If `alpha, beta` are the roots of the equation `lambda(x^(2)-x)+x+5=0` and if `lambda_(1) and lambda_(2)` are two values of `lambda` obtained from `(alpha)/(beta)+(beta)/(alpha)=(4)/(5)`, then `(lambda_(1))/(lambda_(2)^(2))+(lambda_(2))/(lambda_(1)^(2))` equals

Options

A. 4192
B. 4144
C. 4096
D. 4048

Explanation

To solve the given problem, we need to follow these steps:

1. **Rewrite the equation**: The given equation is
\[ \lambda(x^2 - x) + x + 5 = 0. \]
We can rearrange it to:
\[ \lambda x^2 + (1 - \lambda)x + 5 = 0. \]

2. **Identify coefficients**: From the standard form \(Ax^2 + Bx + C = 0\), we have:
- \(A = \lambda\)
- \(B = 1 - \lambda\)
- \(C = 5\)

3. **Use Vieta's formulas**: The roots \(\alpha\) and \(\beta\) of the quadratic equation satisfy:
- \(\alpha + \beta = -\frac{B}{A} = -\frac{1 - \lambda}{\lambda}\)
- \(\alpha \beta = \frac{C}{A} = \frac{5}{\lambda}\)

4. **Set up the equation**: We are given that
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{4}{5}. \]
This can be rewritten using the identity \(\frac{\alpha^2 + \beta^2}{\alpha \beta}\):
\[ \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{4}{5}. \]
We know that \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\). Thus,
\[ \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha \beta} = \frac{4}{5}. \]

5. **Substitute values**: Substitute \(\alpha + \beta\) and \(\alpha \beta\):
\[ \frac{\left(-\frac{1 - \lambda}{\lambda}\right)^2 - 2 \cdot \frac{5}{\lambda}}{\frac{5}{\lambda}} = \frac{4}{5}. \]
Simplifying gives:
\[ \frac{\frac{(1 - \lambda)^2}{\lambda^2} - \frac{10}{\lambda}}{\frac{5}{\lambda}} = \frac{4}{5}. \]
Multiplying both sides by \(\frac{5}{\lambda}\) leads to:
\[ (1 - \lambda)^2 - 10 = 4. \]
Therefore,
\[ (1 - \lambda)^2 = 14. \]

6. **Solve for \(\lambda\)**: Taking the square root gives:
\[ 1 - \lambda = \pm \sqrt{14}. \]
Thus, we have two values for \(\lambda\):
\[ \lambda_1 = 1 - \sqrt{14}, \quad \lambda_2 = 1 + \sqrt{14}. \]

7. **Calculate \(\frac{\lambda_1}{\lambda_2^2} + \frac{\lambda_2}{\lambda_1^2}\)**:
First, compute \(\lambda_1 \lambda_2\):
\[ \lambda_1 \lambda_2 = (1 - \sqrt{14})(1 + \sqrt{14}) = 1 - 14 = -13. \]
Next, compute \(\lambda_1 + \lambda_2\):
\[ \lambda_1 + \lambda_2 = (1 - \sqrt{14}) + (1 + \sqrt{14}) = 2. \]

8. **Use the identity**: We can use the identity:
\[ \frac{\lambda_1^3 + \lambda_2^3}{\lambda_1 \lambda_2^2} = \frac{(\lambda_1 + \lambda_2)(\lambda_1^2 - \lambda_1 \lambda_2 + \lambda_2^2)}{\lambda_1 \lambda_2^2}. \]
We know:
\[ \lambda_1^2 + \lambda_2^2 = (\lambda_1 + \lambda_2)^2 - 2\lambda_1 \lambda_2 = 2^2 - 2(-13) = 4 + 26 = 30. \]
Thus,
\[ \lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)(\lambda_1^2 - \lambda_1 \lambda_2 + \lambda_2^2) = 2(30 + 13) = 2 \times 43 = 86. \]

9. **Final calculation**:
\[ \frac{86}{-13} = -\frac{86}{13}. \]

Thus, the final answer is:
\[ \frac{\lambda_1}{\lambda_2^2} + \frac{\lambda_2}{\lambda_1^2} = -\frac{86}{13}. \]