Let `alpha`, `beta` be two real numbers satisfying the following relations `alpha^(2)+beta^(2)=5`, `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))`
Quadratic equation having roots `alpha` and `beta` is

Options

A. (a) `x^(2)+-x+2=0`
B. (b) `x^(2)+-3x-2=0`
C. (c) `x^(2)+-sqrt(3)x+2=0`
D. (d) none of these

Explanation

To find the quadratic equation having roots \(\alpha\) and \(\beta\) given the conditions \(\alpha^2 + \beta^2 = 5\) and \(3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)\), we will follow these steps:

### Step 1: Use the identity for \(\alpha^5 + \beta^5\)
We can express \(\alpha^5 + \beta^5\) in terms of lower powers of \(\alpha\) and \(\beta\):
\[ \alpha^5 + \beta^5 = (\alpha + \beta)(\alpha^4 + \beta^4) - \alpha\beta(\alpha^3 + \beta^3) \]
We also know that:
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 \]

### Step 2: Express \(\alpha^3 + \beta^3\)
Using the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) \]

### Step 3: Substitute into the given equation
Substituting these identities into the equation \(3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)\):
\[ 3[(\alpha + \beta)(\alpha^4 + \beta^4) - \alpha\beta(\alpha^3 + \beta^3)] = 11(\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) \]

### Step 4: Let \(\alpha + \beta = s\) and \(\alpha\beta = p\)
Now, let \(s = \alpha + \beta\) and \(p = \alpha\beta\). We can rewrite our equations:
1. From \(\alpha^2 + \beta^2 = 5\):
\[ s^2 - 2p = 5 \quad \Rightarrow \quad s^2 = 5 + 2p \]
2. Substitute \(s\) and \(p\) into the equation:
\[ 3[s((5 + 2p) - p)] - 3p\left[s\left(\frac{s^2 - p}{s}\right)\right] = 11s\left(5 - p\right) \]

### Step 5: Simplify the equation
After simplifying, we will arrive at a quadratic equation in terms of \(p\):
\[ 3s(5 + p) - 3ps = 11s(5 - p) \]
This simplifies to:
\[ 15s + 3sp - 3ps = 55s - 11sp \]
Rearranging gives:
\[ (3 + 11)p = 55 - 15 \]
Thus, we can solve for \(p\).

### Step 6: Solve for \(p\)
From the rearranged equation, we find:
\[ 14p = 40 \quad \Rightarrow \quad p = \frac{40}{14} = \frac{20}{7} \]

### Step 7: Find \(s\)
Substituting \(p\) back into the equation for \(s\):
\[ s^2 = 5 + 2\left(\frac{20}{7}\right) = 5 + \frac{40}{7} = \frac{35 + 40}{7} = \frac{75}{7} \]
Thus,
\[ s = \pm \sqrt{\frac{75}{7}} = \pm \frac{5\sqrt{3}}{7} \]

### Step 8: Form the quadratic equation
The quadratic equation with roots \(\alpha\) and \(\beta\) is given by:
\[ x^2 - sx + p = 0 \]
Substituting \(s\) and \(p\):
\[ x^2 - \left(\pm \frac{5\sqrt{3}}{7}\right)x + \frac{20}{7} = 0 \]

### Final Quadratic Equation
Thus, the quadratic equation having roots \(\alpha\) and \(\beta\) is:
\[ 7x^2 \mp 5\sqrt{3}x + 20 = 0 \]