Radical Equations

A radical equation has the variable inside a square root (or another root), like $\sqrt{2x - 1} = 5$. You solve it by undoing the root: square both sides. That step is powerful, but it can sneak in fake answers called extraneous solutions, so checking isn’t optional here. These equations show up in physics formulas (pendulums, falling objects) and regularly on the SAT.

Key ideas

The method

1. Isolate the radical on one side of the equation.
2. Square both sides to remove the square root.
3. Solve the equation that’s left. It’s often linear or quadratic (see solving quadratics by factoring).
4. Check every answer in the original equation. Throw out any that don’t work.

Isolate first. If you square $3 + \sqrt{x} = 8$ as it stands, you get $9 + 6\sqrt{x} + x = 64$, and the square root is still there. Subtract $3$ first instead: $\sqrt{x} = 5$, so $x = 25$.

Why extraneous solutions appear

Squaring loses information about signs. The equation $x = 3$ has one solution, but after squaring, $x^2 = 9$ has two: $3$ and $-3$. The new solution $-3$ is extraneous: it solves the squared equation but not the original.

Radical equations work the same way. The symbol $\sqrt{\phantom{x}}$ always means the non-negative square root, so $\sqrt{\text{something}}$ can never equal a negative number. When you square $\sqrt{x + 3} = x - 3$, you get the same result as squaring $-\sqrt{x + 3} = x - 3$. So the squared equation picks up the solutions of both, and you have to check which ones belong to the original. Example 2 shows this on a graph.

A quick screen: if $\sqrt{A} = B$, then any real solution must make $B \ge 0$ and $A \ge 0$.

Radicals on both sides

If an equation is $\sqrt{A} = \sqrt{B}$, square both sides to get $A = B$. Check that the answer doesn’t make either radicand negative.

If there’s a radical on both sides plus another term, isolate one radical, square, then isolate the remaining radical and square again. Practice question 8 does this.

Rational exponents

A rational exponent is a root in disguise: $x^{1/2} = \sqrt{x}$ and $x^{3/2} = \left(\sqrt{x}\right)^3$. To solve $x^{m/n} = k$, isolate the power, then raise both sides to the reciprocal exponent $\dfrac{n}{m}$:

$$x^{3/2} = 8 \quad\Rightarrow\quad x = 8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 4$$

Watch out for an even numerator. In $x^{2/3} = 9$, the power is a square: $\left(x^{1/3}\right)^2 = 9$, so $x^{1/3} = \pm 3$, giving $x = 27$ or $x = -27$. Both check. (With an even denominator, like $x^{3/2}$, only $x \ge 0$ is allowed.)

Checking with Desmos

Graph each side as its own function, for example y = sqrt(x + 3) and y = x - 3, and click the intersections. Desmos only shows the real solutions of the original equation, so extraneous solutions simply don’t appear. It’s a fast way to see which of your algebra answers to keep.

Worked examples

Example 1: Isolate, square, check

Solve $3 + \sqrt{2x - 1} = 8$.

Solution.

$$\begin{aligned} 3 + \sqrt{2x - 1} &= 8 \\ \sqrt{2x - 1} &= 5 && \text{isolate the radical} \\ 2x - 1 &= 25 && \text{square both sides} \\ x &= 13 \end{aligned}$$

Check: $3 + \sqrt{2(13) - 1} = 3 + \sqrt{25} = 3 + 5 = 8$ ✓. The solution is $x = 13$.

Example 2: An extraneous solution

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

Solution. The radical is already isolated, so square both sides. Remember that $(x - 3)^2$ is a whole binomial squared, not $x^2 - 9$:

$$\begin{aligned} x + 3 &= (x - 3)^2 \\ x + 3 &= x^2 - 6x + 9 \\ 0 &= x^2 - 7x + 6 \\ 0 &= (x - 1)(x - 6) \end{aligned}$$

So $x = 1$ or $x = 6$. Now check both in the original equation:

• $x = 6$: left side $\sqrt{9} = 3$; right side $6 - 3 = 3$. ✓
• $x = 1$: left side $\sqrt{4} = 2$; right side $1 - 3 = -2$. ✗

The only solution is $x = 6$; $x = 1$ is extraneous.

The graph shows why. At $x = 1$, the line $y = x - 3$ meets the lower branch $y = -\sqrt{x + 3}$, which also turns into $x + 3 = (x - 3)^2$ when you square it. In Desmos, graphing y = sqrt(x + 3) and y = x - 3 shows a single intersection, $(6, 3)$.

$y = \sqrt{x + 3}$ meets $y = x - 3$ only at $(6, 3)$. The extraneous $x = 1$ comes from the dashed branch $y = -\sqrt{x + 3}$.

Example 3: Rational exponents

Solve each equation.

• (a) $x^{3/2} = 8$
• (b) $2x^{2/3} - 5 = 13$

Solution.

(a) Raise both sides to the power $\dfrac{2}{3}$, the reciprocal of $\dfrac{3}{2}$:

$$x = 8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$$

Check: $4^{3/2} = \left(\sqrt{4}\right)^3 = 2^3 = 8$ ✓. ($x^{3/2}$ needs $x \ge 0$, so there’s no negative solution.)

(b) Isolate the power first:

$$\begin{aligned} 2x^{2/3} - 5 &= 13 \\ 2x^{2/3} &= 18 \\ x^{2/3} &= 9 \\ \left(x^{1/3}\right)^2 &= 9 \\ x^{1/3} &= \pm 3 && \text{square root of both sides} \\ x &= \pm 27 && \text{cube both sides} \end{aligned}$$

Check: $27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$, and $(-27)^{2/3} = \left(\sqrt[3]{-27}\right)^2 = (-3)^2 = 9$. Both give $2(9) - 5 = 13$ ✓. The solutions are $x = 27$ and $x = -27$.

Example 4: Pendulum length

The time $T$, in seconds, for a pendulum to swing back and forth once is

$$T = 2\pi\sqrt{\frac{L}{9.8}}$$

where $L$ is the length of the pendulum in metres. How long should a pendulum be so that one swing takes $3$ seconds? Round to the nearest hundredth of a metre.

Solution. Substitute $T = 3$, then isolate the radical before squaring:

$$\begin{aligned} 3 &= 2\pi\sqrt{\frac{L}{9.8}} \\ \frac{3}{2\pi} &= \sqrt{\frac{L}{9.8}} && \text{divide by } 2\pi \\ \left(\frac{3}{2\pi}\right)^2 &= \frac{L}{9.8} && \text{square both sides} \\ L &= 9.8\left(\frac{3}{2\pi}\right)^2 \approx 2.23 \end{aligned}$$

The pendulum should be about $2.23$ m long.

Check: $2\pi\sqrt{\dfrac{2.234}{9.8}} \approx 2\pi(0.4775) \approx 3.00$ ✓. In Desmos, you could also graph y = 2pi sqrt(x/9.8) and y = 3 and click the intersection, about $(2.234, 3)$.

Common mistakes

Not checking the answers. Squaring can create extraneous solutions, so every answer has to be substituted into the original equation. On a multiple-choice question, an extraneous value is often one of the wrong choices.

Squaring before isolating the radical. Squaring $3 + \sqrt{x} = 8$ directly leaves a radical behind (the middle term $6\sqrt{x}$). Move everything else to the other side first.

Squaring a binomial term by term. $(x - 3)^2$ is $x^2 - 6x + 9$, not $x^2 + 9$ or $x^2 - 9$. Write it out as $(x - 3)(x - 3)$ if you’re unsure.

Thinking a square root can be negative. $\sqrt{4} = 2$, not $\pm 2$. An equation like $\sqrt{x + 5} = -3$ has no solution, because a square root is never negative. (Squaring would give $x = 4$, which fails the check.)

Losing a solution with an even power. In $x^{2/3} = 9$, remember the $\pm$ when you undo the square: both $27$ and $-27$ work. In contrast, $x^{3/2}$ only makes sense for $x \ge 0$.

Practice

1. (Warm-up) Solve $\sqrt{x} + 4 = 11$.

Solution

$$\sqrt{x} = 7 \quad\Rightarrow\quad x = 49$$

Check: $\sqrt{49} + 4 = 7 + 4 = 11$ ✓.

2. (Warm-up) Solve $\sqrt{5x + 1} = 6$.

Solution

$$5x + 1 = 36 \quad\Rightarrow\quad 5x = 35 \quad\Rightarrow\quad x = 7$$

Check: $\sqrt{36} = 6$ ✓.

3. (Warm-up) Which of the following is a solution of $\sqrt{x + 7} = x + 1$?

• A) $-3$ only
• B) $2$ only
• C) Both $-3$ and $2$
• D) Neither $-3$ nor $2$

Solution

B. Square both sides:

$$x + 7 = x^2 + 2x + 1 \quad\Rightarrow\quad 0 = x^2 + x - 6 = (x + 3)(x - 2)$$

Check $x = 2$: $\sqrt{9} = 3$ and $2 + 1 = 3$ ✓. Check $x = -3$: $\sqrt{4} = 2$ but $-3 + 1 = -2$ ✗, so $-3$ is extraneous.

4. (Core) Solve $\sqrt{4x + 5} = \sqrt{x + 17}$.

Solution

Square both sides:

$$4x + 5 = x + 17 \quad\Rightarrow\quad 3x = 12 \quad\Rightarrow\quad x = 4$$

Check: $\sqrt{4(4) + 5} = \sqrt{21}$ and $\sqrt{4 + 17} = \sqrt{21}$ ✓.

5. (Core) What is the solution of $x - \sqrt{2x + 6} = 1$? (Student-produced response.)

Solution

Isolate the radical, then square:

$$\begin{aligned} x - 1 &= \sqrt{2x + 6} \\ x^2 - 2x + 1 &= 2x + 6 \\ x^2 - 4x - 5 &= 0 \\ (x - 5)(x + 1) &= 0 \end{aligned}$$

Check $x = 5$: $5 - \sqrt{16} = 5 - 4 = 1$ ✓. Check $x = -1$: $-1 - \sqrt{4} = -3 \ne 1$ ✗.

The solution is $5$.

6. (Core) Solve $4x^{3/2} = 108$.

Solution

$$x^{3/2} = 27 \quad\Rightarrow\quad x = 27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$$

Check: $4 \cdot 9^{3/2} = 4\left(\sqrt{9}\right)^3 = 4(27) = 108$ ✓.

7. (Core) When an object is dropped, the time $t$ (in seconds) it takes to fall $h$ metres is about $t = \sqrt{\dfrac{h}{4.9}}$. A stone dropped from a bridge takes $2.5$ s to reach the water. How high is the bridge, to the nearest tenth of a metre?

Solution

$$2.5 = \sqrt{\frac{h}{4.9}} \quad\Rightarrow\quad 6.25 = \frac{h}{4.9} \quad\Rightarrow\quad h = 4.9(6.25) = 30.625$$

The bridge is about $30.6$ m high. Check: $\sqrt{\dfrac{30.625}{4.9}} = \sqrt{6.25} = 2.5$ ✓.

8. (Challenge) Solve $\sqrt{x + 5} + 1 = \sqrt{2x + 8}$.

Solution

Square both sides. The left side is a binomial, so expand it fully:

$$\begin{aligned} (x + 5) + 2\sqrt{x + 5} + 1 &= 2x + 8 \\ 2\sqrt{x + 5} &= x + 2 && \text{isolate the remaining radical} \\ 4(x + 5) &= x^2 + 4x + 4 && \text{square again} \\ 4x + 20 &= x^2 + 4x + 4 \\ 16 &= x^2 \\ x &= 4 \text{ or } x = -4 \end{aligned}$$

Check $x = 4$: $\sqrt{9} + 1 = 4$ and $\sqrt{16} = 4$ ✓. Check $x = -4$: $\sqrt{1} + 1 = 2$ but $\sqrt{0} = 0$ ✗.

The only solution is $x = 4$. (Graphing both sides in Desmos shows a single intersection at $(4, 4)$.)

9. (Challenge) In the equation $\sqrt{x - a} = x - 3$, $a$ is a constant, and $x = 7$ is a solution.

• (a) Find $a$.
• (b) Does the equation have any other solution?

Solution

(a) Substitute $x = 7$: $\sqrt{7 - a} = 4$, so $7 - a = 16$ and $a = -9$.

(b) The equation is $\sqrt{x + 9} = x - 3$. Square both sides:

$$\begin{aligned} x + 9 &= x^2 - 6x + 9 \\ 0 &= x^2 - 7x \\ 0 &= x(x - 7) \end{aligned}$$

So $x = 0$ or $x = 7$. Check $x = 0$: $\sqrt{9} = 3$ but $0 - 3 = -3$ ✗. So $x = 0$ is extraneous, and $x = 7$ is the only solution.