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Extraneous roots, also known as spurious roots, are roots that are not true solutions to an equation. These roots are often introduced during the process of solving an equation and can lead to incorrect results if not properly identified and eliminated. In this article, we will delve deeper into the concept of extraneous roots, provide definitions, examples, an FAQ section, and a quiz to help you test your knowledge.

Definitions

Before diving into the examples and explanations, it is important to define some key terms related to extraneous roots.

Equation: An equation is a mathematical statement that two expressions are equal. For example, 2x + 3 = 7 is an equation.

Solution: A solution is a value that, when substituted into an equation, makes the equation true. For example, x = 2 is a solution to the equation 2x + 3 = 7.

Root: A root of an equation is a solution to the equation. For example, the roots of the equation x^2 – 4 = 0 are x = 2 and x = -2.

Extraneous Root: An extraneous root is a root that is not a true solution to an equation. It is introduced during the process of solving an equation and can lead to incorrect results if not properly identified and eliminated.

Examples

To understand extraneous roots better, let’s take a look at some examples.

Example 1: Solve the equation x + 2 = 5.

Subtracting 2 from both sides of the equation, we get x = 3. This is the only solution to the equation, and it is not extraneous.

Example 2: Solve the equation ?(x + 3) = 2.

Squaring both sides of the equation, we get x + 3 = 4. Solving for x, we get x = 1. This is a valid solution to the equation. However, if we plug in x = 1 back into the original equation, we get ?4 = 2, which is true. But if we plug in x = -4 (which we get after squaring both sides of the equation) back into the original equation, we get ?(-1) = i, which is not a real number and thus not a valid solution to the original equation. Therefore, x = -4 is an extraneous root.

Example 3: Solve the equation 2/x = 4.

Multiplying both sides of the equation by x, we get 2 = 4x. Solving for x, we get x = 1/2. This is the only solution to the equation, and it is not extraneous.

Example 4: Solve the equation x^2 = 4.

Taking the square root of both sides of the equation, we get x = ±2. These are both valid solutions to the equation, and neither is extraneous.

Example 5: Solve the equation x^2 + 5x + 6 = 0.

Factoring the equation, we get (x + 3)(x + 2) = 0. Solving for x, we get x = -3 and x = -2. These are both valid solutions to the equation, and neither is extraneous.

Example 6: Solve the equation x^2 – 2x + 1 = 0.

Factoring the equation, we get (x – 1)^2 = 0. Solving for x, we get x = 1. This is the only solution to the equation, and it is not extraneous.

Example 7: Solve the equation (x – 2)/(x + 3) = (x – 4)/(x – 1).

Multiplying both sides of the equation by the denominators, we get (x – 2)(x – 1) = (x – 4)(x + 3). Expanding both sides, we get x^2 – 3x + 2 = x^2 – x – 12. Simplifying, we get -2x = -14, or x = 7. However, if we plug in x = -3 (which was a denominator in the original equation) back into the original equation, we get 1/0 = undefined, which is not a valid solution to the equation. Therefore, x = -3 is an extraneous root.

Example 8: Solve the equation ?(x – 2) = 3 – x.

Squaring both sides of the equation, we get x – 2 = (3 – x)^2. Expanding, we get x – 2 = 9 – 6x + x^2. Rearranging, we get x^2 – 7x + 11 = 0. Solving for x using the quadratic formula, we get x = (7 ± ?5)/2. However, if we plug in x = (7 – ?5)/2 back into the original equation, we get ?(1 + ?5)/2, which is not a valid solution to the equation. Therefore, x = (7 – ?5)/2 is an extraneous root.

Example 9: Solve the equation log(x) = log(4) + log(3).

Using the product rule of logarithms, we can rewrite the equation as log(x) = log(12). Therefore, x = 12. This is the only solution to the equation, and it is not extraneous.

Example 10: Solve the equation (x + 2)/(x – 3) + (x – 4)/(x + 1) = 0.

Multiplying both sides of the equation by (x – 3)(x + 1), we get (x + 2)(x + 1) + (x – 4)(x – 3) = 0. Expanding and simplifying, we get x^2 – 8x – 11 = 0. Solving for x using the quadratic formula, we get x = 4 + ?27 and x = 4 – ?27. However, if we plug in x = -1 (which was a denominator in the original equation) back into the original equation, we get 1/0 = undefined, which is not a valid solution to the equation. Therefore, x = -1 is an extraneous root.

FAQ

• How are extraneous roots introduced in an equation? A: Extraneous roots are introduced when we perform operations on both sides of an equation that are not valid for all values of the variable. For example, squaring both sides of an equation can introduce extraneous roots.
• How can we identify extraneous roots? A: To identify extraneous roots, we need to plug in each solution we obtained back into the original equation and check if it satisfies the equation. If it does not, then it is an extraneous root.
• Are extraneous roots always introduced when we square both sides of an equation? A: No, extraneous roots are not always introduced when we square both sides of an equation. However, squaring both sides is a common operation that can introduce extraneous roots.

Quiz

1. Solve the equation ?(x + 2) = 4 – x. a) x = -5 b) x = -3 c) x = 2 d) x = 6
2. Solve the equation (x + 3)/(x – 1) = 2. a) x = -1/3 b) x = 0 c) x = 2 d) x = 4
3. Solve the equation log(x) + log(3) = log(15). a) x = 5 b) x = 15 c) x = 45 d) x = 225
4. Solve the equation (x + 2)/(x – 3) = 1/(x + 1). a) x = -1 b) x = -2 c) x = 0 d) x = 4
5. Solve the equation (x – 1)/(x + 2) = 3. a) x = -1/2 b) x = 2 c) x = 5 d) x = 7
6. Solve the equation ?(x – 5) – ?(x – 2) = 1. a) x = 6 b) x = 9 c) x = 11 d) x = 14
7. Solve the equation log(x) – log(5) = log(2). a) x = 5/2 b) x = 2 c) x = 5 d) x = 10
8. Solve the equation (x + 1)/(x – 2) – 2 = (x – 4)/(x + 3). a) x = -1 b) x = 0 c) x = 2 d) x = 4
9. Solve the equation ?(x + 1) + ?(x – 2) = 5. a) x = 1/2 b) x = 2 c) x = 7 d) x = 11
10. Solve the equation log(x – 1) – log(x + 1) = log(2). a) x = -1/2 b) x = 1/2 c) x = 3/2 d) x = 5/2

Conclusion

In conclusion, extraneous roots are roots of an equation that are obtained by solving the equation but are not valid solutions to the original equation. Extraneous roots are introduced when we perform operations on both sides of an equation that are not valid for all values of the variable. To identify extraneous roots, we need to plug in each solution we obtained back into the original equation and check if it satisfies the equation. If it does not, then it is an extraneous root. It is important to check for extraneous roots when solving equations, especially when using operations such as squaring both sides of an equation. By being aware of extraneous roots, we can avoid errors in our calculations and arrive at the correct solutions to the equations we are solving.