#### 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 radial.
• 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.

Problem2.6c:

Solution:

First make a note of the fact that you cannot take the square root of a negative number. Therefore,

Rewrite the radical terms as exponential terms

Raise both sides of the equation to the power 8 and simplify.

The answers are 2.804248 (rounded) and 0.445752 (rounded).

Check the solution by substituting 2.804248 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:
• Right side:
Since the left side of the original equation after we substituted 2.804248 for x, then x = 2.804248 is a solution.

Since the solution x=0.445752 is not (the domain), it cannot be a solution. However, if you forgot this fact, you can discover the same thing by checking the equation.

Check the solution by substituting 0.445752 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:
• Right side:
No solution because one of the terms is undefined over the set of real numbers.

You can also check the answer by graphing the equation:

The graph represents the right side of the original equation minus the left side of the original equation. Since there is one x-intercept at 2.804248, the solution is 2.804248.
If you would like to review the solution for problem 2.6d, click on Solution.

If you would like to return to the problem page, click on Problem.

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