At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Let's go through the solutions step-by-step.
### 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].
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
