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### Problem 5
Evaluate [tex]\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3 \alpha \beta\)[/tex] given that [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the zeroes of the polynomial [tex]\(6x^2 + x - 1\)[/tex].
1. Determine the zeroes:
The polynomial [tex]\(6x^2 + x - 1\)[/tex] can be factored or solved using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
2. Use properties of roots:
For a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], the sum and product of roots are given by:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{1}{6} \][/tex]
[tex]\[ \alpha \beta = \frac{c}{a} = -\frac{1}{6} \][/tex]
3. Simplify the expression:
We need to evaluate [tex]\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3 \alpha \beta\)[/tex].
4. Calculation steps:
[tex]\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = (\alpha^2 + \beta^2)/(\alpha \beta) = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} \][/tex]
Substituting the known values:
[tex]\[ (\alpha + \beta)^2 = \left(-\frac{1}{6}\right)^2 = \frac{1}{36} \][/tex]
[tex]\[ (\alpha + \beta)^2 - 2\alpha \beta = \frac{1}{36} - 2 \left(-\frac{1}{6}\right) = \frac{1}{36} + \frac{1}{3} = \frac{1}{36} + \frac{12}{36} = \frac{13}{36} \][/tex]
[tex]\[ \frac{\frac{13}{36}}{\frac{-1}{6}} = \frac{13}{6} = -2.1666666666666665 \][/tex]
For the second term:
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-\frac{1}{6}}{-\frac{1}{6}} = 1 \][/tex]
For the last term:
[tex]\[ 3 \alpha \beta = 3 \left(-\frac{1}{6}\right) = -0.5 \][/tex]
5. Sum up the parts:
[tex]\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -2.1666666666666665 \][/tex]
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = 1.0 \][/tex]
[tex]\[ 3 \alpha \beta = -0.5 \][/tex]
Adding them together:
[tex]\[ -2.1666666666666665 + 1 + -0.5 = -1.6666666666666667 \][/tex]
So, the final combined value is [tex]\(-1.6666666666666667\)[/tex].
### Problem 6
Given that -3 is one of the zeroes of the quadratic polynomial [tex]\((p-1) x^2 + p x + 1\)[/tex], find the value of [tex]\(p\)[/tex].
1. Zero of the polynomial:
If -3 is a zero, substituting [tex]\(x = -3\)[/tex] in the polynomial [tex]\((p-1)(-3)^2 + p(-3) + 1 = 0\)[/tex]:
[tex]\[ (p-1)(9) + (-3p) + 1 = 0 \][/tex]
Simplify the equation:
[tex]\[ 9p - 9 - 3p + 1 = 0 \][/tex]
Combine like terms:
[tex]\[ 6p - 8 = 0 \][/tex]
Solve for [tex]\(p\)[/tex]:
[tex]\[ 6p = 8 \][/tex]
[tex]\[ p = \frac{8}{6} = \frac{4}{3} \][/tex]
Therefore, the value of [tex]\(p\)[/tex] is [tex]\(\frac{4}{3}\)[/tex].
### Problem 5
Evaluate [tex]\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3 \alpha \beta\)[/tex] given that [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the zeroes of the polynomial [tex]\(6x^2 + x - 1\)[/tex].
1. Determine the zeroes:
The polynomial [tex]\(6x^2 + x - 1\)[/tex] can be factored or solved using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
2. Use properties of roots:
For a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], the sum and product of roots are given by:
[tex]\[ \alpha + \beta = -\frac{b}{a} = -\frac{1}{6} \][/tex]
[tex]\[ \alpha \beta = \frac{c}{a} = -\frac{1}{6} \][/tex]
3. Simplify the expression:
We need to evaluate [tex]\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3 \alpha \beta\)[/tex].
4. Calculation steps:
[tex]\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = (\alpha^2 + \beta^2)/(\alpha \beta) = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} \][/tex]
Substituting the known values:
[tex]\[ (\alpha + \beta)^2 = \left(-\frac{1}{6}\right)^2 = \frac{1}{36} \][/tex]
[tex]\[ (\alpha + \beta)^2 - 2\alpha \beta = \frac{1}{36} - 2 \left(-\frac{1}{6}\right) = \frac{1}{36} + \frac{1}{3} = \frac{1}{36} + \frac{12}{36} = \frac{13}{36} \][/tex]
[tex]\[ \frac{\frac{13}{36}}{\frac{-1}{6}} = \frac{13}{6} = -2.1666666666666665 \][/tex]
For the second term:
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-\frac{1}{6}}{-\frac{1}{6}} = 1 \][/tex]
For the last term:
[tex]\[ 3 \alpha \beta = 3 \left(-\frac{1}{6}\right) = -0.5 \][/tex]
5. Sum up the parts:
[tex]\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -2.1666666666666665 \][/tex]
[tex]\[ \frac{1}{\alpha} + \frac{1}{\beta} = 1.0 \][/tex]
[tex]\[ 3 \alpha \beta = -0.5 \][/tex]
Adding them together:
[tex]\[ -2.1666666666666665 + 1 + -0.5 = -1.6666666666666667 \][/tex]
So, the final combined value is [tex]\(-1.6666666666666667\)[/tex].
### Problem 6
Given that -3 is one of the zeroes of the quadratic polynomial [tex]\((p-1) x^2 + p x + 1\)[/tex], find the value of [tex]\(p\)[/tex].
1. Zero of the polynomial:
If -3 is a zero, substituting [tex]\(x = -3\)[/tex] in the polynomial [tex]\((p-1)(-3)^2 + p(-3) + 1 = 0\)[/tex]:
[tex]\[ (p-1)(9) + (-3p) + 1 = 0 \][/tex]
Simplify the equation:
[tex]\[ 9p - 9 - 3p + 1 = 0 \][/tex]
Combine like terms:
[tex]\[ 6p - 8 = 0 \][/tex]
Solve for [tex]\(p\)[/tex]:
[tex]\[ 6p = 8 \][/tex]
[tex]\[ p = \frac{8}{6} = \frac{4}{3} \][/tex]
Therefore, the value of [tex]\(p\)[/tex] is [tex]\(\frac{4}{3}\)[/tex].
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