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

Given,

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

$\Rightarrow \sqrt{\left(x-3\right)\left(x-1\right)}+\sqrt{\left(x-3\right)\left(x+3\right)}=\sqrt{\left(x-3\right)\left(4x-2\right)}$

$\Rightarrow \sqrt{\left(x-3\right)}\sqrt{\left(x-1\right)}+\sqrt{\left(x-3\right)}\sqrt{\left(x+3\right)}-\sqrt{\left(x-3\right)}\sqrt{\left(4x-2\right)}=0$

$\Rightarrow \sqrt{\left(x-3\right)}=0 \text{or }\left[\sqrt{\left(x-1\right)}+\sqrt{\left(x+3\right)}-\sqrt{\left(4x-2\right)}\right]=0$

So, $x=3$ is one solution and also $x\ge 3$

Now solving $\left[\sqrt{\left(x-1\right)}+\sqrt{\left(x+3\right)}-\sqrt{\left(4x-2\right)}\right]=0$

$\Rightarrow \sqrt{\left(x-1\right)}+\sqrt{\left(x+3\right)}=\sqrt{\left(4x-2\right)}$

Now squaring both side we get,

$\Rightarrow \left(x-1\right)+\left(x+3\right)+2\sqrt{\left(x-1\right)\left(x+3\right)}=\left(4x-2\right)$

$\Rightarrow 2\sqrt{\left(x-1\right)\left(x+3\right)}=\left(2x-4\right)$

$\Rightarrow \left(x-1\right)\left(x+3\right)={\left(x-2\right)}^2$

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

$\Rightarrow 6x=7\Rightarrow x=\frac{7}{6}$ which does not satisfy the given expression and $x\ge 3$

So, there is only one solution which is $x=3$