Problem 520 If \(\alpha\) and \(\beta\) are ... [FREE SOLUTION]

Chapter 17: Problem 520

If \(\alpha\) and \(\beta\) are the roots of \(\mathrm{x}^{2}-\mathrm{px}+\mathrm{q}=0\), find the value of (1) \(\alpha^{2}+\beta^{2}\), (2) \(\alpha^{3}+\beta^{3}\).

Short Answer

Expert verified

The values of \(\alpha^2+\beta^2\) and \(\alpha^3+\beta^3\) are \(p^2 - 2q\) and \(p(p^2 - 3q)\) respectively.

Step by step solution

Apply Vieta's Formulas

By Vieta's formulas, we have: Sum of the roots: \( \alpha + \beta = p\) Product of the roots: \( \alpha\beta = q\) Now, we can use these expressions to find the desired values of \(\alpha^2+\beta^2\) and \(\alpha^3+\beta^3\).

Calculate \(\alpha^2+\beta^2\)

To find the value of \(\alpha^2 + \beta^2\), we can employ the following formula based on the sum and product of the roots: \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) Now, substitute the expressions found in Step 1: \(\alpha^2 + \beta^2 = (p)^2 - 2(q)\) So, the value of \(\alpha^2 + \beta^2\) is equal to \(p^2 - 2q\).

Calculate \(\alpha^3+\beta^3\)

To find the value of \(\alpha^3 + \beta^3\), we can use the following identity: \(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)\) Before we substitute the expressions found in Step 1, we need to find the value of the second factor. We already have the value of \(\alpha^2 + \beta^2\), which is \(p^2 - 2q\). Hence, to calculate \(\alpha^2 - \alpha\beta + \beta^2\) we can use::: \(\alpha^2 - \alpha\beta + \beta^2 = (\alpha^2 + \beta^2) - \alpha\beta\) Now, substitute the expressions found in Step 1 and the result found for (\(\alpha^2+\beta^2\)): \(\alpha^2 - \alpha\beta + \beta^2 = (p^2 - 2q) - q\) This yields the following expression: \(\alpha^2 - \alpha\beta + \beta^2 = p^2 - 3q\) Finally, substitute the expressions found in Step 1 and this result in the formula for \(\alpha^3 + \beta^3\): \) \(\alpha^3 + \beta^3 = (p)(p^2 - 3q)\) So, the value of \(\alpha^3 + \beta^3\) is equal to \(p(p^2 - 3q)\). Therefore, the values of \(\alpha^2+\beta^2\) and \(\alpha^3+\beta^3\) are \(p^2 - 2q\) and \(p(p^2 - 3q)\) respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vieta's Formulas

Vieta's formulas provide a powerful connection between the coefficients of polynomial equations and the sums and products of their roots. Specifically, for a quadratic equation given by \( ax^2 + bx + c = 0 \), the formulas relate the roots \( \alpha \) and \( \beta \) to the coefficients:

Sum of the roots: \( \alpha + \beta = -\frac{b}{a} \)
Product of the roots: \( \alpha\beta = \frac{c}{a} \)

In the provided exercise, the quadratic equation is in the form \( x^2 - px + q = 0 \), which makes it easy to directly apply Vieta's formulas:

The sum of the roots \( \alpha + \beta = p \)
The product of the roots \( \alpha\beta = q \)

These relationships are the foundation for solving more complex expressions involving roots, such as \( \alpha^2 + \beta^2 \) and \( \alpha^3 + \beta^3 \). By re-expressing these higher powers in terms of \( \alpha \) and \( \beta \), we find manageable ways of calculating them.

Quadratic Equations

A quadratic equation is a second-degree polynomial equation that can be represented in the general form \( ax^2 + bx + c = 0 \). Understanding these equations is vital in algebra as they appear frequently in various mathematical problems and real-world situations. Here are key elements of quadratic equations:

Degree: The highest power of the variable is 2, hence the name 'quadratic'.
Standard Form: The equation \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \).
Roots: Solutions to the equation, found using techniques such as factoring, completing the square, or using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Quadratic equations can have either real or complex roots, and the nature of these roots is determined by the discriminant \( b^2 - 4ac \). For the given problem \( x^2 - px + q = 0 \), \( p \) and \( q \) take the roles of coefficients leading directly to root calculations via Vieta's formulas.

Algebraic Identities

Algebraic identities simplify complex expressions and allow us to transform equations in useful ways. These identities include a set of well-known equations that are true for any value, such as:

Square of a binomial: \((a+b)^2 = a^2 + 2ab + b^2 \)
Cube of a binomial: \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)
Difference of squares: \(a^2 - b^2 = (a-b)(a+b) \)

In the context of solving the problem, the algebraic identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \) is essential. This identity helps in expressing the sum of the squares of the roots using their sum and product, known from Vieta's formulas. Similarly, for \( \alpha^3 + \beta^3 \), the identity used is \( \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \). Recognizing and applying these identities is crucial in simplifying calculations and solving algebraic equations effectively.

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91影视

If \(\alpha\) and \(\beta\) are the roots of \(\mathrm{x}^{2}-\mathrm{px}+\mathrm{q}=0\), find the value of (1) \(\alpha^{2}+\beta^{2}\), (2) \(\alpha^{3}+\beta^{3}\).

Short Answer

Step by step solution

Apply Vieta's Formulas

Calculate \(\alpha^2+\beta^2\)

Calculate \(\alpha^3+\beta^3\)

Key Concepts

Vieta's Formulas

Quadratic Equations

Algebraic Identities

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