At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To solve for [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex], where [tex]\(\alpha, \beta, \gamma\)[/tex] are the roots of the polynomial equation [tex]\(x^3 + 2x + 5 = 0\)[/tex], let's start by using the relationships provided by Vieta's formulas for a cubic polynomial.
Given the polynomial [tex]\(x^3 + 2x + 5 = 0\)[/tex], we can identify the coefficients as follows: [tex]\(a = 1\)[/tex], [tex]\(b = 0\)[/tex], [tex]\(c = 2\)[/tex], and [tex]\(d = 5\)[/tex]. Therefore, according to Vieta’s formulas, we have:
1. [tex]\(\alpha + \beta + \gamma = 0\)[/tex],
2. [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha = 2\)[/tex],
3. [tex]\(\alpha\beta\gamma = -5\)[/tex].
Next, we need to express [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex] in terms of these roots. Let's begin by finding the squared roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex]. Using the original equation [tex]\(x^3 + 2x + 5 = 0\)[/tex]:
[tex]\[ \alpha^3 + 2\alpha + 5 = 0 \implies \alpha^3 = -2\alpha - 5, \][/tex]
[tex]\[ \beta^3 + 2\beta + 5 = 0 \implies \beta^3 = -2\beta - 5, \][/tex]
[tex]\[ \gamma^3 + 2\gamma + 5 = 0 \implies \gamma^3 = -2\gamma - 5. \][/tex]
Now, squaring [tex]\(\alpha^2\)[/tex], we have:
[tex]\[ \alpha^4 = (\alpha^2)^2 = (\alpha^3 \cdot \alpha) = (-2\alpha - 5)\alpha = -2\alpha^2 - 5\alpha, \][/tex]
[tex]\[ \beta^4 = (\beta^2)^2 = (\beta^3 \cdot \beta) = (-2\beta - 5)\beta = -2\beta^2 - 5\beta, \][/tex]
[tex]\[ \gamma^4 = (\gamma^2)^2 = (\gamma^3 \cdot \gamma) = (-2\gamma - 5)\gamma = -2\gamma^2 - 5\gamma. \][/tex]
Summing these expressions, we get:
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(\alpha^2 + \beta^2 + \gamma^2) - 5(\alpha + \beta + \gamma). \][/tex]
Since [tex]\(\alpha + \beta + \gamma = 0\)[/tex], the term involving [tex]\(\alpha + \beta + \gamma\)[/tex] vanishes:
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(\alpha^2 + \beta^2 + \gamma^2). \][/tex]
To find [tex]\(\alpha^2 + \beta^2 + \gamma^2\)[/tex], recall the identity for the sum of squares of the roots:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha). \][/tex]
Using Vieta’s results:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = 0^2 - 2(2) = -4. \][/tex]
Thus,
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(-4) = 8. \][/tex]
Therefore, the value of [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex] is [tex]\(\boxed{8}\)[/tex].
Given the polynomial [tex]\(x^3 + 2x + 5 = 0\)[/tex], we can identify the coefficients as follows: [tex]\(a = 1\)[/tex], [tex]\(b = 0\)[/tex], [tex]\(c = 2\)[/tex], and [tex]\(d = 5\)[/tex]. Therefore, according to Vieta’s formulas, we have:
1. [tex]\(\alpha + \beta + \gamma = 0\)[/tex],
2. [tex]\(\alpha\beta + \beta\gamma + \gamma\alpha = 2\)[/tex],
3. [tex]\(\alpha\beta\gamma = -5\)[/tex].
Next, we need to express [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex] in terms of these roots. Let's begin by finding the squared roots [tex]\(\alpha^2, \beta^2, \gamma^2\)[/tex]. Using the original equation [tex]\(x^3 + 2x + 5 = 0\)[/tex]:
[tex]\[ \alpha^3 + 2\alpha + 5 = 0 \implies \alpha^3 = -2\alpha - 5, \][/tex]
[tex]\[ \beta^3 + 2\beta + 5 = 0 \implies \beta^3 = -2\beta - 5, \][/tex]
[tex]\[ \gamma^3 + 2\gamma + 5 = 0 \implies \gamma^3 = -2\gamma - 5. \][/tex]
Now, squaring [tex]\(\alpha^2\)[/tex], we have:
[tex]\[ \alpha^4 = (\alpha^2)^2 = (\alpha^3 \cdot \alpha) = (-2\alpha - 5)\alpha = -2\alpha^2 - 5\alpha, \][/tex]
[tex]\[ \beta^4 = (\beta^2)^2 = (\beta^3 \cdot \beta) = (-2\beta - 5)\beta = -2\beta^2 - 5\beta, \][/tex]
[tex]\[ \gamma^4 = (\gamma^2)^2 = (\gamma^3 \cdot \gamma) = (-2\gamma - 5)\gamma = -2\gamma^2 - 5\gamma. \][/tex]
Summing these expressions, we get:
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(\alpha^2 + \beta^2 + \gamma^2) - 5(\alpha + \beta + \gamma). \][/tex]
Since [tex]\(\alpha + \beta + \gamma = 0\)[/tex], the term involving [tex]\(\alpha + \beta + \gamma\)[/tex] vanishes:
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(\alpha^2 + \beta^2 + \gamma^2). \][/tex]
To find [tex]\(\alpha^2 + \beta^2 + \gamma^2\)[/tex], recall the identity for the sum of squares of the roots:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha). \][/tex]
Using Vieta’s results:
[tex]\[ \alpha^2 + \beta^2 + \gamma^2 = 0^2 - 2(2) = -4. \][/tex]
Thus,
[tex]\[ \alpha^4 + \beta^4 + \gamma^4 = -2(-4) = 8. \][/tex]
Therefore, the value of [tex]\(\alpha^4 + \beta^4 + \gamma^4\)[/tex] is [tex]\(\boxed{8}\)[/tex].
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
