1. Given equation

We need to solve

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

Let

$$A=x^{2}-4x+3=(x-1)(x-3),$$ $$B=x^{2}-9=(x-3)(x+3),$$ $$C=4x^{2}-14x+6=2(2x-1)(x-3).$$

So the equation is

$$\sqrt{A}+\sqrt{B}=\sqrt{C}.$$

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2. Domain conditions

For all square roots to be defined, we need:

$$A\ge 0,\quad B\ge 0,\quad C\ge 0.$$

Now,

• $A=(x-1)(x-3)\ge 0 \implies x\le 1$ or $x\ge 3$,
• $B=(x-3)(x+3)\ge 0 \implies x\le -3$ or $x\ge 3$,
• $C=2(2x-1)(x-3)\ge 0 \implies x\le \tfrac12$ or $x\ge 3$.

Taking intersection:

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

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3. Square the equation

Since both sides are non-negative, squaring is valid:

$$(\sqrt{A}+\sqrt{B})^2=C.$$

So,

$$A+B+2\sqrt{AB}=C.$$

Compute:

$$A+B=(x^2-4x+3)+(x^2-9)=2x^2-4x-6,$$ $$C=4x^2-14x+6.$$

Hence,

$$2\sqrt{AB}=C-(A+B)=4x^2-14x+6-(2x^2-4x-6)=2x^2-10x+12.$$

Thus,

$$\sqrt{AB}=x^2-5x+6=(x-2)(x-3).$$

Now,

$$AB=(x^2-4x+3)(x^2-9)=(x-1)(x-3)^2(x+3).$$

Squaring again,

$$AB=(x^2-5x+6)^2=((x-2)(x-3))^2.$$

So,

$$(x-1)(x-3)^2(x+3)=(x-2)^2(x-3)^2.$$

Therefore,

$$(x-3)^2\big((x-1)(x+3)-(x-2)^2\big)=0.$$

Now simplify the bracket:

$$(x-1)(x+3)=x^2+2x-3,$$ $$(x-2)^2=x^2-4x+4.$$

Thus,

$$(x-1)(x+3)-(x-2)^2=(x^2+2x-3)-(x^2-4x+4)=6x-7.$$

Hence,

$$(x-3)^2(6x-7)=0.$$

Candidate solutions are:

$$x=3,\quad x=\frac76.$$

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4. Check with domain and original equation

From the domain,

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

• $x=\frac76$ is not in the domain, so reject.
• $x=3$ is in the domain.

Check $x=3$ in original equation:

$$\sqrt{9-12+3}+\sqrt{9-9}=\sqrt{36-42+6}$$ $$=\sqrt0+\sqrt0=\sqrt0,$$

which is true.

So there is exactly one real root.

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5. Option matching

The number of real roots is

$$1.$$

So the correct option is B.

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6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.