Polynomial Roots Problems And Solutions Α Β And Coefficients
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Unlocking the secrets of polynomials often involves understanding the relationship between their coefficients and roots. This article delves into some intriguing problems concerning quadratic polynomials and their zeros. We'll explore how to determine the difference between roots, find the value of a coefficient based on root relationships, and analyze the properties of polynomials with specific conditions. Whether you're a student grappling with algebra or a math enthusiast seeking to deepen your understanding, this exploration promises valuable insights and problem-solving techniques.
13. Decoding the Roots of -x² + 8x + 9: Finding α - β
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In this section, we aim to determine the value of (α - β), where α and β are the zeros of the quadratic polynomial -x² + 8x + 9, with the condition that α > β. This problem beautifully combines the concepts of quadratic equations, their roots, and the relationships between these roots and the coefficients of the polynomial. To solve this, we will first need to find the roots α and β of the given polynomial. There are a couple of efficient methods to achieve this: factoring the quadratic expression or using the quadratic formula. Each approach provides a pathway to unveil the values of α and β, allowing us to then calculate their difference.
Factoring the Quadratic Expression
Our initial step involves expressing -x² + 8x + 9 in a factored form, which essentially means rewriting it as a product of two binomials. This is achieved by identifying two numbers that, when multiplied, give the constant term (9) and, when added, give the coefficient of the linear term (8). In this specific polynomial, the factors that satisfy these conditions are 9 and -1, because 9 * -1 = -9 and 9 + (-1) = 8. However, since our leading coefficient is -1, we need to adjust our factoring approach slightly. Let's first rewrite the polynomial by factoring out the -1: -1(x² - 8x - 9). Now, we look for two numbers that multiply to -9 and add to -8. These numbers are -9 and 1. Thus, the factored form becomes -1(x - 9)(x + 1). Setting each factor to zero gives us the roots. Setting (x - 9) = 0 gives x = 9, and setting (x + 1) = 0 gives x = -1. Therefore, the roots of the polynomial are 9 and -1. Since α > β, we assign α = 9 and β = -1. The difference (α - β) is then 9 - (-1) = 10.
Utilizing the Quadratic Formula
Alternatively, we can employ the quadratic formula, a universally applicable method for finding the roots of any quadratic equation. For a polynomial in the form ax² + bx + c, the quadratic formula is given by: x = [-b ± √(b² - 4ac)] / (2a). In our case, a = -1, b = 8, and c = 9. Substituting these values into the formula gives us: x = [-8 ± √(8² - 4(-1)(9))] / (2(-1)). Simplifying this expression, we get x = [-8 ± √(64 + 36)] / (-2), which further simplifies to x = [-8 ± √100] / (-2). This leads to two possible values for x: x = [-8 + 10] / (-2) and x = [-8 - 10] / (-2). The first case yields x = 2 / (-2) = -1, and the second case gives x = -18 / (-2) = 9. Consequently, we find the roots to be -1 and 9, consistent with our factoring method. Again, since α > β, we assign α = 9 and β = -1. The difference (α - β) is then 9 - (-1) = 10.
Conclusion for Problem 13
Both factoring and the quadratic formula have led us to the same roots, confirming the robustness of our solution. We've identified α as 9 and β as -1, and successfully calculated the difference (α - β) to be 10. This solution underscores the importance of mastering fundamental algebraic techniques and understanding the interrelationship between roots and coefficients in polynomial equations. Therefore, the correct answer is (b) 10.
14. Unraveling the Coefficient p in p x² - 2x + 3p When α + β = αβ
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Here, we delve into a problem that requires us to determine the value of the coefficient 'p' in the polynomial px² - 2x + 3p, given the condition that the sum of the roots (α + β) is equal to the product of the roots (αβ). This problem elegantly demonstrates how the relationships between the roots and coefficients of a quadratic polynomial can be utilized to solve for unknown parameters. We will leverage Vieta's formulas, which provide a direct link between the coefficients of a polynomial and the sums and products of its roots. This approach will enable us to set up an equation involving 'p' and subsequently solve for it.
Applying Vieta's Formulas
Vieta's formulas are a cornerstone in the study of polynomial equations, offering a powerful tool for relating the roots of a polynomial to its coefficients. For a general quadratic equation of the form ax² + bx + c = 0, Vieta's formulas state that the sum of the roots (α + β) is equal to -b/a, and the product of the roots (αβ) is equal to c/a. In the context of our problem, we have the polynomial px² - 2x + 3p. Comparing this to the general form, we identify a = p, b = -2, and c = 3p. Applying Vieta's formulas, we find that the sum of the roots, α + β, is equal to -(-2)/p, which simplifies to 2/p. Similarly, the product of the roots, αβ, is equal to 3p/p, which simplifies to 3, assuming p ≠ 0. The problem states that α + β = αβ. We can now set up an equation using the expressions we derived from Vieta's formulas. This equation will involve 'p' as the unknown, allowing us to solve for its value.
Solving for p
Given that α + β = 2/p and αβ = 3, and the problem stipulates that α + β = αβ, we can equate the two expressions: 2/p = 3. This equation is a simple algebraic equation that we can solve for 'p'. To isolate 'p', we can multiply both sides of the equation by 'p', resulting in 2 = 3p. Then, we divide both sides by 3 to obtain p = 2/3. Thus, we have found the value of 'p' that satisfies the given condition. It's crucial to verify that this value of 'p' does not lead to any inconsistencies or contradictions within the original problem. In this case, p = 2/3 is a valid solution, as it doesn't make the leading coefficient zero, ensuring that we indeed have a quadratic polynomial.
Conclusion for Problem 14
By skillfully applying Vieta's formulas and setting up a straightforward algebraic equation, we have successfully determined the value of 'p' to be 2/3. This problem showcases the elegance and utility of Vieta's formulas in connecting the roots and coefficients of polynomials. Our step-by-step approach, from identifying the coefficients to solving for the unknown, underscores the importance of a systematic problem-solving methodology. Therefore, the correct answer is (b) 2/3.
15. Exploring the Polynomial and its Zeros
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(The question is incomplete and needs more information to provide a comprehensive solution. To address this, I will create a hypothetical, similar problem to demonstrate a typical scenario involving zeros of a polynomial. This will allow us to explore related concepts and problem-solving techniques within the context of polynomial zeros.)
Hypothetical Problem: If α and β are the zeros of the polynomial x² + 5x + 6, find the value of α² + β². This hypothetical problem retains the essence of dealing with the zeros of a polynomial but introduces a different challenge: finding an expression involving the squares of the roots. To tackle this, we will again rely on Vieta's formulas to establish relationships between the roots and coefficients. Additionally, we'll employ algebraic manipulation to express the desired quantity (α² + β²) in terms of the known sums and products of the roots. This approach showcases a common strategy in polynomial problems: transforming the target expression into a form that can be easily evaluated using Vieta's formulas.
Applying Vieta's Formulas and Algebraic Manipulation
As before, Vieta's formulas provide us with a direct link between the roots and coefficients of the polynomial. For the polynomial x² + 5x + 6, we identify a = 1, b = 5, and c = 6. Applying Vieta's formulas, we find that the sum of the roots (α + β) is equal to -b/a = -5/1 = -5, and the product of the roots (αβ) is equal to c/a = 6/1 = 6. Now, our goal is to find the value of α² + β². We can't directly obtain this from Vieta's formulas, but we can cleverly manipulate the expression using a common algebraic identity. We know that (α + β)² = α² + 2αβ + β². Rearranging this, we get α² + β² = (α + β)² - 2αβ. This transformation is crucial because it expresses α² + β² in terms of (α + β) and αβ, both of which we know from Vieta's formulas. Now, we can simply substitute the values we found earlier. We have (α + β) = -5 and αβ = 6. Plugging these into our expression, we get α² + β² = (-5)² - 2(6) = 25 - 12 = 13. Thus, we have successfully found the value of α² + β² by strategically combining Vieta's formulas with algebraic manipulation. This problem illustrates the power of these techniques in navigating problems involving polynomial roots.
Conclusion for Hypothetical Problem 15
Through the skillful application of Vieta's formulas and a bit of algebraic finesse, we determined that α² + β² = 13 for the polynomial x² + 5x + 6. This hypothetical problem serves as a valuable illustration of how to approach problems where we need to find expressions involving polynomial roots beyond simple sums and products. The key takeaway is the ability to transform the desired expression into a form that utilizes the information provided by Vieta's formulas. This approach can be generalized to a variety of problems involving roots of polynomials, making it a valuable addition to your problem-solving toolkit.
Conclusion
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In this article, we've tackled several problems involving the roots of quadratic polynomials. From finding the difference between roots to determining coefficients based on root relationships, we've explored a range of techniques and concepts. The importance of understanding Vieta's formulas and mastering algebraic manipulation has been highlighted throughout our exploration. These tools are not only essential for solving specific problems but also for developing a deeper understanding of the relationships within polynomial equations. By practicing these techniques and exploring different types of problems, you can strengthen your grasp of polynomial theory and enhance your problem-solving abilities in mathematics.