<s>[INST] How to solve this irrational equation? How to solve this equation in the set real numbers $$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$ Using Mathematica, I know, this equation has two solutions $x = 1$ and $x = 3.$
 [/INST] Note that, with $x=1$, $y=\sqrt{8x+1}$ takes the value $y=3$ and  with $x=3$, $y=\sqrt{8x+1}$ takes the value $y=5$. Put the point $A(1, 3)$ and $B(3,5)$. The equation of the line passing two points $A$ and $B$ is $y = x + 2.$ And then we write $$\sqrt{8x + 1}-(x+2).$$ Similar to with  $$(x+1- \sqrt{6x - 2}).$$
We write the given equation has the form
$$ \sqrt{8x + 1}-(x+2)  + (x+1- \sqrt{6x - 2})  =2(x^2 - 4x + 3). $$
equavalent to $$ \dfrac{-(x^2 -4x + 3)}{ \sqrt{8x + 1}+(x+2)}+ \dfrac{x^2 -4x + 3}{x+1 +  \sqrt{6x - 2}}=2(x^2 - 4x + 3).$$
Or
$$ (x^2 -4x + 3)\left(\dfrac{1}{ \sqrt{8x + 1}+(x+2)}+ 2- \dfrac{1}{x+1 +  \sqrt{6x - 2}}\right)=0.$$
With $x\geqslant \dfrac{1}{3}$, it is easy to see that
$$  \dfrac{1}{ \sqrt{8x + 1}+(x+2)}+ 2- \dfrac{1}{x+1 +  \sqrt{6x - 2}} >0.$$
Therefore we get $$  x^2 -4x + 3 = 0.$$
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