Solving X^4 - 5x^2 - 14 = 0 By Factoring A Comprehensive Guide

Hey there, math enthusiasts! Today, we're diving into the fascinating world of biquadratic equations, specifically the equation x⁴ - 5x² - 14 = 0. Don't let the fancy name intimidate you; we'll break it down step by step using a familiar technique: factoring. If you've grappled with quadratic equations before, you'll find this process surprisingly straightforward. We'll explore how to transform this seemingly complex equation into a manageable form, identify the key factors, and ultimately, uncover the solutions. So, grab your metaphorical pencils, and let's embark on this mathematical adventure together!

Understanding Biquadratic Equations

Before we jump into the solution, let's clarify what a biquadratic equation actually is. Simply put, it's a polynomial equation where the highest power of the variable is four, and it can be written in the general form ax⁴ + bx² + c = 0. Notice the pattern? The exponents are multiples of two (4 and 2), which is the key characteristic. Our equation, x⁴ - 5x² - 14 = 0, perfectly fits this description. The beauty of these equations lies in their resemblance to quadratic equations. We can leverage this similarity to solve them effectively. Think of it like this: we're going to perform a clever substitution that will turn our biquadratic equation into a quadratic one, which we already know how to handle. This technique is a powerful tool in our mathematical arsenal, allowing us to tackle seemingly complex problems with relative ease. So, with a solid understanding of biquadratic equations under our belts, let's move on to the exciting part – finding the solutions!

The Substitution Trick: Transforming to Quadratic

The first crucial step in solving our biquadratic equation is to transform it into a quadratic equation. We achieve this through a clever substitution. Let's introduce a new variable, say y, and define it as y = x². This might seem like a simple trick, but it's incredibly effective. Now, if y = x², then y² = (x²)² = x⁴. Suddenly, our equation x⁴ - 5x² - 14 = 0 takes on a new form. Replacing x⁴ with y² and x² with y, we get: y² - 5y - 14 = 0. Voila! We've successfully transformed our biquadratic equation into a quadratic equation in terms of y. This transformation is the cornerstone of our solution strategy. It allows us to apply familiar techniques for solving quadratic equations, such as factoring, the quadratic formula, or completing the square. By making this substitution, we've essentially simplified the problem, making it much more approachable. With our equation now in a quadratic form, we're ready to move on to the next step: factoring. Get ready to put your factoring skills to the test!

Factoring the Quadratic Equation

Now that we have our quadratic equation, y² - 5y - 14 = 0, it's time to factor it. Factoring involves expressing the quadratic expression as a product of two binomials. We need to find two numbers that multiply to -14 (the constant term) and add up to -5 (the coefficient of the y term). Think of it like a puzzle: we're searching for the right pieces that fit together perfectly. After a little thought, we can identify the numbers as -7 and 2. Indeed, (-7) * 2 = -14 and (-7) + 2 = -5. Therefore, we can factor the quadratic equation as: (y - 7)(y + 2) = 0. This factored form is incredibly useful because it allows us to easily find the values of y that make the equation true. Remember the zero-product property? It states that if the product of two factors is zero, then at least one of the factors must be zero. Applying this property to our factored equation, we set each factor equal to zero: y - 7 = 0 or y + 2 = 0. Solving these simple equations, we find the solutions for y: y = 7 or y = -2. Great! We've found the values of y, but remember, our original equation was in terms of x. We're not quite done yet. We need to go back and substitute x² for y to find the solutions for x. Let's move on to the next step and uncover the final answers!

Back-Substitution: Finding the Values of x

We've successfully found the values of y, which are y = 7 and y = -2. But remember, our goal is to solve for x. We need to undo the substitution we made earlier, where we defined y as x². So, we'll substitute x² back in for y in each of our solutions. This gives us two equations: x² = 7 and x² = -2. Now, we need to solve each of these equations for x. To do this, we take the square root of both sides of each equation. Remember, when taking the square root, we need to consider both the positive and negative roots. For x² = 7, taking the square root gives us x = ±√7. These are two of our solutions! For x² = -2, taking the square root gives us x = ±√(-2). Since we have a negative number under the square root, we introduce the imaginary unit, i, where i = √(-1). So, √(-2) can be written as √2 * √(-1) = i√2. Therefore, the solutions for this equation are x = ±i√2. And there you have it! We've found all four solutions for our original biquadratic equation. Let's summarize our findings in the next section.

The Solutions: A Comprehensive Overview

After navigating the world of biquadratic equations, employing a clever substitution, and mastering the art of factoring, we've arrived at the solutions for x⁴ - 5x² - 14 = 0. Our journey has led us to four distinct solutions: x = √7, x = -√7, x = i√2, and x = -i√2. Notice that we have two real solutions (√7 and -√7) and two complex solutions (i√2 and -i√2). This is a common characteristic of biquadratic equations; they can have a mix of real and complex roots. To recap, we began by recognizing the equation as a biquadratic equation. We then made the substitution y = x² to transform it into a quadratic equation. We factored the quadratic equation, solved for y, and then back-substituted to find the values of x. This process highlights the power of algebraic manipulation and the interconnectedness of different mathematical concepts. By recognizing patterns and applying appropriate techniques, we can tackle even seemingly challenging equations. So, the final answer, encompassing all the solutions, is x = ±√7 and x = ±i√2. You've successfully conquered this biquadratic equation! Congratulations!

Therefore, the correct answer is:

A. x = ±√7 and x = ±i√2