Unit 8 · Lesson 5 of 6
Solving radical equations
You'll be able to: solve equations with a square root or cube root in them, and throw out answers that don't really work.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.
Rationalizing the denominator of \(\dfrac{12}{\sqrt{3}}\) gives \(k\sqrt{3}\). What number is \(k\)?
Multiply top and bottom by \(\sqrt{3}\): \(\dfrac{12\sqrt{3}}{3}\). Reduce \(12 \div 3 = 4\): \(4\sqrt{3}\). Answer: \(k = 4\).
Solve \(x^2 - 7x + 10 = 0\).
Two numbers that multiply to 10 and add to \(-7\): \(-2\) and \(-5\). So \((x-2)(x-5) = 0\).
Set each factor to 0: \(x = 2\) or \(x = 5\).
Step 2 of 5The idea
A radical equation has the variable inside a radical, like \(\sqrt{x+1} = 3\).
Squaring undoes a square root: \((\sqrt{A})^2 = A\). Cubing undoes a cube root. But first get the radical alone on one side.
Squaring both sides can create an answer that does not work in the original equation. That is an extraneous solution. So the check is not optional here. It is a step.
1. Isolate the radical (get it alone on one side).
2. Square both sides (cube both sides for a cube root).
3. Solve what's left. If it is a quadratic, move everything to one side and factor.
4. Check every answer in the original equation. Cross out any that fail.
Why can squaring add answers? \(\sqrt{x} = -3\) has no solution, because a square root is never negative. But squaring gives \(x = 9\). Check: \(\sqrt{9} = 3\), not \(-3\). So 9 is extraneous.
Why this shows up on the placement test: radical equations are on the MATH 1010 skills list, and a wrong answer choice is often the extraneous one.
Step 3 of 5Worked examples
Example 1: \(\sqrt{2x+3} - 1 = 4\)
- Isolate: add 1 to both sides. \(\sqrt{2x+3} = 5\).
- Square both sides: \(2x + 3 = 25\).
- Solve: \(2x = 22\), so \(x = 11\).
- Check in the original: \(\sqrt{2(11)+3} - 1 = \sqrt{25} - 1 = 5 - 1 = 4\). It works, so \(x = 11\).
Example 2: \(\sqrt{x+7} = x - 5\)
- The radical is already alone. Square both sides: \(x + 7 = (x-5)^2 = x^2 - 10x + 25\).
- Move everything to one side: \(0 = x^2 - 11x + 18\).
- Factor: \((x - 2)(x - 9) = 0\), so \(x = 2\) or \(x = 9\).
- Check \(x = 2\): left \(\sqrt{9} = 3\), right \(2 - 5 = -3\). Not equal, so 2 is extraneous.
- Check \(x = 9\): left \(\sqrt{16} = 4\), right \(9 - 5 = 4\). Equal. The only solution is \(x = 9\).
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. List every solution, separated by commas. If no number works, type none.
Solve \(\sqrt{x} = 7\).
Square both sides: \(x = 49\).
Check: \(\sqrt{49} = 7\). Answer: \(x = 49\).
Solve \(\sqrt{x-4} = 3\).
Square both sides: \(x - 4 = 9\), so \(x = 13\).
Check: \(\sqrt{13 - 4} = \sqrt{9} = 3\). Answer: \(x = 13\).
Solve \(\sqrt{3x+1} + 2 = 6\).
Isolate: \(\sqrt{3x+1} = 4\). Square: \(3x + 1 = 16\), so \(3x = 15\) and \(x = 5\).
Check: \(\sqrt{16} + 2 = 4 + 2 = 6\). Answer: \(x = 5\).
Solve \(\sqrt[3]{2x-1} = 3\).
Cube both sides: \(2x - 1 = 27\), so \(2x = 28\) and \(x = 14\).
Check: \(\sqrt[3]{27} = 3\). Answer: \(x = 14\).
Solve \(\sqrt{x+5} + 3 = 1\).
Isolate: \(\sqrt{x+5} = -2\). A square root is never negative, so there is no solution.
(If you squared anyway, you'd get \(x + 5 = 4\), \(x = -1\). Check: \(\sqrt{4} + 3 = 5\), not 1. Extraneous.) Answer: none.
Solve \(\sqrt{x+10} = x - 2\).
Square: \(x + 10 = x^2 - 4x + 4\). One side: \(0 = x^2 - 5x - 6 = (x-6)(x+1)\), so \(x = 6\) or \(x = -1\).
Check 6: \(\sqrt{16} = 4\) and \(6 - 2 = 4\). Works. Check \(-1\): \(\sqrt{9} = 3\) but \(-1 - 2 = -3\). Extraneous.
Answer: \(x = 6\).
Solve \(\sqrt{5x-6} = x\).
Square: \(5x - 6 = x^2\). One side: \(0 = x^2 - 5x + 6 = (x-2)(x-3)\), so \(x = 2\) or \(x = 3\).
Check 2: \(\sqrt{4} = 2\). Works. Check 3: \(\sqrt{9} = 3\). Works. Sometimes both answers survive.
Answer: \(x = 2\) or \(x = 3\).
An object dropped from rest is falling at about \(v = 8\sqrt{d}\) feet per second after it has fallen \(d\) feet. How many feet has it fallen when its speed is 48 feet per second?
Set \(8\sqrt{d} = 48\). Isolate: divide by 8, \(\sqrt{d} = 6\).
Square: \(d = 36\). Check: \(8\sqrt{36} = 8 \cdot 6 = 48\). Answer: 36 feet.
Step 5 of 5Wrap-up
You can now solve a radical equation in four steps (isolate, square or cube, solve, check) and cross out extraneous solutions.
Read or watch next: OpenStax Intermediate Algebra 2e, 8.6 Solve Radical Equations (free textbook, with more practice at the end of the section, including equations with two radicals).
Also good: Khan Academy: Algebra 2, Square-root equations (short videos). Quick rules: Radicals and complex numbers cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why does squaring both sides sometimes give an answer that doesn't work?" It knows this course and can explain things another way.