Solve Radical Equations – Intermediate Algebra

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Solve Radical Equations

In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation.

Radical Equation

An equation in which a variable is in the radicand of a radical expression is called a radical equation.

As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.

Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.

In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index n to the nth power. This will eliminate the radical.

$\text{For}\phantom{\rule{0.2em}{0ex}}a\ge 0,\phantom{\rule{0.2em}{0ex}}{\left(\sqrt[n]{a}\right)}^{n}=a.$ How to Solve a Radical Equation

Solve: $\sqrt{5n-4}-9=0.$

Step 1 is to isolate the radical on one side of the equation. To isolate the radical add 9 to both sides. The resulting equation is square root of the quantity 5 n minus 4 in parentheses minus 9 plus 9 equals 0 plus 9. This simplifies to square root of the quantity 5 n minus 4 in parentheses equals 9.

Step 2 is to raise both sides of the equation to the power of the index. Since the index of a square root is 2, we square both sides. Remember that the square of the square root of “a” is equal to “a”. The equation that results is the square of the square root of the quantity 5 n minus 4 in parentheses equals 9 squared. This simplifies to 5 n minus 4 equals 81.

Step 3 is to solve the new equation. We get 5 n equals 85 and then n equals 17.

Step 4 is to check the answer in the original equation. Does the square root of the quantity 5 times 17 minus 4 in parentheses minus 9 equal zero? Simplifying the left side we get square root of the quantity 85 minus 4 in parentheses minus 9 and then square root of 81 minus 9 and then 9 minus 9 which does equal 0. This verifies that the solution is n equals 17.

Solve: $\sqrt{3m+2}-5=0.$

$m=\frac{23}{3}$

Solve: $\sqrt{10z+1}-2=0.$

$z=\frac{3}{10}$

Solve a radical equation with one radical.

Isolate the radical on one side of the equation.
Raise both sides of the equation to the power of the index.
Solve the new equation.
Check the answer in the original equation.

When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.

Solve: $\sqrt{9k-2}+1=0.$


To isolate the radical, subtract 1 to both sides.
Simplify.

Because the square root is equal to a negative number, the equation has no solution.

Solve: $\sqrt{2r-3}+5=0.$

$\text{no solution}$

Solve: $\sqrt{7s-3}+2=0.$

$\text{no solution}$

If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.

Binomial Squares $\begin{array}{}\\ \\ {\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}\\ {\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}\end{array}$

Don’t forget the middle term!

Solve: $\sqrt{p-1}+1=p.$


To isolate the radical, subtract 1 from both sides.
Simplify.
Square both sides of the equation.
Simplify, using the Product of Binomial Squares Pattern on the right. Then solve the new equation.
It is a quadratic equation, so get zero on one side.
Factor the right side.
Use the Zero Product Property.
Solve each equation.
Check the answers.

The solutions are $p=1,\phantom{\rule{0.5em}{0ex}}p=2.$

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

Solve: $\sqrt{y-5}+5=y.$

When the index of the radical is 3, we cube both sides to remove the radical.

${\left(\sqrt[3]{a}\right)}^{3}=a$

Solve: $\sqrt[3]{5x+1}+8=4.$

$\sqrt[3]{5x+1}+8=4\phantom{\rule{1.8em}{0ex}}$
To isolate the radical, subtract 8 from both sides.	$\sqrt[3]{5x+1}=-4\phantom{\rule{1.2em}{0ex}}$
Cube both sides of the equation.	${\left(\sqrt[3]{5x+1}\right)}^{3}={\left(-4\right)}^{3}$
Simplify.	$5x+1=-64\phantom{\rule{0.6em}{0ex}}$
Solve the equation.	$5x=-65\phantom{\rule{0.6em}{0ex}}$
$x=-13\phantom{\rule{0.6em}{0ex}}$
Check the answer.

The solution is $x=-13.\phantom{\rule{0.35em}{0ex}}$

Solve: $\sqrt[3]{4x-3}+8=5$

Solve: $\sqrt[3]{6x-10}+1=-3$

Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical. We raise each side of the equation to the power of the denominator of the rational exponent. Since ${\left({a}^{m}\right)}^{n}={a}^{m·n},$ we have for example,

${\left({x}^{\frac{1}{2}}\right)}^{2}=x,\phantom{\rule{0.5em}{0ex}}{\left({x}^{\frac{1}{3}}\right)}^{3}=x$

Remember, ${x}^{\frac{1}{2}}=\sqrt{x}$ and ${x}^{\frac{1}{3}}=\sqrt[3]{x}.$

Solve: ${\left(3x-2\right)}^{\frac{1}{4}}+3=5.$

${\left(3x-2\right)}^{\frac{1}{4}}+3=5\phantom{\rule{1.1em}{0ex}}$
To isolate the term with the rational exponent, subtract 3 from both sides.	${\left(3x-2\right)}^{\frac{1}{4}}=2\phantom{\rule{1.1em}{0ex}}$
Raise each side of the equation to the fourth power.	${\left({\left(3x-2\right)}^{\frac{1}{4}}\right)}^{4}={\left(2\right)}^{4}$
Simplify.	$3x-2=16\phantom{\rule{0.6em}{0ex}}$
Solve the equation.	$3x=18\phantom{\rule{0.6em}{0ex}}$
$x=6\phantom{\rule{1.17em}{0ex}}$
Check the answer.

The solution is $x=6.\phantom{\rule{0.95em}{0ex}}$

Solve: ${\left(9x+9\right)}^{\frac{1}{4}}-2=1.$

Solve: ${\left(4x-8\right)}^{\frac{1}{4}}+5=7.$

Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution!

Solve: $\sqrt{r+4}-r+2=0.$

$\sqrt{r+4}-r+2=0\phantom{\rule{4.8em}{0ex}}$
Isolate the radical.	$\sqrt{r+4}=r-2\phantom{\rule{3.2em}{0ex}}$
Square both sides of the equation.	${\left(\sqrt{r+4}\right)}^{2}={\left(r-2\right)}^{2}\phantom{\rule{2em}{0ex}}$
Simplify and then solve the equation	$r+4={r}^{2}-4r+4\phantom{\rule{0.7em}{0ex}}$
It is a quadratic equation, so get zero on one side.	$0={r}^{2}-5r\phantom{\rule{2.4em}{0ex}}$
Factor the right side.	$0=r\left(r-5\right)\phantom{\rule{2.1em}{0ex}}$
Use the Zero Product Property.	$0=r\phantom{\rule{1.5em}{0ex}}0=r-5$
Solve the equation.	$r=0\phantom{\rule{1em}{0ex}}r=5\phantom{\rule{1.5em}{0ex}}$
Check your answer.
	The solution is r = 5.
is an extraneous solution.

Solve: $\sqrt{m+9}-m+3=0.$

Solve: $\sqrt{n+1}-n+1=0.$

When there is a coefficient in front of the radical, we must raise it to the power of the index, too.

Solve: $\text{3}\phantom{\rule{0.2em}{0ex}}\sqrt{3x-5}-8=4.$

$\text{3}\phantom{\rule{0.2em}{0ex}}\sqrt{3x-5}-8=4\phantom{\rule{1.05em}{0ex}}$
Isolate the radical term.	$3\sqrt{3x-5}=12\phantom{\rule{0.6em}{0ex}}$
Isolate the radical by dividing both sides by 3.	$\sqrt{3x-5}=4\phantom{\rule{1.05em}{0ex}}$
Square both sides of the equation.	${\left(\sqrt{3x-5}\right)}^{2}={\left(4\right)}^{2}$
Simplify, then solve the new equation.	$3x-5=16\phantom{\rule{0.6em}{0ex}}$
$3x=21\phantom{\rule{0.6em}{0ex}}$
Solve the equation.	$x=7\phantom{\rule{1.05em}{0ex}}$
Check the answer.

The solution is $x=7.\phantom{\rule{1.05em}{0ex}}$