What are extraneous solutions to rational and radical equations? how do they arise? how do we check if a solution is extraneous? use these equations as examples to you explain:

9514 1404 393

Explanation:

Extraneous solutions to equations are ones that arise in the solution process but do not satisfy the original equation.

They often arise when clearing denominators or radicals. Multiplying both sides of an equation by expressions that may be zero may introduce that zero as a solution to the equation. Clearing even-index radicals may introduce solutions that would require the radical to take on a negative value. In these cases (an extra zero, or a negative radical value), the "solutions" introduced will be extraneous.

A solution can be identified as being extraneous if it does not satisfy the original equation.

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By way of example, consider the equation ...

  x = √(x+18) +2

A graphing calculator shows the solution to this to be x=7. That solution is easily verified as satisfying the equation:

  7 = √(7+18) +2 = √25 +2 = 5 +2

The usual solution process isolates the radical, then squares both sides of the equation.

  (x -2) = √(x +18)

  x^2 -4x +4 -(x +18) = 0 . . . . square and subtract the right side

  x^2 -5x -14 = 0 . . . . . . . . . . simplify

  (x -7)(x +2) = 0 . . . . . . . . . . factor

This has solutions x = 7 and x = -2. The attached graph shows how the extraneous solution of x=-2 arises from the negative branch of the square root function. It is a solution to the equation x = ±√(x+18) +2, but that is not the original equation.

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This problem statement suggests that an extraneous solution might arise in the solution of the equation ...

  2/x +1/2 = 5/4

The usual process for solution would be to multiply both sides of the equation by 4x (the least common denominator). That multiplier is zero when x=0, so has the potential of introducing x=0 as an extraneous solution. However, in this case, it does not seem to:

  4x(2/x +1/2) = 4x(5/4)

  8 +2x = 5x . . . . . . . . . . . eliminate parentheses

  8 = 3x . . . . . . . . subtract 2x

  x = 8/3  . . . . divide by 3

As a check, we can substitute this into the original equation:

  2/(8/3) + 1/2 = 5/4

  3/4 + 1/2 = 5/4 . . . . . true

It seems that extraneous solutions to rational equations can often be avoided by  recasting the equation to the form f(x) = 0. This will require the numerator of the rational function f(x) to be zero. Any numerator factors in common with those in the denominator can be cancelled, so do not contribute extraneous solutions.

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