The point `(alpha^(2)+2lamda+5,lamda^(2)+1)` lies on the line `x+y=10` for:

Options

A. All real value of `lamda`
B. Some real value of `lamda`
C. `lamda=-1`
D. `lamda=2`

Answer

To determine for which values of \( \lambda \) the point \( ( \lambda^2 + 2\lambda + 5, \lambda^2 + 1 ) \) lies on the line \( x + y = 10 \), we can follow these steps:

### Step 1: Substitute the point into the line equation
We start by substituting the coordinates of the point into the equation of the line.

\[ x + y = 10 \]

Substituting \( x = \lambda^2 + 2\lambda + 5 \) and \( y = \lambda^2 + 1 \):

\[ (\lambda^2 + 2\lambda + 5) + (\lambda^2 + 1) = 10 \]

### Step 2: Simplify the equation
Combine like terms:

\[ 2\lambda^2 + 2\lambda + 6 = 10 \]

### Step 3: Rearrange the equation
Now, we will rearrange the equation to set it to zero:

\[ 2\lambda^2 + 2\lambda + 6 - 10 = 0 \]

This simplifies to:

\[ 2\lambda^2 + 2\lambda - 4 = 0 \]

### Step 4: Divide the equation by 2
To simplify further, divide the entire equation by 2:

\[ \lambda^2 + \lambda - 2 = 0 \]

### Step 5: Solve the quadratic equation
Now, we will use the quadratic formula to find the values of \( \lambda \). The quadratic formula is given by:

\[ \lambda = \frac{-b \pm \sqrt{D}}{2a} \]

Where \( a = 1, b = 1, c = -2 \). First, we need to calculate the discriminant \( D \):

\[ D = b^2 - 4ac = 1^2 - 4(1)(-2) = 1 + 8 = 9 \]

### Step 6: Calculate the roots
Now substituting \( D \) back into the quadratic formula:

\[ \lambda = \frac{-1 \pm \sqrt{9}}{2 \cdot 1} = \frac{-1 \pm 3}{2} \]

This gives us two potential solutions:

1. \( \lambda = \frac{-1 + 3}{2} = \frac{2}{2} = 1 \)
2. \( \lambda = \frac{-1 - 3}{2} = \frac{-4}{2} = -2 \)

### Conclusion
The values of \( \lambda \) for which the point lies on the line \( x + y = 10 \) are \( \lambda = 1 \) and \( \lambda = -2 \).