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

Options

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

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Rewrite the given quadratic equation
The given equation is:
\[ \lambda (x^2 - x) + x + 5 = 0 \]
This can be rewritten as:
\[ \lambda x^2 - \lambda x + x + 5 = 0 \]
which simplifies to:
\[ \lambda x^2 + (1 - \lambda)x + 5 = 0 \]

### Step 2: Identify coefficients
From the quadratic equation \(ax^2 + bx + c = 0\), we have:
- \(a = \lambda\)
- \(b = 1 - \lambda\)
- \(c = 5\)

### Step 3: Use Vieta's formulas
According to Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{1 - \lambda}{\lambda}\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{5}{\lambda}\)

### Step 4: Use the given condition
We are given that:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 4 \]
This can be rewritten as:
\[ \frac{\alpha^2 + \beta^2}{\alpha \beta} = 4 \]
Using the identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\), we substitute:
\[ \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = 4 \]
Substituting the values from Vieta's:
\[ \frac{\left(-\frac{1 - \lambda}{\lambda}\right)^2 - 2\left(\frac{5}{\lambda}\right)}{\frac{5}{\lambda}} = 4 \]

### Step 5: Simplify the equation
This simplifies to:
\[ \frac{\frac{(1 - \lambda)^2}{\lambda^2} - \frac{10}{\lambda}}{\frac{5}{\lambda}} = 4 \]
Multiplying through by \(\frac{5}{\lambda}\):
\[ \frac{(1 - \lambda)^2}{\lambda^2} - \frac{10}{\lambda} = 20 \]
Rearranging gives:
\[ \frac{(1 - \lambda)^2}{\lambda^2} = 20 + \frac{10}{\lambda} \]
Multiplying through by \(\lambda^2\):
\[ (1 - \lambda)^2 = 20\lambda^2 + 10\lambda \]

### Step 6: Expand and rearrange
Expanding the left side:
\[ 1 - 2\lambda + \lambda^2 = 20\lambda^2 + 10\lambda \]
Rearranging gives:
\[ \lambda^2 - 22\lambda + 1 = 0 \]

### Step 7: Find the roots \(\lambda_1\) and \(\lambda_2\)
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
Calculating the discriminant:
\[ \sqrt{484 - 4} = \sqrt{480} = 4\sqrt{30} \]
Thus:
\[ \lambda_{1,2} = \frac{22 \pm 4\sqrt{30}}{2} = 11 \pm 2\sqrt{30} \]

### Step 8: Calculate \(\frac{\lambda_1}{\lambda_2^2} + \frac{\lambda_2}{\lambda_1^2}\)
Let \(\lambda_1 = 11 + 2\sqrt{30}\) and \(\lambda_2 = 11 - 2\sqrt{30}\).
We need to find:
\[ \frac{\lambda_1}{\lambda_2^2} + \frac{\lambda_2}{\lambda_1^2} \]
Using the identity:
\[ \frac{a}{b^2} + \frac{b}{a^2} = \frac{a^3 + b^3}{a^2b^2} \]
where \(a = \lambda_1\) and \(b = \lambda_2\):
\[ \lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)(\lambda_1^2 - \lambda_1\lambda_2 + \lambda_2^2) \]
Calculating:
- \(\lambda_1 + \lambda_2 = 22\)
- \(\lambda_1 \lambda_2 = 1\)
- \(\lambda_1^2 + \lambda_2^2 = (\lambda_1 + \lambda_2)^2 - 2\lambda_1\lambda_2 = 484 - 2 = 482\)

Thus:
\[ \lambda_1^3 + \lambda_2^3 = 22(482 - 1) = 22 \times 481 = 10682 \]

### Step 9: Calculate \(\lambda_1^2 \lambda_2^2\)
\[ \lambda_1^2 \lambda_2^2 = (\lambda_1 \lambda_2)^2 = 1^2 = 1 \]

### Step 10: Final Calculation
Now substituting back:
\[ \frac{\lambda_1^3 + \lambda_2^3}{\lambda_1^2 \lambda_2^2} = \frac{10682}{1} = 10682 \]

Thus, the final answer is:
\[ \boxed{10682} \]