Solving For U: A Real Number Equation Explained

by Alex Johnson 48 views

Let's dive into solving an equation where we need to find the real number value of u. The equation we're tackling is:

u = √(10u - 21)

This involves a square root, so we'll need to be a bit careful with our steps to make sure we don't introduce any extraneous solutions. Extraneous solutions are basically solutions that pop up during the solving process but don't actually satisfy the original equation. Think of them as imposters! To start, we'll square both sides of the equation to get rid of the square root.

Step-by-Step Solution

1. Squaring Both Sides

Squaring both sides of the equation u = √(10u - 21) gives us:

u² = (√(10u - 21))²

Which simplifies to:

u² = 10u - 21

2. Rearranging the Equation

Now, let's rearrange this into a quadratic equation by moving all the terms to one side:

u² - 10u + 21 = 0

3. Factoring the Quadratic

We need to factor this quadratic equation. We're looking for two numbers that multiply to 21 and add up to -10. Those numbers are -3 and -7. So, we can factor the equation as:

(u - 3)(u - 7) = 0

4. Finding Potential Solutions

Setting each factor equal to zero gives us our potential solutions:

u - 3 = 0 or u - 7 = 0

Which means:

u = 3 or u = 7

5. Checking for Extraneous Solutions

This is a crucial step. We need to plug each potential solution back into the original equation u = √(10u - 21) to see if it actually works.

Checking u = 3

Substitute u = 3 into the original equation:

3 = √(10(3) - 21)

3 = √(30 - 21)

3 = √9

3 = 3

This is true, so u = 3 is a valid solution.

Checking u = 7

Substitute u = 7 into the original equation:

7 = √(10(7) - 21)

7 = √(70 - 21)

7 = √49

7 = 7

This is also true, so u = 7 is also a valid solution.

Conclusion

Therefore, the solutions to the equation u = √(10u - 21) are u = 3 and u = 7. Both solutions satisfy the original equation, so we don't have any extraneous solutions in this case.

Why Checking for Extraneous Solutions is Important

When solving equations involving radicals (like square roots, cube roots, etc.), you must always check your solutions. Squaring both sides of an equation can sometimes introduce solutions that don't actually work in the original equation. Think about it like this: if a = b, then a² = b². However, if a² = b², it doesn't necessarily mean that a = b (it could be that a = -b). That's why the checking step is so vital. Skipping it can lead to incorrect answers.

Let's Consider a Similar Example (With an Extraneous Solution)

Suppose we had the equation:

x = √(x + 6)

Squaring both sides gives us:

x² = x + 6

Rearranging, we get:

x² - x - 6 = 0

Factoring, we find:

(x - 3)(x + 2) = 0

So, our potential solutions are x = 3 and x = -2.

Let's check them:

  • If x = 3: 3 = √(3 + 6) which simplifies to 3 = √9, so 3 = 3. This solution works!
  • If x = -2: -2 = √(-2 + 6) which simplifies to -2 = √4, so -2 = 2. This is not true! Therefore, x = -2 is an extraneous solution. The only valid solution is x = 3.

This example highlights why the checking step is non-negotiable.

Tips for Solving Radical Equations

  • Isolate the Radical: Before squaring (or cubing, etc.), make sure the radical term is isolated on one side of the equation. This simplifies the process and reduces the chance of making errors.
  • Square Both Sides: Square both sides of the equation to eliminate the square root. If you have a cube root, cube both sides, and so on.
  • Solve the Resulting Equation: After eliminating the radical, you'll be left with a polynomial equation. Solve this equation using standard techniques (factoring, quadratic formula, etc.).
  • Check Your Solutions: Always, always, always check your solutions in the original equation to identify and discard any extraneous solutions. This is the most important step!
  • Be Mindful of the Domain: Remember that the expression inside a square root must be non-negative. This can help you anticipate potential extraneous solutions even before you start checking them.

By following these steps carefully, you can confidently solve radical equations and avoid the pitfalls of extraneous solutions. Remember, practice makes perfect!

More Complex Scenarios

What happens if you have more than one radical in the equation? The general strategy remains the same, but you might need to repeat the process of isolating a radical and squaring both sides multiple times. For example, consider the equation:

√(x + 5) + √(x) = 5

  1. Isolate one radical: √(x + 5) = 5 - √(x)

  2. Square both sides: (√(x + 5))² = (5 - √(x))² x + 5 = 25 - 10√(x) + x

  3. Simplify and isolate the remaining radical: 10√(x) = 20 √(x) = 2

  4. Square both sides again: x = 4

  5. Check the solution: √(4 + 5) + √(4) = √9 + √4 = 3 + 2 = 5 The solution x = 4 is valid.

The key takeaway is to strategically isolate and eliminate radicals one at a time until you arrive at a solvable equation.

Advanced Techniques

In some cases, you might encounter equations that are difficult or impossible to solve algebraically. In such situations, numerical methods can be employed to approximate the solutions. Techniques like the Newton-Raphson method or bisection method can provide accurate approximations of the roots of the equation.

Furthermore, graphical methods can be used to visualize the equation and estimate the solutions. By plotting the functions on both sides of the equation, the points of intersection represent the solutions.

Understanding these advanced techniques can be valuable when dealing with complex radical equations that defy traditional algebraic approaches.

In conclusion, solving radical equations requires a systematic approach, careful attention to detail, and a thorough understanding of the potential for extraneous solutions. By mastering these techniques, you can confidently tackle a wide range of problems involving radicals.

For more information on solving equations, you can visit Khan Academy's Algebra Section.