The number of real roots of the equation $$\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$$, is:

To find the number of real roots of the given radical equation, we first factorize each quadratic expression under the square roots:

$$\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$$

Factoring the expressions yields:

$$\sqrt{(x - 1)(x - 3)} + \sqrt{(x + 3)(x - 3)} = \sqrt{2(2x - 1)(x - 3)}$$

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Step 1: Determine the domain of the equation

For the functions to be defined in the real number system, the terms inside the square roots must be non-negative:

$$(x - 1)(x - 3) \ge 0 \implies x \in (-\infty, 1] \cup [3, \infty)$$

$$(x + 3)(x - 3) \ge 0 \implies x \in (-\infty, -3] \cup [3, \infty)$$

$$2(2x - 1)(x - 3) \ge 0 \implies x \in \left(-\infty, \frac{1}{2}\right] \cup [3, \infty)$$

Taking the intersection of all three intervals gives the valid domain:

$$x \in (-\infty, -3] \cup [3, \infty)$$

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Step 2: Solve the equation

From the factored equation, it is clear that $$x = 3$$ makes every term zero:

$$\sqrt{0} + \sqrt{0} = \sqrt{0} \implies 0 = 0$$

Thus, $$x = 3$$ is a valid real root. We now look for other potential roots by assuming $$x \neq 3$$.

Case 1: For $$x > 3$$

Since $$x - 3 > 0$$, we can divide the entire equation by $$\sqrt{x - 3}$$:

$$\sqrt{x - 1} + \sqrt{x + 3} = \sqrt{2(2x - 1)}$$

Squaring both sides of the equation:

$$(x - 1) + (x + 3) + 2\sqrt{(x - 1)(x + 3)} = 2(2x - 1)$$

$$2x + 2 + 2\sqrt{x^2 + 2x - 3} = 4x - 2$$

$$2\sqrt{x^2 + 2x - 3} = 2x - 4$$

$$\sqrt{x^2 + 2x - 3} = x - 2$$

Squaring both sides again:

$$x^2 + 2x - 3 = (x - 2)^2$$

$$x^2 + 2x - 3 = x^2 - 4x + 4$$

$$6x = 7 \implies x = \frac{7}{6}$$

Since $$x = \frac{7}{6}$$ does not satisfy the condition $$x > 3$$, it is an extraneous solution.

Case 2: For $$x \le -3$$

Since $$x - 3 < 0$$, we can rewrite the terms as $$-(3 - x)$$ where $$3 - x > 0$$. Dividing the original equation by $$\sqrt{3 - x}$$ gives:

$$\sqrt{1 - x} + \sqrt{-x - 3} = \sqrt{2(1 - 2x)}$$

Squaring both sides of the equation:

$$(1 - x) + (-x - 3) + 2\sqrt{(1 - x)(-x - 3)} = 2(1 - 2x)$$

$$-2x - 2 + 2\sqrt{x^2 + 2x - 3} = 2 - 4x$$

$$2\sqrt{x^2 + 2x - 3} = 4 - 2x$$

$$\sqrt{x^2 + 2x - 3} = 2 - x$$

Squaring both sides again:

$$x^2 + 2x - 3 = (2 - x)^2$$

$$x^2 + 2x - 3 = x^2 - 4x + 4$$

$$6x = 7 \implies x = \frac{7}{6}$$

Since $$x = \frac{7}{6}$$ does not satisfy the condition $$x \le -3$$, it is also an extraneous solution.

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Step 3: Conclusion

The only value that satisfies the equation within the valid domain is $$x = 3$$.

Therefore, the number of real roots of the equation is equal to 1.