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EQUATIONS CONTAINING VARIABLES UNDER ONE OR MORE RADICALS

Note:

In order to solve for x, you must isolate x.
In order to isolate x, you must remove it from under the radical.
If there is just one radical in the equation, isolate the radical.
Then raise both sides of the equation to a power equal to the index of the radical.
With these types of equations, sometimes there are extraneous solutions; therefore, you must check your answers.
If the index of the radical is even, many times there will be a restriction on the values of x.

Example 4:. $\sqrt[5]{2x+3}+7=8$

Isolate the radical term

$\begin{eqnarray*}&&\\ \sqrt[5]{2x+3} &=&1 \\ && \end{eqnarray*}$

Raise both sides of the equation to the power 5 and simplify.

$\begin{eqnarray*}\left( \sqrt[5]{2x+3}\right) ^{5} &=&\left( 1\right) ^{5} \\ &... ... && \\ && \\ 2x &=&-2 \\ && \\ && \\ x &=&-1 \\ && \\ && \end{eqnarray*}$

The answer is x=-1

Check the solution by substituting -1 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.

Left side: $\sqrt[5]{2\left( -1\right) +3}+7=1+7=8$

Right Side: 8

Since the left side of the original equation equals the right side of the original equation after you substitute -1 for x, then -1 is a solution.

You can also check your solution by graphing the function

$\begin{eqnarray*}f(x) &=&\sqrt[5]{2x+3}-1 \\ && \\ && \end{eqnarray*}$

The above function is formed by subtracting the right side of the original equation from the left side of the original equation. The x-intercept of the graph is the solution to the original equation. As you can see, there is one x-intercept at -1. We have verified our solutions graphically.

If you would like to test yourself by working some problems similar to this example, click on problem.

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Author:Nancy Marcus