The number of real roots of the equation `sqrt(x^(2)-4x+3)+sqrt(x^(2)-9)=sqrt(4x^(2)-14x+6)`,is :

Options

A. `2`
B. `3`
C. `1`
D. `0`

Answer

To solve the equation

\[ \sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}, \]

we will follow these steps:

### Step 1: Simplify the Square Roots
First, we simplify the expressions under the square roots.

1. The first square root:
\[ x^2 - 4x + 3 = (x-1)(x-3). \]
2. The second square root:
\[ x^2 - 9 = (x-3)(x+3). \]
3. The right-hand side:
\[ 4x^2 - 14x + 6 = 4(x^2 - \frac{14}{4}x + \frac{6}{4}) = 4(x^2 - \frac{7}{2}x + \frac{3}{2}). \]

### Step 2: Set the Equation
Now, we can rewrite the equation as:
\[ \sqrt{(x-1)(x-3)} + \sqrt{(x-3)(x+3)} = \sqrt{4(x^2 - \frac{7}{2}x + \frac{3}{2})}. \]

### Step 3: Analyze the Domain
For the square roots to be defined, we need:
1. \(x^2 - 4x + 3 \geq 0\) which gives \(x \leq 1\) or \(x \geq 3\).
2. \(x^2 - 9 \geq 0\) which gives \(x \leq -3\) or \(x \geq 3\).
3. \(4x^2 - 14x + 6 \geq 0\) which gives the roots of the quadratic equation.

### Step 4: Solve for Roots
We can find the roots of the quadratic \(4x^2 - 14x + 6 = 0\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 4 \cdot 6}}{2 \cdot 4}. \]
Calculating the discriminant:
\[ 196 - 96 = 100 \implies \sqrt{100} = 10. \]
Thus, the roots are:
\[ x = \frac{14 \pm 10}{8} \implies x = 3 \text{ or } x = \frac{1}{2}. \]

### Step 5: Check Validity of Roots
Now we check which of these roots are valid in the context of the original equation:
- For \(x = 3\):
\[ \sqrt{(3-1)(3-3)} + \sqrt{(3-3)(3+3)} = 0 + 0 = 0. \]
- For \(x = \frac{1}{2}\):
\[ \sqrt{(\frac{1}{2}-1)(\frac{1}{2}-3)} + \sqrt{(\frac{1}{2}-3)(\frac{1}{2}+3)} \text{ (not valid since square roots are not defined)}. \]

### Conclusion
The only valid root is \(x = 3\). Therefore, the number of real roots of the equation is:

\[ \boxed{1}. \]