Solving Radical Equations
Get the square root by itself on one side, square both sides, and solve. Then check every answer in the original equation, because squaring can produce answers that do not work. A square root is never negative, so any answer that would make it negative has to go.
And if the square root ends up equal to a negative number before you square, you can stop right there. There is no solution.
Radical equation questions ask you to solve, to find a constant from a given solution, to count the solutions, and to recognize an extraneous one. Cube roots and fractional exponents come up as well.
What you will be asked
| Step | What to do |
|---|---|
| 1. Isolate the radical | Move everything else to the other side before squaring. Squaring $\sqrt{x} + 3$ as a whole creates a messy cross term. |
| 2. Square both sides | Square the whole side: $(x - 5)^2 = x^2 - 10x + 25$, not $x^2 - 25$. |
| 3. Solve | The result is usually linear or quadratic. |
| 4. Check | Substitute each answer into the original equation. Reject any that make the two sides differ. |
| Cube roots | Cube both sides. Cube roots can be negative, so cubing never creates extraneous answers. |
| Fractional exponents | $x^{\frac{1}{2}} = \sqrt{x}$ and $x^{\frac{3}{2}} = \left(\sqrt{x}\right)^3$. To undo an exponent of $\dfrac{3}{2}$, raise both sides to $\dfrac{2}{3}$. |
Mistakes to watch for
- Squaring before isolating the radical. Squaring $\sqrt{x} + 3 = 7$ as it stands gives a harder equation. Subtract $3$ first.
- Squaring a binomial term by term. $(x - 5)^2$ is $x^2 - 10x + 25$. The middle term is easy to forget.
- Skipping the check. The extraneous answer is very often one of the choices.
- Allowing a square root to be negative. $\sqrt{x} = -4$ has no solution. Only cube roots and other odd roots can be negative.
- Mishandling fractional exponents. If $x^{\frac{3}{2}} = 27$, then $x = 27^{\frac{2}{3}} = 9$, not $27 \cdot \dfrac{2}{3}$.
Worked examples
Example 1: finding a constant
The equation $\sqrt{3x + k} - 2 = 5$ has the solution $x = 6$. What is $k$?
- Substitute $x = 6$: $\sqrt{18 + k} - 2 = 5$.
- Isolate the radical: $\sqrt{18 + k} = 7$.
- Square: $18 + k = 49$, so $k = 31$.
Example 2: an extraneous solution
Solve $\sqrt{x + 7} = x - 5$.
- Square both sides: $x + 7 = x^2 - 10x + 25$.
- Rearrange: $x^2 - 11x + 18 = 0$, so $(x - 2)(x - 9) = 0$ and $x = 2$ or $x = 9$.
- Check $x = 2$: $\sqrt{9} = 3$, but $2 - 5 = -3$. It does not work.
- Check $x = 9$: $\sqrt{16} = 4$ and $9 - 5 = 4$. The only solution is $x = 9$.
Example 3: no solution at a glance
Solve $\sqrt{2x - 1} + 3 = 0$.
- Isolate the radical: $\sqrt{2x - 1} = -3$.
- A square root is never negative, so no value of $x$ works.
- If you square anyway, you get $x = 5$, and the check shows $\sqrt{9} + 3 = 6$, not $0$.
Example 4: a cube root
Solve $\sqrt[3]{x - 4} = 3$.
- Cube both sides: $x - 4 = 27$.
- $x = 31$. Check: $\sqrt[3]{27} = 3$.
- A cube root can equal a negative number, so $\sqrt[3]{x - 4} = -3$ would have the solution $x = -23$.
Practice questions
Solve each one and check your answers in the original equation before you open the solution.
Easy
What value of $x$ satisfies $\sqrt{x} + 4 = 9$?
- A$5$
- B$13$
- C$25$
- D$169$
Show solution
Answer: C. Isolate the radical: $\sqrt{x} = 5$. Square: $x = 25$. Choice D squares $13$ after adding $4$ instead of subtracting.
EasyStudent-produced response
What is the solution to $\sqrt{2x + 1} = 5$?
Enter your answer.
Show solution
Answer: $12$. Square both sides: $2x + 1 = 25$, so $2x = 24$ and $x = 12$. Check: $\sqrt{25} = 5$.
MediumStudent-produced response
The equation $\sqrt{3x + k} - 2 = 5$ has the solution $x = 6$, where $k$ is a constant. What is the value of $k$?
Enter your answer.
Show solution
Answer: $31$. Substitute $x = 6$: $\sqrt{18 + k} = 7$. Square: $18 + k = 49$, so $k = 31$.
Medium
The equation $\sqrt{kx - 3} = x - 1$ has the solution $x = 2$, where $k$ is a constant. What is the value of $k$?
- A$1$
- B$3$
- C$4$
- D$2$
Show solution
Answer: D. Substitute $x = 2$: $\sqrt{2k - 3} = 1$. Square: $2k - 3 = 1$, so $k = 2$.
Medium
What are all the solutions to $\sqrt{x + 7} = x - 5$?
- A$x = 2$ only
- B$x = 9$ only
- C$x = 2$ and $x = 9$
- DNo solution
Show solution
Answer: B. Square: $x + 7 = x^2 - 10x + 25$, so $x^2 - 11x + 18 = 0$ and $x = 2$ or $x = 9$. At $x = 2$ the right side is $-3$ while the square root is $3$, so $x = 2$ is extraneous. Only $x = 9$ works.
HardStudent-produced response
The equation $\sqrt{x^2 - kx} = x - 3$ has the solution $x = 9$, where $k$ is a constant. What is the value of $k$?
Enter your answer.
Show solution
Answer: $5$. Substitute $x = 9$: $\sqrt{81 - 9k} = 6$. Square: $81 - 9k = 36$, so $9k = 45$ and $k = 5$.
Hard
How many real solutions does the equation $\sqrt{x - 3} + 5 = 2$ have?
- AExactly one
- BExactly two
- CZero
- DInfinitely many
Show solution
Answer: C. Isolate the radical: $\sqrt{x - 3} = -3$. A square root is never negative, so there are no real solutions. Squaring would give $x = 12$, but $\sqrt{9} + 5 = 8$, not $2$.
Hard
What is the solution to $\sqrt[3]{2x + 5} = -3$?
- A$-16$
- B$-4$
- C$11$
- DThere is no real solution.
Show solution
Answer: A. Cube both sides: $2x + 5 = -27$, so $2x = -32$ and $x = -16$. A cube root can be negative, so this answer is valid. Choice D would be right for a square root, not a cube root.
HardStudent-produced response
What is the positive solution to $x = \sqrt{x + 12}$?
Enter your answer.
Show solution
Answer: $4$. Square: $x^2 = x + 12$, so $x^2 - x - 12 = 0$ and $(x - 4)(x + 3) = 0$. The value $x = -3$ fails because the left side would be negative while the square root is $3$. The solution is $x = 4$: $\sqrt{16} = 4$.
Hard
If $x^{\frac{3}{2}} = 27$ and $x > 0$, what is the value of $x$?
- A$3$
- B$18$
- C$9$
- D$81$
Show solution
Answer: C. Raise both sides to the power $\dfrac{2}{3}$: $x = 27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$. Check: $9^{\frac{3}{2}} = \left(\sqrt{9}\right)^3 = 27$. Choice B multiplies $27$ by $\dfrac{2}{3}$.
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