Solve X^4 - 5x^2 - 14 = 0 By Factoring: Step-by-Step

by Mei Lin 53 views

Hey everyone! Today, let's dive into solving a quartic equation using factoring. Specifically, we're going to tackle the equation x^4 - 5x^2 - 14 = 0. This might look intimidating at first, but don't worry, we'll break it down step by step. Factoring is a powerful technique, and mastering it will help you solve many polynomial equations. So, grab your pencils, and let's get started!

1. Understanding the Problem and the Factoring Strategy

Before we jump into the solution, let's understand what we're dealing with. The equation x^4 - 5x^2 - 14 = 0 is a quartic equation because the highest power of x is 4. Quartic equations can be tricky to solve directly, but in many cases, we can use clever substitutions and factoring techniques to simplify them. In this particular case, we can notice a pattern: the powers of x are even (4 and 2). This suggests we can use a substitution to transform it into a quadratic equation, which we know how to solve.

Our main keyword here is factoring, so we'll focus on that. The general strategy involves the following steps:

  1. Substitution: Replace x² with a new variable (like y) to turn the quartic equation into a quadratic equation.
  2. Factor the Quadratic: Factor the resulting quadratic equation into two binomials.
  3. Solve for the New Variable: Find the roots of the factored quadratic equation.
  4. Back-Substitute: Replace the new variable with x² again.
  5. Solve for x: Solve the resulting quadratic equations for x. This will give us the solutions to the original quartic equation.

This might sound like a lot of steps, but it's a systematic approach that will make the problem much easier to handle. The beauty of this method is that it transforms a complex-looking equation into a familiar form that we can easily solve. Keep in mind, factoring relies heavily on recognizing patterns and applying algebraic rules correctly. So, let's put this strategy into action and see how it works!

2. The Substitution Step: Transforming the Equation

The first step in our journey to solve the equation x^4 - 5x^2 - 14 = 0 is to make a clever substitution. This substitution will transform our quartic equation into a more manageable quadratic equation. The key observation here is that x⁴ can be written as (x²)². This allows us to see a quadratic pattern. Let's introduce a new variable, say y, such that:

y = x²

Now, we can rewrite our original equation in terms of y. Substituting y for x² and y² for x⁴, we get:

y² - 5y - 14 = 0

Wow! Look at that! Our quartic equation has magically transformed into a quadratic equation. This is a significant simplification. Quadratic equations are much easier to factor and solve compared to quartics. This substitution technique is a common trick in algebra, and it's incredibly useful for dealing with equations that have a quadratic-like structure. By recognizing the pattern and making the right substitution, we've made our problem much more approachable. The goal here is to transform a complex problem into a simpler one that we already know how to solve. So, give yourself a pat on the back, guys – we're making great progress!

Now that we have a quadratic equation, the next step is to factor it. This is where our factoring skills will come into play. Remember, factoring involves finding two binomials that multiply together to give us our quadratic expression. Let's move on to the next section and see how we can crack this quadratic!

3. Factoring the Quadratic Equation: Finding the Right Factors

We've successfully transformed our quartic equation into a quadratic equation: y² - 5y - 14 = 0. Now comes the fun part – factoring! Factoring a quadratic equation involves finding two binomials that, when multiplied together, give us the quadratic expression. To factor y² - 5y - 14, we need to find two numbers that:

  • Multiply to -14 (the constant term)
  • Add up to -5 (the coefficient of the y term)

Let's think about the factors of -14. We have the following possibilities:

  • -1 and 14
  • 1 and -14
  • -2 and 7
  • 2 and -7

Out of these pairs, which one adds up to -5? Bingo! It's 2 and -7. So, we can rewrite our quadratic equation as:

(y + 2)(y - 7) = 0

That's it! We've factored the quadratic equation. Factoring is like a puzzle – you need to find the right pieces that fit together. In this case, the pieces are the binomials (y + 2) and (y - 7). Now that we have our factored equation, we can move on to the next step: solving for y. This will lead us closer to finding the solutions for our original equation in terms of x. You guys are doing awesome! We're steadily marching towards the final answer.

4. Solving for y: Finding the Roots of the Quadratic

We've successfully factored our quadratic equation into (y + 2)(y - 7) = 0. Now, we need to solve for y. This step is based on a fundamental principle in algebra: the zero-product property. The zero-product property states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if ab = 0, then either a = 0 or b = 0 (or both). Applying this property to our factored equation, we have:

(y + 2) = 0 or (y - 7) = 0

Now, we can solve each of these linear equations separately:

  • For y + 2 = 0, subtract 2 from both sides to get y = -2.
  • For y - 7 = 0, add 7 to both sides to get y = 7.

So, we have found two solutions for y: y = -2 and y = 7. These are the roots of our quadratic equation. We're halfway there! But remember, we're not trying to solve for y; we're trying to solve for x. We introduced y as a temporary variable to make our equation easier to handle. Now, we need to go back to our original variable, x. This is where the back-substitution step comes in. We'll replace y with x² and solve the resulting equations for x. Let's move on to the next section and see how this works!

5. Back-Substitution: Returning to the Original Variable

We've found the solutions for y: y = -2 and y = 7. Now, it's time to bring back our original variable, x. Remember, we made the substitution y = x². So, to get back to x, we need to replace y with x² in our solutions. This gives us two new equations:

  • x² = -2
  • x² = 7

These are now simple quadratic equations in terms of x. We can solve each of them by taking the square root of both sides. But remember, when we take the square root, we need to consider both the positive and negative roots. This is because both the positive and negative values, when squared, will give us the same result. For example, both 2² and (-2)² are equal to 4. Let's solve each equation separately.

First, let's consider x² = 7. Taking the square root of both sides, we get:

x = ±√7

So, we have two solutions: x = √7 and x = -√7. These are real solutions. Now, let's tackle the second equation: x² = -2. Taking the square root of both sides, we get:

x = ±√(-2)

Wait a minute! We have the square root of a negative number. This means we'll be dealing with imaginary numbers. Remember that √(-1) is defined as the imaginary unit, denoted by i. So, we can rewrite √(-2) as √(2 * -1) = √(2) * √(-1) = √2 * i. Therefore, the solutions for this equation are:

x = ±i√2

So, we have two more solutions: x = i√2 and x = -i√2. These are imaginary solutions. We're almost there! We've found all the solutions for x. Let's gather them together in the final step.

6. The Final Solutions: Putting It All Together

We've gone through all the steps, and now it's time to present our final solutions for the equation x^4 - 5x^2 - 14 = 0. We found four solutions in total:

  • x = √7
  • x = -√7
  • x = i√2
  • x = -i√2

These are the roots of the equation. We found two real solutions (√7 and -√7) and two imaginary solutions (i√2 and -i√2). This makes sense because a quartic equation can have up to four solutions, and they can be real or complex (imaginary). We successfully used factoring and substitution to solve this equation. You guys did a fantastic job following along! This problem demonstrates the power of algebraic techniques in simplifying complex equations. By recognizing patterns, making substitutions, and applying factoring rules, we can tackle even the most intimidating-looking problems. Remember, practice makes perfect. The more you practice these techniques, the more comfortable you'll become with them. So, keep solving equations, and keep exploring the fascinating world of algebra! I believe this also corresponds to option C in your original problem options. Great job, everyone!

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