Menyelesaikan dengan menggunakan radikal $x^n = c$

(Video) Exponents in Algebra

by Math Antics

This video by Math Antics shows you how to solve equations in the form $x^n=c$.

Summary

We solve any algebraic equation by isolating a variable. How do we isolate a variable being raised to a power?

$$x^3=8$$

The answer is radicals! Just like addition can undo subtraction, radicals can undo exponents. If you have an exponent raised to the third power, you take the cube root of both sides.

$$\begin{align} x^{\color{#8925AE}{3}}&=8\\ \sqrt[\color{#8925AE}{3}]{x^{\color{#8925AE}{3}}}&=\sqrt[\color{#8925AE}{3}]{8}\\ x&=\sqrt[3]{8}\\ x&=2 \end{align}$$

Right now, we are only worrying about the real solutions to an equation, like 2 was for the previous example. Actually, $x^3=8$ also has two complex solutions, which you will learn about later.

Let's do another example!

$$x^2=36$$

We solve this equation by taking the square root of both sides.

$$\begin{align} \sqrt{x^2}&=\sqrt{36}\\ x&=6 \end{align}$$

But, $-6$ is also an answer since $(-6)\cdot(-6)=36$. So, when taking square roots, it's best to put a plus or minus sign, $\pm$, in front. $$\begin{align} x^2&=36\\ \sqrt{x^2}&=\pm\sqrt{36}\\ x&=\pm6 \end{align}$$

In summary, some equations have multiple solutions. Variables raised to the second power have a positive and negative solution! Remember, every time you take a square root, put a $\pm$ in front!

Solving $x^n = c$ with prime factorization

Suppose $c$ is a real number. We want to solve the equation and find the real solutions of $$x^n = c$$

If $n$ is odd, then there is always a unique real solution for $x$: $$x = \sqrt[n]{c}$$ the $n$th root of $c$.

If $n$ is even, we have to be more careful. First, there are no real solutions if $c$ is negative. If $c=0$, there will just be the solution $x=0$. If $c>0$, there will be a positive real solution and a negative real solution. $$x = \pm\sqrt[n]{c}$$

To simplify the $n$th root further when $c$ is an integer (or sometimes a fraction), it helps to further break down $c$. Write $c$ in terms of its prime factorization, we can then simplify.

For example, say $c=240$. We can use a prime factorization method to find: $$240=2\cdot2\cdot2\cdot2\cdot3\cdot5$$

Depending on what $n$ is, $x$ will be different.

If $n=2,$ then:

$$\begin{align} x^2 &= 240\\ \sqrt{x^2} &= \pm \sqrt{240}\\ x &= \pm\sqrt{240}\\ x&=\pm \sqrt{2\cdot 2 \cdot 2\cdot 2\cdot 3\cdot5} \\ x&= \pm4\sqrt{15} \end{align}$$

So, the solutions of $x^2 = 240$ are $x= 4\sqrt{15}$ and $x= -4\sqrt{15}$. Note that both of these are real, not complex, solutions.

If $n=3,$ then:

$$\begin{align} x^3 &= 240\\ \sqrt[3]{x^3} &= \sqrt[3]{240}\\ x &= \sqrt[3]{240}\\ x&=\sqrt[3]{2\cdot 2 \cdot 2\cdot 2\cdot 3\cdot5} \\ x&= 2 \sqrt[3]{30} \end{align}$$

So, the real solution of $x^3 = 240$ is $x= 2 \sqrt[3]{30}$. (If you're curious, you can learn more about complex solutions on our site.)

Here is a graphic with the general method of solving by square roots and cube roots.

Try the following problem to test your understanding.

Solution

$$\begin{align} x^3 &= 162\\ \sqrt[3]{x^3} &= \sqrt[3]{162}\\ x &= \sqrt[3]{162}\\ x&=\sqrt[3]{2\cdot 3 \cdot 3\cdot 3\cdot 3} \\ x&= 3 \sqrt[3]{6} \end{align}$$

If you want to check your work, you can plug in the value for $x$ that you solved for back into the original equation. Solve as normal to check your answer.

Solve by Radical: $x^{n}=c$

This concept can get kind of confusing, so let's think about how we could rewrite the following equation:

$$x = \sqrt[2]{4}$$

We would start by squaring both sides.

$$x^2=4$$

All we did was isolate both sides and rewrite the equation to make it a little cleaner. This back-and-forth is how we will tackle this new way of solving radical expressions.

To solve an equation $x^{n}=c$, we can take the $n$-th root of both sides of the equation. Assume that $n$ is positive. If $n$ is odd, then there is only one real solution.

$$x = \sqrt[n]{c}$$

If $n$ is even, then there are two real solutions (positive and negative).

$$x = \pm\sqrt[n]{c}$$

The right hand side looks really complex. Let’s take a look at $x^3=27$. To solve an equation in the form $x^{n}=c$, we can take the $n$-th root of both sides of the equation. In this example, $n=3$ and $c=27$. Since $n$ is odd, there is only one real solution. $$\begin{align} x^{3}&=27\\ \sqrt[3]{x^{3}}&=\sqrt[3]{27}\\ x&= \sqrt[3]{27}\\ x &=3 \end{align}$$

Let’s look at another example: $x^4=16$.

We can take the $n$-th root of both sides of the equation. In this example, $n=4$ and $c=16$. Since $n$ is even, there are two real solutions that are positive and negative.

$$\begin{align} x^4&=16 \\ \sqrt[4]{x^{4}}&=\pm \sqrt[4]{16}\\ x&= \pm2 \\ \end{align}$$

We can take the $n$-th root of both sides of the equation. Since $n=3$ and $c=64$ and $n$ is odd, there is only one real solution. The other solutions are complex!

$$\begin{align} x^3&=64\\ \sqrt[3]{x^3}&=\sqrt[3]{64}\\ x&=4 \end{align}$$

It's important to double check your work because you may come across extraneous solutions. When you plug the answer back into the original equation, if you have a true statement, the solution is valid. You'll learn more about this concept later!