If Α And Β Are The Zeroes Of The Quadratic Polynomial F(x) = X² - 2x - 8, Then Find The Value Of: (i) Α - Β (ii) Α² + Β² (iii) Α⁴ + Β⁴ (iv) Αβ² + Α²β (v) 1/α + 1/β (vi) 1/α + 1/β - Αβ (vii) 1/α - 1/β (viii) Α³ + Β³ (ix) 1/α² + 1/β² (x) Α²/β² + Β²/α². Discussion Category: Mathematics.
In mathematics, quadratic polynomials hold a fundamental position, serving as the building blocks for more complex algebraic expressions and equations. Understanding the properties of their roots, the values of the variable that make the polynomial equal to zero, is crucial for solving various mathematical problems and gaining deeper insights into the behavior of these functions. This comprehensive guide delves into the fascinating relationship between the roots and coefficients of a quadratic polynomial, using the specific example of f(x) = x² - 2x - 8 to illustrate key concepts and techniques. Let's embark on this mathematical journey and unlock the secrets hidden within the roots of quadratic polynomials.
Understanding Quadratic Polynomials and Their Roots
A quadratic polynomial is a polynomial of degree two, generally expressed in the form f(x) = ax² + bx + c, where a, b, and c are constants and a ≠ 0. The roots (or zeroes) of a quadratic polynomial are the values of x that satisfy the equation f(x) = 0. These roots represent the points where the graph of the quadratic polynomial intersects the x-axis. For a quadratic polynomial, there can be at most two distinct real roots, one repeated real root, or two complex roots.
The relationship between the roots and coefficients of a quadratic polynomial is a cornerstone concept in algebra. For a quadratic polynomial f(x) = ax² + bx + c, if α and β are the roots, then the following relationships hold:
- Sum of roots: α + β = -b/a
- Product of roots: αβ = c/a
These relationships provide a powerful tool for analyzing and solving quadratic equations, as they allow us to determine the roots without explicitly solving the equation. Furthermore, they enable us to find various expressions involving the roots, as we will explore in detail in the following sections.
H2: Case Study: f(x) = x² - 2x - 8
Let's consider the specific quadratic polynomial f(x) = x² - 2x - 8. This polynomial serves as an excellent example to demonstrate the application of the relationships between roots and coefficients. By analyzing this polynomial, we can gain a deeper understanding of how these relationships work in practice and how they can be used to solve various problems.
First, we need to identify the coefficients of the polynomial. In this case, a = 1, b = -2, and c = -8. Now, let α and β be the zeroes (roots) of the polynomial f(x). Using the relationships between roots and coefficients, we have:
- Sum of roots: α + β = -(-2)/1 = 2
- Product of roots: αβ = -8/1 = -8
These two equations, α + β = 2 and αβ = -8, are fundamental to solving the subsequent problems. They provide us with essential information about the roots, even without explicitly finding their values. By manipulating these equations and applying algebraic identities, we can determine the values of various expressions involving α and β. In the following sections, we will explore how to use these relationships to find the values of different expressions, showcasing the power and versatility of this approach.
H2: Finding the Values of Expressions Involving α and β
Now that we have established the sum and product of the roots, α + β = 2 and αβ = -8, we can proceed to find the values of various expressions involving α and β. This section will systematically address each expression, demonstrating the techniques and algebraic manipulations required to arrive at the solutions. Each expression offers a unique challenge and highlights the importance of understanding and applying the relationships between roots and coefficients.
H3: (i) α - β
To find α - β, we can use the following algebraic identity:
(α - β)² = (α + β)² - 4αβ
Substituting the values of α + β and αβ, we get:
(α - β)² = (2)² - 4(-8) = 4 + 32 = 36
Taking the square root of both sides, we obtain:
α - β = ±√36 = ±6
Therefore, the value of α - β is either 6 or -6. The sign depends on which root is larger. This result demonstrates how we can find the difference between the roots without explicitly solving the quadratic equation, using only the sum and product of the roots.
H3: (ii) α² + β²
To find α² + β², we can use another algebraic identity:
α² + β² = (α + β)² - 2αβ
Substituting the values of α + β and αβ, we get:
α² + β² = (2)² - 2(-8) = 4 + 16 = 20
Therefore, the value of α² + β² is 20. This calculation showcases the usefulness of the sum and product of roots in determining the sum of their squares, a common expression in various mathematical contexts.
H3: (iii) α⁴ + β⁴
To find α⁴ + β⁴, we can utilize the result from the previous calculation (α² + β²) and apply the following identity:
α⁴ + β⁴ = (α² + β²)² - 2(αβ)²
Substituting the values of α² + β² and αβ, we get:
α⁴ + β⁴ = (20)² - 2(-8)² = 400 - 2(64) = 400 - 128 = 272
Therefore, the value of α⁴ + β⁴ is 272. This demonstrates how we can build upon previous results to find expressions involving higher powers of the roots, showcasing the interconnectedness of these calculations.
H3: (iv) αβ² + α²β
To find αβ² + α²β, we can factor out αβ:
αβ² + α²β = αβ(β + α)
Substituting the values of αβ and α + β, we get:
αβ² + α²β = (-8)(2) = -16
Therefore, the value of αβ² + α²β is -16. This calculation highlights the importance of factoring and recognizing common factors in simplifying expressions involving roots.
H3: (v) 1/α + 1/β
To find 1/α + 1/β, we can find a common denominator:
1/α + 1/β = (β + α) / αβ
Substituting the values of α + β and αβ, we get:
1/α + 1/β = (2) / (-8) = -1/4
Therefore, the value of 1/α + 1/β is -1/4. This expression is the sum of the reciprocals of the roots, and it can be easily calculated using the sum and product of the roots.
H3: (vi) 1/α + 1/β - αβ
To find 1/α + 1/β - αβ, we can use the result from the previous calculation (1/α + 1/β) and the value of αβ:
1/α + 1/β - αβ = (-1/4) - (-8) = -1/4 + 8 = 31/4
Therefore, the value of 1/α + 1/β - αβ is 31/4. This expression combines the sum of the reciprocals of the roots with their product, demonstrating how we can combine previous results to solve more complex expressions.
H3: (vii) 1/α - 1/β
To find 1/α - 1/β, we can find a common denominator:
1/α - 1/β = (β - α) / αβ
We already know α - β = ±6, so β - α = ∓6. Substituting the values of β - α and αβ, we get:
1/α - 1/β = (∓6) / (-8) = ±3/4
Therefore, the value of 1/α - 1/β is either 3/4 or -3/4. This expression is the difference of the reciprocals of the roots, and it utilizes the result we obtained for α - β.
H3: (viii) α³ + β³
To find α³ + β³, we can use the following algebraic identity:
α³ + β³ = (α + β)(α² - αβ + β²)
We already know α + β = 2 and α² + β² = 20. We can rewrite the expression as:
α³ + β³ = (α + β)((α² + β²) - αβ)
Substituting the values, we get:
α³ + β³ = (2)(20 - (-8)) = 2(28) = 56
Therefore, the value of α³ + β³ is 56. This calculation demonstrates how we can use algebraic identities and previously calculated values to find the sum of the cubes of the roots.
H3: (ix) 1/α² + 1/β²
To find 1/α² + 1/β², we can find a common denominator:
1/α² + 1/β² = (β² + α²) / (α²β²)
We already know α² + β² = 20 and αβ = -8, so (αβ)² = 64. Substituting the values, we get:
1/α² + 1/β² = (20) / (64) = 5/16
Therefore, the value of 1/α² + 1/β² is 5/16. This expression is the sum of the reciprocals of the squares of the roots.
H3: (x) α²/β² + β²/α²
To find α²/β² + β²/α², we can find a common denominator:
α²/β² + β²/α² = (α⁴ + β⁴) / (α²β²)
We already know α⁴ + β⁴ = 272 and αβ = -8, so (αβ)² = 64. Substituting the values, we get:
α²/β² + β²/α² = (272) / (64) = 17/4
Therefore, the value of α²/β² + β²/α² is 17/4. This expression involves the ratio of the squares of the roots, and it utilizes the results we obtained for α⁴ + β⁴ and (αβ)².
H3: Conclusion
Through this comprehensive exploration of the quadratic polynomial f(x) = x² - 2x - 8, we have demonstrated the power and versatility of the relationships between roots and coefficients. By utilizing the sum and product of the roots, we were able to find the values of various expressions involving α and β without explicitly solving the quadratic equation. These techniques are essential tools in algebra and provide valuable insights into the behavior of quadratic polynomials. Mastering these concepts will undoubtedly enhance your problem-solving abilities and deepen your understanding of mathematical relationships.