#### xaktly | Algebra

Root equations

In order to best understand root functions, you should know how to use rational exponents (like $x^{1/2}$), and you should know what they mean: Rational and negative exponents

You should also be adept with the laws of exponents, given in the same section. You can download a table of the laws of exponents here.

You should also be familiar with the function concept and domain & range.

### Solving root equations

This page consists of a couple of examples of solving root equations — here are some examples:

\begin{align} &\cdot \phantom{00} \sqrt{2x - 4} + 2 = x - 8 \\[5pt] &\cdot \phantom{00} \sqrt[3]{x-2} = x^2 + 5 \\[5pt] &\cdot \phantom{00} (2x + 1)^{\frac{1}{4}} = 3 \end{align}

The general idea for solving such equations is to isolate the radical(s) on one (or either) side, then raise both sides to the power that is the inverse of the root. It's important to recall that roots can be written using rational exponents. For example,

\begin{align} \sqrt{x} &= x^{\frac{1}{2}} \\[5pt] \sqrt[3]{x} &= x^{\frac{1}{3}} \\[5pt] \sqrt[4]{x^3} &= x^{\frac{3}{4}} \; \dots \end{align}

In the rational exponent, the numerator is the power of the independent variable, and the denominator is the root.

If we can translate radicals like $\sqrt{x + 6}$ to rational exponent expressions (that one is $(x+6)^{\frac{1}{2}}$), then the inverse operation becomes much more clear. Here are some examples:

 radical rationalexp. inverse $\sqrt{x}$ $x^{\frac{1}{2}}$ $\left(\sqrt{x}\right)^2$ $\sqrt{x^3}$ $x^{\frac{3}{2}}$ $\left(\sqrt{x^3}\right)^{\frac{2}{3}}$ $\sqrt[3]{x}$ $x^{\frac{1}{3}}$ $\left(\sqrt[3]{x}\right)^3$ $\sqrt[3]{x^2}$ $x^{\frac{2}{3}}$ $\left(\sqrt[3]{x^2}\right)^{\frac{3}{2}}$ $\sqrt[4]{x}$ $x^{\frac{1}{4}}$ $\left(\sqrt[4]{x}\right)^4$

### Example 1

Solve for x: $3x = \sqrt{9(x+20)}$

Solution: First, square both sides of the equation to get

$$9x^2 = 9(x + 20)$$

Then use good algebra moves to solve for x:

\require{cancel} \begin{align} \cancel{9}x^2 &= \cancel{9}(x + 20) \\[5pt] x^2 - x - 20 &= 0 \\[5pt] (x - 5)(x + 4) &= 0 \\[5pt] x &= 5, \, -4 \end{align}

Now we should check both of those solutions for compatibility with the original problem. First $x = 5$:

\begin{align} 3(5) &= \sqrt{9(5 + 20)} \\[5pt] 15 &= \sqrt{9 \cdot 25} \\[5pt] 15 &= 3 \cdot 5 \; \; \color{green}{\checkmark} \end{align}

Then $x = -4$:

$$3(-4) = \sqrt{9(-4+20)}$$

This can't be true. A square root of a positive number $(20-4)$ can't be negative, so we have to toss the $x = -4$ solution and conclude that this equation has only one solution, $x = 5$.

Here's a graph of $y = 3x$ vs. $y = \sqrt{9(x+20)}$ so you can see the situation visually. The red curve is the root side and the black is the line $y = 3x$. There is no intersection at $x = -4$.

### Example 2

Solve for x: $\sqrt[3]{x^2+6} = 2$

Solution: First, cube both sides of the equation, then solve for $x$.

\begin{align} \left( (x^2 + 6)^{\frac{1}{3}} \right)^3 &= 2^3 \\[5pt] x^2 + 6 &= 8 \\[5pt] x^2 &= 2 \\[5pt] x &= \pm \sqrt{2} \end{align}

Now we can check both of those solutions by sketching graphs of $y = 2$ (black) and $y = \sqrt[3]{x^2+6}$ (red). We can see that there are two intersections at $\pm \sqrt{2}$, so these are both good solutions — there are no extraneous solutions here.

### Example 3

Solve for x: $\sqrt[3]{x^3 - 3x^2 - 6x} = -2$

Solution: First, cube both sides of the equation:

\begin{align} \left( \sqrt[3]{x^3 - 3x^2 - 6x} \right)^3 &= (-2)^3 \\[5pt] x^3 - 3x^2 - 6x &= -8 \\[5pt] x^3 - 3x^2 - 6x + 8 &= 0 \end{align}

Now we can use the rational root theorem to make a list of possible rational roots, $x \in \{\pm 1, \pm 2, \pm 4, \pm 8 \}$. If we test $x = 1$ by synthetic substitution, we get:

$$\begin{array}{c|rrrr} 1 & 1 & -3 & -6 & 8 \\[5pt] \phantom{0} & \downarrow & 1 & -2 & -8 \\[5pt] \hline & 1 & -2 & -8 & 0\\[5pt] \end{array}$$

So $x = 1$ is a root, which means that our equation can be written as

$$(x - 1)(x^2 - 2x - 8) = 0$$

The quadratic is easy to factor, so we have

$$(x - 1)(x - 4)(x + 2) = 0$$

So our solutions are $x = 1, 4, -2$.

Note: That was a pretty easy rational-root theorem problem. Often one has to guess at a few more root candidates in order to find one that works. Remember that once you've reduced such an equation to one or more binomials times a quadratic, that you have other tools to factor the quadratic — use those. Also remember that the rational root theorem only provides candidates for roots. Any given polynomial function may not even have any rational roots.

Here are graphs of $y = x^3 - 3x^2 - 6x$ and $y = -8$. You can see that these functions intersect at $x = 1, 4, -2$.

### Practice problems

Solve for the variable:

1. $3 = \sqrt{x - 1}$

Solution

Start out by squaring both sides to clear the root, then use good algebra moves to solve for x:

\begin{align} 3^2 &= \left( \sqrt{x - 1} \right)^2 \\[5pt] 9 &= x - 1 \\[5pt] x &= 10 \end{align}

2. $\sqrt{x + 5} = 0$

Solution

Start out by squaring both sides to clear the root, then use good algebra moves to solve for x:

\begin{align} \left( \sqrt{x + 5} \right)^2 &= 0^2\\[5pt] x + 5 &= 0 \\[5pt] x &= -5 \end{align}

3. $\sqrt{2x - 6} = \sqrt{3x - 14}$

Solution

Start out by squaring both sides to clear the roots, then use good algebra moves to solve for x:

\begin{align} \left( \sqrt{2x - 6} \right)^2 &= \left( \sqrt{3x - 14} \right)^2\\[5pt] 2x - 6 &= 3x - 14 \\[5pt] -x &= -8 \\[5pt] x &= 8 \end{align}

Here's a graph of $\sqrt{2x-6}$ and $\sqrt{3x - 14}$. They intersect at $x = 8$.

4. $\sqrt{8x} = x$

Solution

Start out by squaring both sides to clear the root, then use good algebra moves to solve for x:

\begin{align} \left( \sqrt{8x} \right)^2 &= x^2 \\[5pt] 8x &= x^2 \\[5pt] x &= 0, \, 8 \\[5pt] \end{align}

Notice that in the second-to-last step, we divided by x, thus "dropping" a root, $x = 0.$ We always have to be wary of that.

5. $\sqrt{3 - 2x} = \sqrt{1 - 3x}$

Solution

Start out by squaring both sides to clear the root, then use good algebra moves to solve for x:

\begin{align} \left( \sqrt{3 - 2x} \right)^2 &= \left( \sqrt{1 - 3x} \right)^2\\[5pt] 3 - 2x &= 1 - 3x \\[5pt] x &= -2 \end{align}

6. $\sqrt{x - 6} - \sqrt{x + 9} + 3 = 0$

Solution

This equation contains both roots and numbers without any exponent (recall that a root is just a rational exponent). A good way to tackle this problem is to arrange to:

$$\sqrt{x - 6} = \sqrt{x + 9} - 3$$

Now square both sides:

\begin{align} \left( \sqrt{x - 6} \right)^2 &= \left( \sqrt{x + 9} - 3 \right)^2 \\[5pt] x - 6 &= \left( \sqrt{x + 9} \right)^2 \\[5pt] &\phantom{000} - 2(3)\sqrt{x + 9} + 3^2 \tag{1} \\[5pt] x - 6 &= x + 9 - 6 \sqrt{x + 9} + 9 \\[5pt] 6 \sqrt{x + 9} &= 24 \tag{2} \\[5pt] \sqrt{x + 9} &= 4 \\[5pt] \left( \sqrt{x + 9} \right)^2 &= 4^2 \tag{3}\\[5pt] x + 9 &= 16 \\[5pt] x &= 7 \end{align}

In step (1) we multiplied binomials. Remember that (a+b)2 = a2 + 2ab + b2. In step (2) we just gathered like terms and put the root on the left side. In step (3) we need to square both sides again. That's common in these kinds of problems. The rest is just a few simple algebra moves. Here is a graph of $y = \sqrt{x - 6}$ (black) and $y = \sqrt{x+9}+3$ (red). They intersect at $x = 7$.

7. $4 + \sqrt[3]{x - 6} = 2$

Solution

First move the constant (4) to the right and cube both sides. From then it's just some algebra moves:

\begin{align} \sqrt[3]{x - 6} &= -2 \\[5pt] \left( \sqrt[3]{x - 6} \right)^3 &= (-2)^3 \\[5pt] x - 5 &= -8 \\[5pt] x &= -2 \end{align}

Here is a graph of $y = \sqrt[3]{x - 6}$ (black) and $y = -2$ (red). They intersect at $x = -2$.

8. $\sqrt[3]{(x - 3)^2} = 4$

Solution

With a problem like this – a mix of roots and powers, it's often better to express the root and the power together as a rational exponent, like this:

$$(x - 3)^{\frac{2}{3}} = 4$$

Now raise both sides of the equation to the reciprocal power, $\frac{3}{2}.$

\begin{align} \left((x - 3)^{\frac{2}{3}}\right)^{\frac{3}{2}} &= 4^{\frac{3}{2}} \\[5pt] x - 3 &= \left( 4^{\frac{1}{2}} \right)^3 \\[5pt] x - 3 &= 2^3 \\[5pt] x &= 8 + 3 \\[5pt] x &= 11 \end{align}

9. $\sqrt[4]{2x + 2} = 3$

Solution

Raise both sides to the fourth power, then solve for $x$:

\begin{align} \left( \sqrt[4]{2x+2} \right)^4 &= 3^4 \\[5pt] 2x + 2 &= 3^4 \\[5pt] 2x + 2 &= 81 \\[5pt] 2x &= 79 \\[5pt] x &= \frac{79}{2} \end{align}

10. $x = 5 + (3x - 11)^{\frac{1}{2}}$

Solution

Move the 5 to the left, then square both sides and solve for $x$:

\begin{align} (x - 5)^2 &= \left[ (3x - 11)^{\frac{1}{2}} \right]^2 \\[5pt] x^2 - 10 x + 25 &= 3x - 11 \\[5pt] x^2 -13 x + 36 &= 0 \\[5pt] (x-9)(x-4) &= 0 \\[5pt] x &= 9, \, 4 \end{align}

If we plug the $x = 4$ solution back into the original equation we get $4-5 = \sqrt{3x - 11}$, but that would imply a negative square root, so we have to toss that solution. The only solution is $x = 9$. Here's a graph of the solution

1. $\sqrt{2x + 102} = x + 11$

Solution

Square both sides, then solve for $x$:

\require{cancel} \begin{align} \left(\sqrt{2x + 102} \right)^2 &= x + 11 \\[5pt] 2x + 102 &= (x + 11)^2 \\[5pt] 2x + 102 &= x^2 + 22x + 121 \\[5pt] x^2 + 20x + 19 &= 0 \\[5pt] (x + 19)(x + 1) &= 0 \\[5pt] x &= \cancel{-19}, \, -1 \end{align}

We have to strike the $x = -19$ solution because the original problem would then read,

\begin{align} \sqrt{2(-19) + 102} &= -19 + 11 \\[5pt] \sqrt{64} &= -8, \end{align}

which can't be true, so $x = -19$ is an extraneous solution that we have to toss out.

2. $\sqrt{-x + 33} = -x + 3$

Solution

Square both sides, then solve for $x$:

\begin{align} -x + 33 &= x^2 - 6x + 9 \\[5pt] x^2 - 5x -24 &= 0 \\[5pt] (x - 8)(x + 3) &= 0 \\[5pt] x &= \cancel{8}, \, -3 \end{align}

The solution $x = 8$ must be put aside because it would cause our root in the original problem to equal a negative number:

$$\sqrt{-8 + 33} = -5$$

3. $\sqrt{-4x + 12} = -2x + 6$

Solution

Square both sides, then solve for $x$:

\begin{align} -4x + 12 &= 4x^2 - 24x + 36 \\[5pt] 4x^2 - 20x + 24 &= 0 \\[5pt] x^2 - 5x + 6 &= 0 \\[5pt] (x - 6)(x + 1) &= 0 \\[5pt] x &= \cancel{6}, \, -1 \end{align}

$x = 6$ is an extraneous solution.

4. $\sqrt{3-x} = \sqrt{x+7} + 2$

Solution

Square both sides, then solve for $x$:

\begin{align} \left( \sqrt{3-x} \right)^2 = \left( \sqrt{x+7} + 2 \right)^2 \\[5pt] 3 - x &= x + 7 + 4 \sqrt{x+7} + 4 \\[5pt] 3 - x - x - 7 - 4 &= 4 \sqrt{x+7} \\[5pt] \frac{-2-8}{4} &= \sqrt{x+7} \\[5pt] -\frac{x}{2} - 2 &= \sqrt{x+7} \end{align}

Now square both sides again:

\begin{align} \left( -\frac{x}{2} - 2 \right)^2 = \left( \sqrt{x+7} \right)^2 \\[5pt] \frac{x^2}{4} + 2x + 4 &= x + 7 \\[5pt] \frac{x^2}{4} + x - 3 &= 0 \\[5pt] x^2 + 4x - 12 &= 0 \\[5pt] (x+6)(x-2) &= 0 \\[5pt] x &= -6, \, 2 \end{align}

Now we need to check these possible solutions:

$$\sqrt{3+6} = \sqrt{1} + 3 \; \color{green}{\checkmark}$$

$$\sqrt{3-2} \ne \sqrt{9} + 2 \; \color{red}{X}$$

The only solution is $x = -6$

5. $\sqrt{4x} = -x + 8$

Solution

\begin{align} 4x &= x^2 - 16x + 64 \\[5pt] x^2 - 20x + 64 &= 0 \\[5pt] (x-16)(x-4) &= 0 \\[5pt] x &= 4, \, \cancel{16} \end{align}

$x = 16$ is an extraneous solution.

6. $(x-2)^{\frac{1}{2}} = (28-2x)^{\frac{1}{4}}$

Solution

In this case, let's raise each side of the equation to the 4th power:

\begin{align} (x-2)^{\frac{4}{2}} &= (28-2x)^{\frac{4}{4}} \\[5pt] (x-2)^2 &= 28 - 2x \\[5pt] x^2 - 4x + 4 + 2x -28 &= 0 \\[5pt] x^2 -2x -24 &= 0 \\[5pt] (x-6)(x+4) &= 0 \\[5pt] x = 6, \, -4 \end{align}

Now to check these solutions in the orignal equation:

\begin{align} \sqrt{6-2} &= (28 - 12)^{\frac{1}{4}} \\[5pt] \sqrt{4} &= \sqrt[4]{16} \\[5pt] 2 &= 2 \; \color{green}{\checkmark} \end{align}

Now -4:

\begin{align} \sqrt{-4-2} &= (28 - (-8))^{\frac{1}{4}} \\[5pt] \sqrt{-6} &= \sqrt[4]{36} \end{align}

The solution $x = -4$ does not check out. We can't make an imaginary root equal to a real root.

7. $\sqrt[3]{x^2+4} -2 = 0$

Solution

Let's move the -2 to the right, then cube both sides of the equation:

\begin{align} \sqrt[3]{x^2+4} &= 2\\[5pt] x^2+4 &= 2^3 \\[5pt] x^2 + 4 - 8 &= 0 \\[5pt] x^2 &= 4 \\[5pt] x &= \pm 2 \end{align}

We should check these solutions. First $x = 2$:

\begin{align} \sqrt[3]{2^2 + 4} - 2 &= 0 \\[5pt] \sqrt[3]{8} - 2 &= 0 \\[5pt] 2 - 2 &= 0 \; \color{green}{\checkmark} \end{align}

And $x = -2$:

$$\sqrt[3]{(-2)^2 + 4} - 2 = 0$$

That checks out in the same way; both roots are good.

8. $\sqrt[3]{x^2 + x - 12} = \sqrt[3]{x + 52}$

Solution

First cube both sides:

\begin{align} x^2 + \cancel{x} - 12 &= \cancel{x} + 52 \\[5pt] x^2 - 64 &= 0 \\[5pt] x^2 &= 64 \\[5pt] x^2 &= \pm 8 \end{align}

We should check these solutions. First $x = 8$:

\begin{align} \sqrt[3]{8^2 + 8 - 12} &= \sqrt[3]{8 + 52} \\[5pt] \sqrt[3]{64 + 8 - 12} &= \sqrt[3]{60} \\[5pt] \sqrt[3]{60} &= \sqrt[3]{60} \; \color{green}{\checkmark} \end{align}

And $x = -8$:

\begin{align} \sqrt[3]{(-8)^2 + 8 - 12} &= \sqrt[3]{-8 + 52} \\[5pt] \sqrt[3]{60} &= \sqrt[3]{46} \; \color{red}{X} \end{align}

So only $x = 8$ is a solution.

9. $2 \sqrt[3]{3x - 8} + 6 = 2$

Solution

Move the 6 to the right, divide by 2 and cube both sides:

\begin{align} \sqrt[3]{3x - 8} &= \frac{2-6}{2} \\[5pt] 3x - 8 &= (-2)^3 \\[5pt] 3x - 8 &= -8 \\[5pt] x &= 0 \end{align}

We should check this solution:

\begin{align} 2 \sqrt[3]{3(0)-8} + 6 &= 2 \\[5pt] 2(-2) + 6 &= 2 \\[5pt] 2 &= 2 \; \; \color{green}{\checkmark} \end{align}

So $x = 0$ is a good solution.

10. $\sqrt{8x + 9} = x$

Solution

First square both sides:

\begin{align} 8x + 9 &= x^2 \\[5pt] x^2 - 8x - 9 &= 0 \\[5pt] (x - 9)(x + 1) &= 0 \\[5pt] x &= 9, \; -1 \end{align}

We should check these solutions:

\begin{align} \sqrt{8(9) + 9} &= 9 \\[5pt] \sqrt{81} &= 9 \; \color{green}{\checkmark} \\[5pt] \sqrt{8(-1) + 9} &= -1 \\[5pt] \sqrt{1} \ne -1 \; \; \color{red}{X} \end{align}

So only the $x = 9$ solution is good.

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