what are extraneous solutions of an equation? Why do they sometimes occur in the process of solving rational equations? examples

When you have final answers, always go back and check them. If the original problem had a constraint as to what the solutions could be, then that is a tip off.

Extraneous solutions occur in the process of solving equations, particularly rational (fractional) equations, where one or more of the solutions obtained are not valid solutions to the original equation. These solutions may arise due to the presence of restrictions or limitations on the values of variables involved in the equation.

To understand why extraneous solutions occur, let's consider an example of solving a rational equation:

Example 1:
Let's solve the equation (x + 2)/(x - 1) = 2.

Step 1: Cross-multiply to eliminate the fraction.
(x + 2) = 2(x - 1)

Step 2: Expand and simplify.
x + 2 = 2x - 2
0 = x - 4

Step 3: Solve for x.
x = 4

In this case, we obtained a single solution: x = 4. However, we are not done yet.

Step 4: Check for extraneous solutions by substituting the obtained solution back into the original equation.

(x + 2)/(x - 1) = 2
(4 + 2)/(4 - 1) = 2
6/3 = 2
2 = 2

The obtained solution x = 4 satisfies the original equation, so it is a valid solution, and there are no extraneous solutions in this case.

Example 2:
Now, let's solve a similar equation involving square roots: √(x + 1) + 1 = 3.

Step 1: Isolate the square root term.
√(x + 1) = 3 - 1
√(x + 1) = 2

Step 2: Square both sides to eliminate the square root.
(x + 1) = 2²
x + 1 = 4
x = 4 - 1
x = 3

Step 3: Verify the obtained solution.

√(x + 1) + 1 = 3
√(3 + 1) + 1 = 3
√4 + 1 = 3
2 + 1 = 3
3 = 3

In this case, the obtained solution x = 3 satisfies the original equation. Hence, there are no extraneous solutions.

However, consider if we had squared both sides again without checking for extraneous solutions:

Step 2 (incorrect approach): Square both sides again.
(x + 1)² = (2²)²
(x + 1)² = 4²
(x + 1)² = 16

Step 3 (incorrect assumption): Solve for x.
x + 1 = 16
x = 16 - 1
x = 15

Now, if we check this solution:

√(x + 1) + 1 = 3
√(15 + 1) + 1 = 3
√16 + 1 = 3
4 + 1 = 3
5 = 3

This clearly does not satisfy the original equation, so x = 15 is an extraneous solution. It arose due to squaring both sides of the equation twice, which may introduce additional roots that are not valid in the original equation.

Therefore, it is essential to check all solutions obtained by substituting them back into the original equation to identify any extraneous solutions.