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11.3: Radical Equations

Difficulty Level: At Grade Created by: CK-12
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Solving radical equations is no different from solving linear or quadratic equations. Before you can begin to solve a radical equation, you must know how to cancel the radical. To do that, you must know its inverse.

Original Operation Inverse Operation
Cube Root Cubing (to the third power)
Square Root Squaring (to the second power)
Fourth Root Fourth power
\begin{align*}n\end{align*}nth” Root \begin{align*}n\end{align*}nth” power

To solve a radical equation, you apply the solving equation steps you learned in previous chapters, including the inverse operations for roots.

Example 1: Solve \begin{align*}\sqrt{2x-1}=5\end{align*}2x1=5.

Solution: The first operation that must be removed is the square root. Square both sides.

\begin{align*}\left ( \sqrt{2x-1} \right )^2&=5^2\\ 2x-1&=25\\ 2x&=26\\ x&=13\end{align*}(2x1)22x12xx=52=25=26=13

Remember to check your answer by substituting it into the original problem to see if it makes sense.

Example: Solve \begin{align*}\sqrt{x+15}=\sqrt{3x-3}\end{align*}x+15=3x3.

Solution: Begin by canceling the square roots by squaring both sides.

\begin{align*}\left ( \sqrt{x+15} \right )^2&=\left ( \sqrt{3x-3} \right )^2\\ x+15&=3x-3\\ \text{Isolate the} \ x-\text{variable}: \qquad \qquad 18 & = 2x\\ x&=9\end{align*}(x+15)2x+15Isolate the xvariable:18x=(3x3)2=3x3=2x=9

Check the solution: \begin{align*}\sqrt{9+15}=\sqrt{3(9)-3} \rightarrow \sqrt{24}=\sqrt{24}\end{align*}. The solution checks.

Extraneous Solutions

Not every solution of a radical equation will check in the original problem. This is called an extraneous solution. This means you can find a solution using algebra, but it will not work when checked. This is because of the rule in Lesson 11.2.

Even roots of negative numbers are undefined.

Example: Solve \begin{align*}\sqrt{x-3}-\sqrt{x}=1\end{align*}.


\begin{align*}\text{Isolate one of the radical expressions}. \qquad \ \sqrt{x-3}&=\sqrt{x}+1\\ \text{Square both sides}. \quad \left ( \sqrt{x-3} \right )^2 & = \left ( \sqrt{x}+1 \right )^2\\ \text{Remove parentheses}. \qquad \quad \ x-3&=\left ( \sqrt{x} \right )^2 + 2\sqrt{x}+1\\ \text{Simplify}. \qquad \quad \ x-3&=x+2\sqrt{x}+1\\ \text{Now isolate the remaining radical}. \qquad \qquad -4&=2\sqrt{x}\\ \text{Divide all terms by} \ 2. \qquad \qquad -2 &= \sqrt{x}\\ \text{Square both sides}. \qquad \qquad \quad \ x&=4\end{align*}

Check: \begin{align*}\sqrt{4-3} - \sqrt{4} = \sqrt{1}-2=1-2=-1\end{align*}. The solution does not check out. The equation has no real solutions. Therefore, \begin{align*}x=4\end{align*} is an extraneous solution.

Radical Equations in Real Life

Example: A sphere has a volume of \begin{align*}456 \ cm^3\end{align*}. If the radius of the sphere is increased by 2 cm, what is the new volume of the sphere?


  1. Define variables. Let \begin{align*}R=\end{align*} the radius of the sphere.
  2. Find an equation. The volume of a sphere is given by the formula: \begin{align*}V=\frac{4}{3}\pi r^3\end{align*}.

By substituting 456 for the volume variable, the equation becomes \begin{align*}456=\frac{4}{3} \pi r^3\end{align*}

\begin{align*}\text{Multiply by} \ 3. \qquad \quad 1368 &= 4\pi r^3\\ \text{Divide by} \ 4\pi. \qquad 108.92&=r^3\\ \text{Take the cube root of each side.} \qquad \qquad \ r&=\sqrt[3]{108.92} \Rightarrow r = 4.776 \ cm\\ \text{The new radius is 2 centimeters more.} \qquad \qquad \ r &= 6.776 \ cm\\ \text{The new volume is}: \qquad \qquad V &= \frac{4}{3}\pi (6.776)^3 = 1302.5 \ cm^3\end{align*}

Check by substituting the values of the radii into the volume formula.

\begin{align*}V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (4.776)^3=456 \ cm^3\end{align*}. The solution checks out.

Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set.  However, the practice exercise is the same in both.

CK-12 Basic Algebra: Extraneous Solutions to Radical Equations (11:10)

CK-12 Basic Algebra: Radical Equation Examples (5:16)

CK-12 Basic Algebra: More Involved Radical Equation Example (11:54)

In 1-16, find the solution to each of the following radical equations. Identify extraneous solutions.

  1. \begin{align*}\sqrt{x+2}-2=0\end{align*}
  2. \begin{align*}\sqrt{3x-1}=5\end{align*}
  3. \begin{align*}2\sqrt{4-3x}+3=0\end{align*}
  4. \begin{align*}\sqrt[3]{x-3}=1\end{align*}
  5. \begin{align*}\sqrt[4]{x^2-9}=2\end{align*}
  6. \begin{align*}\sqrt[3]{-2-5x}+3=0\end{align*}
  7. \begin{align*}\sqrt{x}=x-6\end{align*}
  8. \begin{align*}\sqrt{x^2-5x}-6=0\end{align*}
  9. \begin{align*}\sqrt{(x+1)(x-3)}=x\end{align*}
  10. \begin{align*}\sqrt{x+6}=x+4\end{align*}
  11. \begin{align*}\sqrt{x}=\sqrt{x-9}+1\end{align*}
  12. \begin{align*}\sqrt{3x+4}=-6\end{align*}
  13. \begin{align*}\sqrt{10-5x}+\sqrt{1-x}=7\end{align*}
  14. \begin{align*}\sqrt{2x-2}-2\sqrt{x}+2=0\end{align*}
  15. \begin{align*}\sqrt{2x+5}-3\sqrt{2x-3}=\sqrt{2-x}\end{align*}
  16. \begin{align*}3\sqrt{x}-9=\sqrt{2x-14}\end{align*}
  17. The area of a triangle is \begin{align*}24 \ in^2\end{align*} and the height of the triangle is twice as long and the base. What are the base and the height of the triangle?
  18. The volume of a square pyramid is given by the formula \begin{align*}V=\frac{A(h)}{3}\end{align*}, where \begin{align*}A=\end{align*} area of the base and \begin{align*}h=\end{align*} height of the pyramid. The volume of a square pyramid is 1,600 cubic meters. If its height is 10 meters, find the area of its base.
  19. The volume of a cylinder is \begin{align*}245 \ cm^3\end{align*} and the height of the cylinder is one-third the diameter of the cylinder's base. The diameter of the cylinder is kept the same, but the height of the cylinder is increased by two centimeters. What is the volume of the new cylinder? \begin{align*}(\text{Volume} = \pi r^2 \cdot h)\end{align*}
  20. The height of a golf ball as it travels through the air is given by the equation \begin{align*}h=-16t^2+256\end{align*}. Find the time when the ball is at a height of 120 feet.

Mixed Review

  1. Joy sells two types of yarn: wool and synthetic. Wool is $12 per skein and synthetic is $9 per skein. If Joy sold 16 skeins of synthetic and collected a total of 432, how many skeins of wool did she sell?
  2. Solve \begin{align*}16 \ge |x-4|\end{align*}.
  3. Graph the solution \begin{align*}\begin{cases}y \le 2x-4\\ y>-\frac{1}{4} x+6\end{cases}\end{align*}.
  4. You randomly point to a day in the month of February 2011. What is the probability your finger lands on a Monday?
  5. Carbon-14 has a half life of 5,730 years. Your dog dug a bone from your yard. It had 93% carbon-14 remaining. How old is the bone?
  6. What is true about solutions to inconsistent systems?

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