex].
Solution
Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite [latex]{8}^{\frac{2}{3}}[/latex] as [latex]{\left({8}^{\frac{1}{3}}\right)}^{2}[/latex].
[latex]\begin{array}{l}{\left({8}^{\frac{1}{3}}\right)}^{2}\hfill&={\left(2\right)}^{2}\hfill \\ \hfill&=4\hfill \end{array}[/latex]
Try It
Evaluate [latex]{64}^{-\frac{1}{3}}[/latex].
Solution
[latex]\frac{1}{4}[/latex]
Example: Solve the Equation Including a Variable Raised to a Rational Exponent
Solve the equation in which a variable is raised to a rational exponent: [latex]{x}^{\frac{5}{4}}=32[/latex].
Solution
The way to remove the exponent on
x is by raising both sides of the equation to a power that is the reciprocal of [latex]\frac{5}{4}[/latex], which is [latex]\frac{4}{5}[/latex].
[latex]\begin{array}{ll}{x}^{\frac{5}{4}}\hfill&=32\hfill & \hfill \\ {\left({x}^{\frac{5}{4}}\right)}^{\frac{4}{5}}\hfill&={\left(32\right)}^{\frac{4}{5}}\hfill & \hfill \\ x\hfill&={\left(2\right)}^{4}\hfill & \text{The fifth root of 32 is 2}.\hfill \\ \hfill&=16\hfill & \hfill \end{array}[/latex]
Try It
Solve the equation [latex]{x}^{\frac{3}{2}}=125[/latex].
Solution
[latex]25[/latex]
Example: Solving an Equation Involving Rational Exponents and Factoring
Solve [latex]3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}[/latex].
Solution
This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.
[latex]\begin{array}{l}3{x}^{\frac{3}{4}}-\left({x}^{\frac{1}{2}}\right)\hfill&={x}^{\frac{1}{2}}-\left({x}^{\frac{1}{2}}\right)\hfill \\ 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}}\hfill&=0\hfill \end{array}[/latex]
Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite [latex]{x}^{\frac{1}{2}}[/latex] as [latex]{x}^{\frac{2}{4}}[/latex]. Then, factor out [latex]{x}^{\frac{2}{4}}[/latex] from both terms on the left.
[latex]\begin{array}{l}3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}}\hfill&=0\hfill \\ {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)\hfill&=0\hfill \end{array}[/latex]
Where did [latex]{x}^{\frac{1}{4}}[/latex] come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply [latex]{x}^{\frac{2}{4}}[/latex] back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to [latex]\frac{2}{4}[/latex] equals [latex]\frac{3}{4}[/latex]. Thus, the exponent on x in the parentheses is [latex]\frac{1}{4}[/latex].
Let us continue. Now we have two factors and can use the zero factor theorem.
[latex]\begin{array}{ll}{x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)\hfill&=0\hfill & \hfill \\ {x}^{\frac{2}{4}}\hfill&=0\hfill & \hfill \\ x=0\hfill & \hfill \\ 3{x}^{\frac{1}{4}}-1\hfill&=0\hfill & \hfill \\ 3{x}^{\frac{1}{4}}\hfill&=1\hfill & \hfill \\ {x}^{\frac{1}{4}}\hfill&=\frac{1}{3}\hfill & \text{Divide both sides by 3}.\hfill \\ {\left({x}^{\frac{1}{4}}\right)}^{4}\hfill&={\left(\frac{1}{3}\right)}^{4}\hfill & \text{Raise both sides to the reciprocal of }\frac{1}{4}.\hfill \\ x\hfill&=\frac{1}{81}\hfill & \hfill \end{array}[/latex]
The two solutions are [latex]x=0[/latex], [latex]x=\frac{1}{81}[/latex].
Try It
Solve: [latex]{\left(x+5\right)}^{\frac{3}{2}}=8[/latex].
Solution
[latex]\{-1\}[/latex]
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