Roots of the equation `(x^2-4x+3)+lamda(x^2-6x+8)=0` , `lamda in R` will be

Options

A. always real
B. real only when `lambda` is positive
C. real only when `lambda` is negative
D. always is magnary

Explanation

To solve the equation \((x^2 - 4x + 3) + \lambda (x^2 - 6x + 8) = 0\) for the roots, we can follow these steps:

### Step 1: Rewrite the Equation
Start by rewriting the given equation:
\[ x^2 - 4x + 3 + \lambda (x^2 - 6x + 8) = 0 \]
This can be rearranged as:
\[ (1 + \lambda)x^2 + (-4 + 6\lambda)x + (3 + 8\lambda) = 0 \]

### Step 2: Identify Coefficients
From the rearranged equation, we can identify the coefficients:
- \(a = 1 + \lambda\)
- \(b = -4 + 6\lambda\)
- \(c = 3 + 8\lambda\)

### Step 3: Calculate the Discriminant
The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (-4 + 6\lambda)^2 - 4(1 + \lambda)(3 + 8\lambda) \]

### Step 4: Expand the Discriminant
Now, we will expand \(D\):
1. Calculate \(b^2\):
\[ (-4 + 6\lambda)^2 = 16 - 48\lambda + 36\lambda^2 \]
2. Calculate \(4ac\):
\[ 4(1 + \lambda)(3 + 8\lambda) = 4(3 + 8\lambda + 3\lambda + 8\lambda^2) = 12 + 44\lambda + 32\lambda^2 \]
3. Combine to find \(D\):
\[ D = (16 - 48\lambda + 36\lambda^2) - (12 + 44\lambda + 32\lambda^2) \]
\[ D = 16 - 48\lambda + 36\lambda^2 - 12 - 44\lambda - 32\lambda^2 \]
\[ D = 4 + 4\lambda^2 - 92\lambda \]

### Step 5: Analyze the Discriminant
For the roots to be real, the discriminant must be greater than or equal to zero:
\[ 4 + 4\lambda^2 - 92\lambda \geq 0 \]
Dividing the entire inequality by 4:
\[ 1 + \lambda^2 - 23\lambda \geq 0 \]
Rearranging gives:
\[ \lambda^2 - 23\lambda + 1 \geq 0 \]

### Step 6: Find the Roots of the Quadratic
To find the roots of the quadratic equation \(\lambda^2 - 23\lambda + 1 = 0\), we use the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -23\), and \(c = 1\):
\[ \lambda = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{23 \pm \sqrt{529 - 4}}{2} = \frac{23 \pm \sqrt{525}}{2} \]
Calculating \(\sqrt{525}\):
\[ \sqrt{525} = \sqrt{25 \cdot 21} = 5\sqrt{21} \]
Thus, the roots are:
\[ \lambda = \frac{23 \pm 5\sqrt{21}}{2} \]

### Step 7: Determine the Intervals
The quadratic \(\lambda^2 - 23\lambda + 1\) opens upwards (since the coefficient of \(\lambda^2\) is positive). Therefore, it is non-negative outside the roots:
\[ \lambda \leq \frac{23 - 5\sqrt{21}}{2} \quad \text{or} \quad \lambda \geq \frac{23 + 5\sqrt{21}}{2} \]

### Conclusion
The roots of the equation \((x^2 - 4x + 3) + \lambda (x^2 - 6x + 8) = 0\) will be real for values of \(\lambda\) outside the interval:
\[ \lambda \in \left(-\infty, \frac{23 - 5\sqrt{21}}{2}\right] \cup \left[\frac{23 + 5\sqrt{21}}{2}, +\infty\right) \]