Solve
$$\sqrt{2x^2 - 7x + 1} - \sqrt{2x^2 - 9x + 4} = 1 \tag1$$
I tried to do the following:
$$(2x^2 - 7x + 1) - (2x^2 - 9x + 4) = 2x-3\tag2$$
Dividing $(2)$ by $(1)$ yields
$$\sqrt{2x^2 - 7x + 1} + \sqrt{2x^2 - 9x + 4} = 2x-3 \tag3$$
Now, adding $(1)$ and $(3)$
$$2\sqrt{2x^2 - 7x + 1} = 2x-4$$
$$\implies \sqrt{2x^2 - 7x + 1} = x-2$$
Squaring both sides and simplifying, yields
$$x^2 - 3x - 3 = 0$$
Solving this quadratic yields an irrational solution. But the solution given in my textbook are $0, 5$ which do work in $(1)$. Where did I go wrong?
When you add (1) and (3), the right side should be $2x-2$. Then it simplifies to $x^2 - 5x = 0$ which has the solutions 0 and 5.