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 $x = 3.$

Write:
$$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$ 
As:
$$(\sqrt{8x + 1} - \sqrt{6x - 2}) = (2x^2 - 8x + 7).$$ 
Squaring both sides yields:
$$14 x-2 \sqrt{6 x-2}\sqrt{8 x+1}-1 = 4x^4-32x^3+92x^2-112x+49$$
Simplifying:
$$-2 \sqrt{6 x-2}\sqrt{8 x+1} = 4x^4-32x^3+92x^2-112x+49$$
Squaring both sides again yields:
$$ 4(6x-2)(8x+1) = 16 x^8-256 x^7+1760 x^6-6896 x^5+16928 x^4-26384 x^3+25076 x^2-12600 x+2500$$
Simplifying:
$$16 x^8-256 x^7+1760 x^6-6896 x^5+16928 x^4-26384 x^3+24884 x^2-12560 x+2508 = 0$$
This gives us more roots, so care must be taken, but indeed, we get back two of the $8$ roots as $x = 1$ and $x = 3$.
Numerical methods would also have worked earlier and guessing too.
Regards
