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 A radical equation is an equation that has a variable in a radicand (or a variable with a rational exponent).
On this page, all radicals and expressions with rational exponents represent real numbers.

The "radical" in a radical equation may be of any root value: square root, cube root, fourth root, etc. This page will concentrate on working with "square root" equations.

Remember that a radical equation has the variable "under" the radical, not simply a numerical radical value within the equation.
 is a radical equation is NOT a radical equation

The solution to a radical equation is a real number which, when substituted for the variable, yields a true equation. Radical equations with square roots can be solved by "squaring", with cube roots by "cubing", and so on. This solution process is an application of the principle:
 If a and b are real numbers, n is a positive integer, and a = b, then an = bn.
The solutions, however, get significantly "messier" to find as the indexes increase.

To solve radical equations:
 1 Isolate the radical (or one of the radicals) to one side of the equal sign. 2 If the radical is a square root, square each side of the equations. If the radical is not a square root, raise each side to a power equal to the index of the root. 3 Solve the resulting equation. 4 Check your answer(s) to avoid extraneous roots.
 Square "sides", not "terms". SIDES: 2 + 4 = 6 (2 + 4)² = 6² 36 = 36 TERMS: 2 + 4 = 6 2² + 4² ≠ 6² 4 + 16 ≠36
 How to use your TI-83+/84+ calculator with radical equations. Click here.

There may be a problem with solutions to these types of equations as some of the "perceived" solutions may not actually work when plugged back into the original equations.

This problem arises from the fact that the converse of the statement:
If a and b are real numbers, n is a positive integer, and a = b, then an = bn.
is true when n is odd, but not necessarily true when n is even.

Consider the square root situation shown below. We can square both sides of the first radical equation and obtain a true quadratic equation. This new quadratic equation contains not only the square of (2x - 15), but can also contain the square of (-2x + 15), which we do not want. If we take the square root of both sides of the new quadratic equation, we will get two interpretations, . The new equation may yield more information than we need by offering up solutions that belong to the interpretation we do not want (called extraneous roots).

 Notice how the graph of the sides of the original radical equation show ONE solution. The graph of the new equation shows TWO solutions. The solution of x = 6.25 will not check in the radical equation. x = 4x2 - 60x + 225

The process of squaring the sides of an equation creates a "derived" equation which may not be equivalent to the original radical equation. Consequently, solving this new derived equation may create solutions that are not related to the original equation. These "extra" roots that are not true solutions of the original radical equation are called extraneous roots and are rejected as answers.
CHECKING will be essential to finding correct solutions
(and catching these pesky extraneous roots).

Example where the answer checks!
 ANSWER: x = 19

Example where the answer does NOT check!

Graphs do not intersect.
No solution.
x = 5 is an extraneous root.

ANSWER: no solution
 Once the radical was set = -3, it became obvious that there would be NO solution, since the principal square root cannot be a negative value.

Example where BOTH answers check!
 The graph and the TABLE show the answers. ANSWERS: x = {2,4}

Example where ONLY ONE answer checks!
 Graph shows only one intersection point. x = -3 is extraneous root. ANSWER: x = 7

Example with radicals on BOTH SIDES!
 ANSWER: x = 2

Example with BOTH radicals multiplied by constants!
 ANSWER: x = 10

Example involving x2 !
 ANSWER: x = 3

Example with radicals (and more) on BOTH sides!
 Graph shows only one point of intersection. x = 2 is extraneous root. ANSWER: x = -6

 Let's try solving a cube root equation.

The concept is the same as we used when working with the square roots, except that now we are cubing instead of squaring.
 ANSWER: x = 1

 Let's try solving a rational exponent equation.

 x = -4 is an extraneous root ANSWER: x = 6

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