Extraneous Solutions

When solving mathematical equations, especially those involving complex operations like square roots, logarithms, or rational expressions, it's possible to encounter solutions that appear valid during the solving process but do not satisfy the original equation. These are known as extraneous solutions.

In this article, we will learn the technique for solving Extraneous Equations.

What are Extraneous Equations?

Extraneous equations are equations that arise during the process of solving an algebraic equation but are not true solutions to the original equation. These false solutions typically occur when both sides of an equation are raised to an even power or when operations like squaring both sides of an equation are performed. This process can introduce solutions that do not satisfy the original equation.

How to Solve Extraneous Equations

Follow these steps to solve Extraneous Equations:

• Isolate Radical
• Square both Sides
• Simplify
• Rearrange
• Factorize
• Solve
• Check

All these steps are shown in the problems.

Extraneous Equations: Solved Problems

Problem 1: Solve \sqrt{x + 3} = x - 1.

Solution:

Isolate radical: \sqrt{x + 3} = x - 1.

Square both sides: (\sqrt{x + 3})^2 = (x - 1)^2.

Simplify: x+3 = x² - 2x+1.

Rearrange: x² - 3x - 2 = 0.

Factorize: (x - 4)(x+1) = 0.

Solve: x = 4 and x = - 1

Check:

For x = 4: \sqrt{4 + 3} = 4 - 1 \rightarrow 3 = 3 (Valid)

For x = - 1: \sqrt{-1 + 3} = -1 - 1 \rightarrow \sqrt{2} \neq -2−1+3​=−1 (Invalid)

Extraneous Solution: x = - 1

True Solution: x = 4

Problem 2: Solve x + \sqrt{x} = 6.

Solution:

Isolate the square root: \sqrt{x} = 6-x.

Square both sides: x = (6 - x)².

Expand: x = 36 - 12x+x².

Rearrange: x² - 13x+36 = 0.

Factorize: (x - 9)(x - 4) = 0.

Solve: x = 9 and x = 4.

Check:

For x = 9: 9 + \sqrt{9} = 6 \rightarrow 12 = 6 (Invalid)

For x = 4: 4 + \sqrt{4} = 6 \rightarrow 6 = 6 (Valid)

Extraneous Solution: x = 9

True Solution: x = 4

Problem 3: Solve \sqrt{2x + 3} = x - 2.

Solution:

Isolate the radical:\sqrt{2x + 3} = x - 2.

Square both sides:(\sqrt{2x + 3})^2 = (x - 2)^2.

Simplify: 2x+3 = x² - 4x+4.

Rearrange: x² - 6x+1 = 0.

Use quadratic formula: x = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}​.

Check:

For x = 3 + 2\sqrt{2}​: \sqrt{2(3 + 2\sqrt{2}) + 3} = (3 + 2\sqrt{2}) (Valid)

For x = 3 - 2\sqrt{2}​: \sqrt{2(3 - 2\sqrt{2}) + 3} = (3 - 2\sqrt{2}) (Invalid)

Extraneous Solution: x = 3 - 2\sqrt{2}​

True Solution: x = 3 + 2\sqrt{2}

Problem 4: Solve \sqrt{x + 4} = x + 2.

Solution:

Isolate the radical: \sqrt{x + 4} = x + 2.

Square both sides: (\sqrt{x + 4})^2 = (x + 2)^2.

Simplify: x+4 = x²+4x+4.

Rearrange: x²+3x = 0.

Factorize: x(x+3) = 0.

Solve: x = 0 and x = - 3.

Check:

For x=0: \sqrt{0 + 4} = 0 + 2 \rightarrow 2 = 2 (Valid)

For x=−3: \sqrt{-3 + 4} = -3 + 2 \rightarrow 1 \neq -1 (Invalid)

Extraneous Solution: x = - 3

True Solution: x = 0

Problem 5: Solve \sqrt{x - 2} + 1 = x.

Solution:

Isolate the radical: \sqrt{x - 2} = x - 1.

Square both sides: (\sqrt{x - 2})^2 = (x - 1)^2.

Simplify: x - 2 = x² - 2x+1.

Rearrange: x² - 3x+3 = 0.

Quadratic formula gives complex roots, indicating no real solutions.

Extraneous Solutions: None

True Solution: None

Problem 6: Solve \sqrt{x + 6} = 2 - x.

Solution:

Isolate the radical: \sqrt{x + 6} = 2 - x.

Square both sides: (\sqrt{x + 6})^2 = (2 - x)^2.

Simplify: x+6 = 4 - 4x+x².

Rearrange: x² - 5x - 2 = 0.

Solve: x = \frac{5 \pm \sqrt{33}}{2}.

Check:

For x = \frac{5 + \sqrt{33}}{2}​​: (Valid)

For x = \frac{5 - \sqrt{33}}{2}: (Invalid)

Extraneous Solution: x = \frac{5 - \sqrt{33}}{2}

True Solution: x = \frac{5 + \sqrt{33}}{2}

Extraneous Equations: Worksheet

Q1. Equation: \sqrt{x + 5} = x - 3

Q2. Equation: \frac{1}{x - 2} + 2 = \frac{2}{x - 2}

Q3. Equation: x = \frac{2x}{x - 1}

Q4. Equation: \log(x - 2) + \log(x + 3) = 1

Q5. Equation: \sqrt{2x + 3} + \sqrt{x - 1} = 5

Q6. Equation: 3x - 4 = \frac{2x}{x - 3} + 13x - 4 = \frac{2x}{x - 3} + 1

Q7. Equation: \frac{3}{x} = x + 1

Q8. Equation: \\sqrt{x + 2} - \sqrt{x - 1} = 1

Q9. Equation: 9x - 1 = \sqrt{x + 9}

Q10. Equation: \frac{x - 1}{x + 1} = \frac{3x + 5}{2x - 1}

Conclusion

When solving equations, especially those involving square roots, logarithms, fractions, or other complex operations, it’s crucial to check each potential solution in the original equation. Extraneous solutions often arise when the steps to solve the equation involve squaring both sides, multiplying by expressions that could be zero, or other manipulations that might introduce solutions not valid in the original context. Always verify your solutions to ensure they are legitimate.