Polynomial Roots: How Many For 7 + 5x⁴ - 3x²?

Hey everyone! Today, we're going to tackle a fascinating question in the realm of mathematics: how many roots, both real and complex, does the polynomial 7 + 5x⁴ - 3x² have in total? This might sound intimidating at first, but trust me, we'll break it down step by step and make it crystal clear. So, buckle up and let's dive into the world of polynomials and their roots!

Understanding the Fundamentals of Polynomial Roots

Before we jump into the specifics of our polynomial, let's establish a solid foundation by understanding what roots actually are and the different types we encounter. In simple terms, a root of a polynomial is a value that, when plugged into the polynomial, makes the entire expression equal to zero. These roots are also known as solutions or zeros of the polynomial equation. For example, if we have a polynomial like x² - 4, the roots would be 2 and -2 because when you substitute either of these values for x, the expression becomes zero. Now, roots can be classified into two main categories: real roots and complex roots. Real roots are those that can be represented on the number line – they are the familiar numbers we use every day. Think of integers, fractions, and even irrational numbers like the square root of 2. On the other hand, complex roots involve the imaginary unit 'i', which is defined as the square root of -1. These roots pop up when we try to solve equations that don't have solutions within the realm of real numbers. They take the form of a + bi, where 'a' and 'b' are real numbers and 'i' is the imaginary unit. The beauty of complex numbers is that they allow us to solve any polynomial equation, regardless of its complexity. The Fundamental Theorem of Algebra plays a crucial role in our quest to find the number of roots. This theorem, a cornerstone of algebra, states that a polynomial of degree 'n' has exactly 'n' roots, counting multiplicity, in the complex number system. This means that if we have a polynomial with the highest power of x being 4 (a quartic polynomial), we know for sure that it will have exactly 4 roots, some of which may be real, some may be complex, and some may even be repeated. Multiplicity refers to the number of times a particular root appears as a solution. For instance, in the polynomial (x - 2)² = 0, the root 2 has a multiplicity of 2 because it appears twice as a solution. Understanding the Fundamental Theorem of Algebra is key to solving our initial question about the roots of the polynomial 7 + 5x⁴ - 3x². It provides us with the assurance that we know how many roots to expect, guiding our exploration and analysis of the equation. Now that we have a solid grasp of the basics, let's move on to analyzing our specific polynomial and uncovering its roots.

Analyzing the Polynomial 7 + 5x⁴ - 3x²

Alright, let's get our hands dirty with the polynomial in question: 7 + 5x⁴ - 3x². The first thing we need to do is identify the degree of the polynomial. Remember, the degree is simply the highest power of the variable 'x'. In this case, we see that the highest power is 4 (in the term 5x⁴), so we're dealing with a quartic polynomial, meaning it's a polynomial of degree 4. Now, according to our trusty companion, the Fundamental Theorem of Algebra, we know that this polynomial must have exactly 4 roots, counting multiplicity. This is a crucial piece of information because it sets the stage for our search. We know we need to find four roots in total, but the question remains: how many of these roots are real, and how many are complex? To answer this, we need to dive a bit deeper into the structure of the polynomial. Notice that our polynomial is missing the x³ and x terms. This might seem like a minor detail, but it actually hints at a potential simplification strategy. We can use a clever trick called substitution to transform this quartic polynomial into a quadratic one, which we know how to solve easily. Let's make the substitution y = x². This means that x⁴ becomes y² (since x⁴ = (x²)²). Now, if we replace x² with y and x⁴ with y² in our original polynomial, we get a new expression: 7 + 5y² - 3y. Rearranging the terms, we have a quadratic equation in terms of y: 5y² - 3y + 7 = 0. Isn't that neat? We've successfully transformed a degree 4 polynomial into a degree 2 polynomial! Now, we can use our knowledge of quadratic equations to find the roots of this new equation in terms of y. Once we have the values of y, we can then substitute back x² for y and solve for x, which will give us the roots of our original polynomial. This substitution technique is a powerful tool in polynomial analysis, allowing us to simplify complex expressions and make them more manageable. It's like having a mathematical Swiss Army knife in your toolkit!

Solving the Transformed Quadratic Equation

Now that we've transformed our quartic polynomial into a quadratic equation, 5y² - 3y + 7 = 0, it's time to find its roots. We have a few options for solving quadratic equations, but one of the most reliable and widely used methods is the quadratic formula. This formula provides a direct way to calculate the roots of any quadratic equation in the form ay² + by + c = 0. The quadratic formula is given by: y = (-b ± √(b² - 4ac)) / (2a). In our case, we have a = 5, b = -3, and c = 7. Let's plug these values into the quadratic formula and see what we get. Substituting the values, we have: y = (3 ± √((-3)² - 4 * 5 * 7)) / (2 * 5). Simplifying the expression under the square root, we get: y = (3 ± √(9 - 140)) / 10. This simplifies further to: y = (3 ± √(-131)) / 10. Uh oh, we've encountered something interesting! We have a negative number under the square root, which means we're dealing with complex roots. The square root of -131 can be written as √131 * √-1, which is √131 * i, where 'i' is the imaginary unit. So, our roots for y are: y = (3 ± √131 * i) / 10. This gives us two complex roots: y₁ = (3 + √131 * i) / 10 and y₂ = (3 - √131 * i) / 10. These roots are complex conjugates of each other, meaning they have the same real part but opposite imaginary parts. This is a common occurrence when solving quadratic equations with negative discriminants (the part under the square root, b² - 4ac). Now, remember that we made a substitution y = x². We've found the values of y, but we still need to find the values of x, which are the roots of our original polynomial. To do this, we'll substitute back x² for y and solve for x. This will lead us to the final piece of the puzzle.

Finding the Roots of the Original Polynomial

Okay, we've made excellent progress! We've solved for y in our transformed quadratic equation, and now it's time to bring it all back to our original variable, x. Remember our substitution: y = x². We found two complex values for y: y₁ = (3 + √131 * i) / 10 and y₂ = (3 - √131 * i) / 10. To find the values of x, we need to substitute back and solve for x in the equations x² = y₁ and x² = y₂. Let's start with x² = y₁. This means we need to find the square root of the complex number (3 + √131 * i) / 10. Finding the square root of a complex number might seem daunting, but it's a standard procedure in complex number arithmetic. In general, to find the square root of a complex number a + bi, we can express the square root as another complex number c + di and solve for c and d. However, for the sake of brevity and to avoid getting bogged down in complex number calculations, we'll acknowledge that x² = y₁ will yield two complex roots. Similarly, for x² = y₂, we need to find the square root of the complex number (3 - √131 * i) / 10. Again, this will involve finding the square root of a complex number, which will result in two more complex roots. So, from x² = y₂, we'll also get two complex roots. Now, let's take stock of what we've found. We started with a quartic polynomial, which we knew had 4 roots thanks to the Fundamental Theorem of Algebra. We transformed it into a quadratic equation, solved for y, and found two complex roots for y. Then, we substituted back to find x and realized that each value of y would give us two complex roots for x. Therefore, we have a total of 4 complex roots for our original polynomial. This means that the polynomial 7 + 5x⁴ - 3x² has no real roots and four complex roots. And that, my friends, is the answer to our initial question! We've successfully navigated the world of polynomials, transformations, and complex numbers to uncover the roots of this fascinating equation. Remember, the key is to break down complex problems into smaller, manageable steps and utilize the tools and theorems at our disposal. You got this!

Conclusion

So, there you have it, folks! We've successfully determined that the polynomial 7 + 5x⁴ - 3x² has a total of four roots, and all of them are complex. This journey took us through the fundamentals of polynomial roots, the power of substitution, and the intricacies of complex numbers. We saw how the Fundamental Theorem of Algebra guarantees the number of roots and how the quadratic formula can be a trusty tool for solving equations. Remember, math can seem intimidating at first, but by breaking it down into smaller steps and understanding the underlying concepts, you can conquer any problem. Keep exploring, keep questioning, and keep learning! You've got the power to unravel the mysteries of mathematics, one root at a time.