Solving √x-2 - 5 = 0 And Identifying Extraneous Solutions

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Introduction

In this article, we will delve into the process of solving the equation √x-2 - 5 = 0. This equation involves a square root, which means we need to be particularly careful about extraneous solutions. An extraneous solution is a value that we obtain during the solving process that satisfies the transformed equation but not the original equation. Therefore, checking our solution is crucial in this type of problem. We will go through each step meticulously, explaining the rationale behind each operation and finally verify whether our solution is valid or extraneous.

Step-by-Step Solution

Isolating the Square Root

Our first goal is to isolate the square root term on one side of the equation. This makes it easier to eliminate the square root. The given equation is:

√x-2 - 5 = 0

To isolate the square root, we add 5 to both sides of the equation:

√x-2 = 5

Squaring Both Sides

Now that we have isolated the square root, we can eliminate it by squaring both sides of the equation. This is a standard technique for solving equations involving square roots. Squaring both sides gives us:

(√x-2)² = 5²

This simplifies to:

x - 2 = 25

Solving for x

Next, we solve for x by adding 2 to both sides of the equation:

x = 25 + 2

Thus,

x = 27

Checking the Solution

Now comes the critical part: checking whether our solution is valid or extraneous. We substitute x = 27 back into the original equation:

√x-2 - 5 = 0

Substitute x = 27:

√27-2 - 5 = 0

Simplify:

√25 - 5 = 0

5 - 5 = 0

0 = 0

Since the equation holds true, our solution x = 27 is valid.

Discussion on Extraneous Solutions

Extraneous solutions can arise when we perform operations that are not reversible in the same way, such as squaring both sides of an equation. Squaring both sides can introduce solutions that did not exist in the original equation because it treats both positive and negative roots similarly. For instance, consider the equation:

√x = -3

This equation has no real solution because the square root of a number cannot be negative. However, if we square both sides, we get:

x = 9

But if we substitute x = 9 back into the original equation:

√9 = -3

3 = -3

This is clearly false, so x = 9 is an extraneous solution. This example underscores the necessity of checking solutions in equations involving radicals or rational exponents.

Why Extraneous Solutions Occur

Extraneous solutions occur because squaring both sides of an equation can introduce additional solutions that do not satisfy the original equation. When we square both sides, we are essentially saying that if A = B, then A² = B². However, the converse is not necessarily true. If A² = B², it does not always follow that A = B. It could also be the case that A = -B.

Consider a more general case:

Suppose we have the equation:

A(x) = B(x)

Squaring both sides gives:

A(x)² = B(x)²

This is equivalent to:

A(x)² - B(x)² = 0

Which can be factored as:

(A(x) - B(x))(A(x) + B(x)) = 0

This implies that either:

A(x) - B(x) = 0 or A(x) + B(x) = 0

The first equation, A(x) - B(x) = 0, is our original equation, A(x) = B(x). However, the second equation, A(x) + B(x) = 0, or A(x) = -B(x), introduces potential solutions that do not satisfy the original equation. These are the extraneous solutions.

Strategies to Avoid Extraneous Solutions

  1. Isolate the Radical: Always isolate the radical term before squaring. This simplifies the equation and reduces the chances of introducing extraneous solutions.
  2. Check Your Solutions: After solving the equation, substitute each solution back into the original equation to verify its validity. Discard any solutions that do not satisfy the original equation.
  3. Understand the Domain: Consider the domain of the radical expression. For example, the square root of a negative number is not a real number, so solutions that lead to a negative number under the radical are extraneous.

Examples of Extraneous Solutions in Different Contexts

Example 1: A More Complex Radical Equation

Solve: √(3x + 1) = x - 1

  1. Square both sides: 3x + 1 = (x - 1)²
  2. Expand: 3x + 1 = x² - 2x + 1
  3. Rearrange: x² - 5x = 0
  4. Factor: x(x - 5) = 0
  5. Solutions: x = 0 or x = 5

Check x = 0:

√(3(0) + 1) = 0 - 1

√1 = -1

1 = -1 (False)

So, x = 0 is an extraneous solution.

Check x = 5:

√(3(5) + 1) = 5 - 1

√16 = 4

4 = 4 (True)

So, x = 5 is a valid solution. In this case, x = 0 is an extraneous solution introduced by squaring both sides of the equation.

Example 2: Rational Equations

Extraneous solutions can also occur in rational equations, where variables appear in the denominator. These occur when a solution makes the denominator zero, which is undefined.

Solve: (1/x) = (x - 2)/(2x)

  1. Multiply both sides by 2x: 2 = x - 2
  2. Solve for x: x = 4

Check x = 4 in the original equation:

(1/4) = (4 - 2)/(2(4))

(1/4) = (2/8)

(1/4) = (1/4) (True)

So, x = 4 is a valid solution.

Now, consider the equation:

(1/(x - 2)) = 3/(x - 2)

If we cross-multiply, we get:

x - 2 = 3(x - 2)

x - 2 = 3x - 6

2x = 4

x = 2

However, if we substitute x = 2 into the original equation, we get:

(1/(2 - 2)) = 3/(2 - 2)

(1/0) = (3/0)

This is undefined, so x = 2 is an extraneous solution. It arises because x = 2 makes the denominator zero.

Real-World Applications

The concept of extraneous solutions is not just a mathematical curiosity; it has real-world applications, especially in physics and engineering. For example, when modeling physical phenomena using equations, extraneous solutions might represent scenarios that are not physically possible.

Projectile Motion

Consider a problem involving projectile motion, where the height of a projectile is given by a quadratic equation. Solving the equation to find the time when the projectile hits the ground might yield two solutions. However, a negative time value would be an extraneous solution because time cannot be negative in this context. Therefore, understanding and identifying extraneous solutions is crucial for making accurate predictions in real-world scenarios.

Electrical Circuits

In electrical circuit analysis, solving equations for current or voltage might produce extraneous solutions. For example, a negative current value might be mathematically valid but physically impossible in a particular circuit configuration. Therefore, engineers need to be mindful of extraneous solutions when designing and analyzing circuits.

Conclusion

Solving equations involving square roots requires careful attention to detail, especially when it comes to extraneous solutions. In the case of the equation √x-2 - 5 = 0, we found that x = 27 is indeed a valid solution after verifying it in the original equation. The potential for extraneous solutions highlights the importance of always checking our solutions, particularly when we perform operations like squaring both sides of an equation. By understanding why extraneous solutions occur and how to identify them, we can ensure the accuracy and validity of our mathematical results. Remember to isolate the radical, square both sides (if necessary), solve for the variable, and, most importantly, check your solutions in the original equation.

In summary, solving radical equations is a multi-step process that requires careful algebraic manipulation and a thorough understanding of extraneous solutions. By mastering these concepts, you can confidently tackle a wide range of mathematical problems and ensure the accuracy of your solutions.