Equations With Radicals and Rational Exponents

Learning Outcomes

Solve a radical equation, identify extraneous solution.
Solve an equation with rational exponents.

Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as

\begin{matrix} \sqrt{3 x + 18} = x \sqrt{x + 3} = x - 3 \sqrt{x + 5} - \sqrt{x - 3} = 2 \end{matrix}

$\begin{array}{ccc} \sqrt{3x+18}=x & \\ \sqrt{x+3}=x-3 & \\ \sqrt{x+5}-\sqrt{x - 3}=2\end{array}$

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. 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 resulting equation.
If a radical term still remains, repeat steps 1–2.
Check solutions by substituting them into the original equation.

Example: Solving an Equation with One Radical

Solve $\sqrt{15 - 2x}=x$ .

Show Solution

The radical is already isolated on the left side of the equal sign, so proceed to square both sides.

\begin{matrix} \sqrt{15 - 2 x} = x {(\sqrt{15 - 2 x})}^{2} = {(x)}^{2} 15 - 2 x = x^{2} \end{matrix}

$\begin{array}{lll}\sqrt{15 - 2x}=x & \\ {\left(\sqrt{15 - 2x}\right)}^{2}={\left(x\right)}^{2} & \\ 15 - 2x={x}^{2}\end{array}$

We see that the remaining equation is a quadratic. Set it equal to zero and solve.

\begin{matrix} 0 = x^{2} + 2 x - 15 0 = (x + 5) (x - 3) x = - 5 x = 3 \end{matrix}

$\begin{array}{llll}0={x}^{2}+2x - 15 & \\ 0=\left(x+5\right)\left(x - 3\right) & \\ x=-5 & \\ x=3 \end{array}$

The proposed solutions are $x=-5$ and $x=3$ . Let us check each solution back in the original equation. First, check $x=-5$ .

\begin{matrix} \sqrt{15 - 2 x} = x \sqrt{15 - 2 (- 5)} = - 5 \sqrt{25} = - 5 5 \neq - 5 \end{matrix}

$\begin{array}{llll}\sqrt{15 - 2x}=x & \\ \sqrt{15 - 2\left(-5\right)}=-5 & \\ \sqrt{25}=-5 & \\ 5\ne -5\end{array}$

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 $x=3$ .

\begin{matrix} \sqrt{15 - 2 x} = x \sqrt{15 - 2 (3)} = 3 \sqrt{9} = 3 3 = 3 \end{matrix}

$\begin{array}{llll}\sqrt{15 - 2x}=x & \\ \sqrt{15 - 2\left(3\right)}=3 & \\ \sqrt{9}=3 & \\ 3=3\end{array}$

The solution is $x=3$ .

Try It

Solve the radical equation: $\sqrt{x+3}=3x - 1$

Show Solution

$x=1$ ; extraneous solution $x=-\frac{2}{9}$

Example: Solving a Radical Equation Containing Two Radicals

Solve $\sqrt{2x+3}+\sqrt{x - 2}=4$ .

Show Solution

As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.

\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} = 4 \sqrt{2 x + 3} = 4 - \sqrt{x - 2} & Subtract \sqrt{x - 2} from both sides . {(\sqrt{2 x + 3})}^{2} = {(4 - \sqrt{x - 2})}^{2} & Square both sides . \end{matrix}

$\begin{array}{llllll}\sqrt{2x+3}+\sqrt{x - 2}=4\hfill & \hfill & \\ \sqrt{2x+3}=4-\sqrt{x - 2}\hfill & \text{Subtract }\sqrt{x - 2}\text{ from both sides}.\hfill & \\ {\left(\sqrt{2x+3}\right)}^{2}={\left(4-\sqrt{x - 2}\right)}^{2}\hfill & \text{Square both sides}.\hfill \end{array}$

Use the perfect square formula to expand the right side: ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$ .

\begin{matrix} 2 x + 3 = {(4)}^{2} - 2 (4) \sqrt{x - 2} + {(\sqrt{x - 2})}^{2} 2 x + 3 = 16 - 8 \sqrt{x - 2} + (x - 2) 2 x + 3 = 14 + x - 8 \sqrt{x - 2} & Combine like terms . x - 11 = - 8 \sqrt{x - 2} & Isolate the second radical . {(x - 11)}^{2} = {(- 8 \sqrt{x - 2})}^{2} & Square both sides . x^{2} - 22 x + 121 = 64 (x - 2) \end{matrix}

$\begin{array}{lllllllllll}2x+3={\left(4\right)}^{2}-2\left(4\right)\sqrt{x - 2}+{\left(\sqrt{x - 2}\right)}^{2}\hfill & \hfill & \\ 2x+3=16 - 8\sqrt{x - 2}+\left(x - 2\right)\hfill & \hfill & \\ 2x+3=14+x - 8\sqrt{x - 2}\hfill & \text{Combine like terms}.\hfill & \\ x - 11=-8\sqrt{x - 2}\hfill & \text{Isolate the second radical}.\hfill & \\ {\left(x - 11\right)}^{2}={\left(-8\sqrt{x - 2}\right)}^{2}\hfill & \text{Square both sides}.\hfill & \\ {x}^{2}-22x+121=64\left(x - 2\right)\hfill & \hfill \end{array}$

Now that both radicals have been eliminated, set the quadratic equal to zero and solve.

\begin{matrix} x^{2} - 22 x + 121 = 64 x - 128 x^{2} - 86 x + 249 = 0 (x - 3) (x - 83) = 0 & Factor and solve . x = 3 x = 83 \end{matrix}

$\begin{array}{llllllllll}{x}^{2}-22x+121=64x - 128\hfill & \hfill & \\ {x}^{2}-86x+249=0\hfill & \hfill & \\ \left(x - 3\right)\left(x - 83\right)=0\hfill & \text{Factor and solve}.\hfill & \\ x=3\hfill & \hfill & \\ x=83\hfill & \hfill \end{array}$

The proposed solutions are $x=3$ and $x=83$ . Check each solution in the original equation.

\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} = 4 \sqrt{2 x + 3} = 4 - \sqrt{x - 2} \sqrt{2 (3) + 3} = 4 - \sqrt{(3) - 2} \sqrt{9} = 4 - \sqrt{1} 3 = 3 \end{matrix}

$\begin{array}{lllll}\sqrt{2x+3}+\sqrt{x - 2}=4\hfill & \\ \sqrt{2x+3}=4-\sqrt{x - 2}\hfill & \\ \sqrt{2\left(3\right)+3}=4-\sqrt{\left(3\right)-2}\hfill & \\ \sqrt{9}=4-\sqrt{1}\hfill \\ 3=3\hfill \end{array}$

One solution is $x=3$ .

Check $x=83$ .

\begin{matrix} \sqrt{2 x + 3} + \sqrt{x - 2} = 4 \sqrt{2 x + 3} = 4 - \sqrt{x - 2} \sqrt{2 (83) + 3} = 4 - \sqrt{(83 - 2)} \sqrt{169} = 4 - \sqrt{81} 13 \neq - 5 \end{matrix}

$\begin{array}{lllll}\sqrt{2x+3}+\sqrt{x - 2}=4\hfill & \\ \sqrt{2x+3}=4-\sqrt{x - 2}\hfill & \\ \sqrt{2\left(83\right)+3}=4-\sqrt{\left(83 - 2\right)}\hfill & \\ \sqrt{169}=4-\sqrt{81}\hfill & \\ 13\ne -5\hfill \end{array}$

The only solution is $x=3$ . We see that $x=83$ is an extraneous solution.

Try It

Solve the equation with two radicals: $\sqrt{3x+7}+\sqrt{x+2}=1$ .

Show Solution

$x=-2$ ; extraneous solution $x=-1$

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, ${16}^{\frac{1}{2}}$ is another way of writing $\sqrt{16}$ ; ${8}^{\frac{1}{3}}$ is another way of writing $\text{ }\sqrt[3]{8}$ . 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, $\frac{2}{3}\left(\frac{3}{2}\right)=1$ , $3\left(\frac{1}{3}\right)=1$ , 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:

a^{\frac{m}{n}} = {(a^{\frac{1}{n}})}^{m} = {(a^{m})}^{\frac{1}{n}} = \sqrt[n]{a^{m}} = {(\sqrt[n]{a})}^{m}

${a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}={\left(\sqrt[n]{a}\right)}^{m}$

Example: Evaluating a Number Raised to a Rational Exponent

Evaluate ${8}^{\frac{2}{3}}$ .

Show 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 ${8}^{\frac{2}{3}}$ as ${\left({8}^{\frac{1}{3}}\right)}^{2}$ .

\begin{matrix} {(8^{\frac{1}{3}})}^{2} & = {(2)}^{2} = 4 \end{matrix}

$\begin{array}{l}{\left({8}^{\frac{1}{3}}\right)}^{2}\hfill&={\left(2\right)}^{2}\hfill \\ \hfill&=4\hfill \end{array}$

Try It

Evaluate ${64}^{-\frac{1}{3}}$ .

Show Solution

$\frac{1}{4}$

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: ${x}^{\frac{5}{4}}=32$ .

Show 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 $\frac{5}{4}$ , which is $\frac{4}{5}$ .

\begin{matrix} x^{\frac{5}{4}} = 32 {(x^{\frac{5}{4}})}^{\frac{4}{5}} = {(32)}^{\frac{4}{5}} x = {(2)}^{4} & The fifth root of 32 is 2 . x = 16 \end{matrix}

$\begin{array}{llllllll}{x}^{\frac{5}{4}}=32\hfill & \hfill & \\ {\left({x}^{\frac{5}{4}}\right)}^{\frac{4}{5}}={\left(32\right)}^{\frac{4}{5}}\hfill & \hfill & \\ x={\left(2\right)}^{4}\hfill & \text{The fifth root of 32 is 2}.\hfill & \\ x=16\hfill & \hfill \end{array}$

Try It

Solve the equation ${x}^{\frac{3}{2}}=125$ .

Show Solution

$25$

Example: Solving an Equation Involving Rational Exponents and Factoring

Solve $3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}$ .

Show 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.

\begin{matrix} 3 x^{\frac{3}{4}} - (x^{\frac{1}{2}}) = x^{\frac{1}{2}} - (x^{\frac{1}{2}}) 3 x^{\frac{3}{4}} - x^{\frac{1}{2}} = 0 \end{matrix}

$\begin{array}{ll}3{x}^{\frac{3}{4}}-\left({x}^{\frac{1}{2}}\right)={x}^{\frac{1}{2}}-\left({x}^{\frac{1}{2}}\right)\hfill & \\ 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}}=0\hfill \end{array}$

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 ${x}^{\frac{1}{2}}$ as ${x}^{\frac{2}{4}}$ . Then, factor out ${x}^{\frac{2}{4}}$ from both terms on the left.

\begin{matrix} 3 x^{\frac{3}{4}} - x^{\frac{2}{4}} = 0 x^{\frac{2}{4}} (3 x^{\frac{1}{4}} - 1) = 0 \end{matrix}

$\begin{array}{ll}3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}}=0\hfill & \\ {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)=0\hfill \end{array}$

Where did ${x}^{\frac{1}{4}}$ come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply ${x}^{\frac{2}{4}}$ 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 $\frac{2}{4}$ equals $\frac{3}{4}$ . Thus, the exponent on x in the parentheses is $\frac{1}{4}$ .

Let us continue. Now we have two factors and can use the zero factor theorem.

\begin{matrix} x^{\frac{2}{4}} (3 x^{\frac{1}{4}} - 1) = 0 x^{\frac{2}{4}} = 0 x = 0 3 x^{\frac{1}{4}} - 1 = 0 3 x^{\frac{1}{4}} = 1 x^{\frac{1}{4}} = \frac{1}{3} & Divide both sides by 3 . {(x^{\frac{1}{4}})}^{4} = {(\frac{1}{3})}^{4} & Raise both sides to the reciprocal of \frac{1}{4} . x = \frac{1}{81} \end{matrix}

$\begin{array}{llllllllllllll}{x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)=0\hfill & \hfill & \\ {x}^{\frac{2}{4}}=0\hfill & \hfill & \\ x=0\hfill & \hfill & \\ 3{x}^{\frac{1}{4}}-1=0\hfill & \hfill & \\ 3{x}^{\frac{1}{4}}=1\hfill & \hfill & \\ {x}^{\frac{1}{4}}=\frac{1}{3}\hfill & \text{Divide both sides by 3}.\hfill & \\ {\left({x}^{\frac{1}{4}}\right)}^{4}={\left(\frac{1}{3}\right)}^{4}\hfill & \text{Raise both sides to the reciprocal of }\frac{1}{4}.\hfill & \\ x=\frac{1}{81}\hfill & \hfill \end{array}$

The two solutions are $x=0$ , $x=\frac{1}{81}$ .

Try It

Solve: ${\left(x+5\right)}^{\frac{3}{2}}=8$ .

Show Solution

$-1$

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Module 4: Equations and Inequalities