EQUATIONS CONTAINING VARIABLES UNDER ONE OR MORE RADICALS

Recall the following:

• In order to solve for the unknown variable, you must isolate the variable.
• In the order of operations, multiplication and division are completed before addition and subtraction.
• 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 are three radicals in the equation,isolate one of the radicals.
• Then raise both sides of the equation to a power equal to the index of the isolated radical.
• Isolate the remaining radical.
• Raise both sides of the equation to a power equal to the index of the isolated radical.
• You should now have a polynomial equation. Solve it.
• Remember that you did not start out with a polynomial; therefore, there may be extraneous solutions. Therefore, you must check your answers.

Solve for x in the following equation.

Problem 2.5b:

Answer: x=5.

Solution:

First make a note of the fact that you cannot take the square root of a negative number. Therefore, the term is valid only if , the term is valid only if , and the term is valid only if . The equation is valid if all three terms are valid, therefore the domain is restricted to the common domain of the three terms or the set of real numbers

Square both sides of the equation and simplify.

Isolate the term and simplify.

Square both sides of the equation and simplify.

Use the quadratic formula to solve for x.

Discard the second answer because 0.391676 is not in the domain.

The answers is x=5.

Check the solution by substituting 5 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 equals the right side of the original equation after we substituted 5 for x, then x=5 is a solution.

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. The x-intercept(s) of this graph is(are) the solution(s). Since there is just one x-intercept at 5, then the only solution is x=5.

If you would like to review the solution to problem 2.5c, click on solution.

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