Solution of an equation involving a single variable

$\begin{aligned} \sqrt{2x-3}+\sqrt{5x-1}&=\sqrt{3x-2}\\ \sqrt{5x-1}&=\sqrt{3x-2}-\sqrt{2x-3}\\ 5x-1&=(3x-2)-2(\sqrt{3x-2})(\sqrt{2x-3})+(2x-3)\\ 4&=-2\sqrt{(3x-2)(2x-3)}\\ -2&=\sqrt{6x^2-13x+6}\\ 6x^2-13x+6&=4\\ 6x^2-13x+2&=0\\ (x-2)(6x-1)&=0\\ x&=2\text{ (rej.) or }x=\frac16\text{ (rej.)}\\ \therefore\boxed{\text{no solution.}} \end{aligned}$

It's tempting to square both sides immediately, but rearranging first leads to some helpful cancellation:

$\sqrt{5x-1}=\sqrt{3x-2}-\sqrt{2x-3}$

Squaring:

$5x-1=3x-2-2\sqrt{3x-2} \cdot \sqrt{2x-3} + 2x-3=5x-5-2\sqrt{3x-2} \cdot \sqrt{2x-3}$

Cancelling and rearranging again:

$4=-2\sqrt{3x-2} \cdot \sqrt{2x-3}$

Dividing through by $2$ and squaring again:

$4=(3x-2)(2x-3)$

This is just a quadratic; expanding and rearranging we have $6x^2-13x+2=0$ , with roots $x=\frac16$ and $x=2$ .

Note the order of the logic above: we've shown that if $x$ is a root of the original equation, it must be either $\frac16$ or $2$ . This doesn't mean that these are roots; we need to check by substituting back in to the original equation, and we find neither of these values works. Since the above chain of reasoning showed that these were the only two possibilities we had to check, we can conclude that the equation has no roots .

$(A+B)^2 = C^2$

Where $A = \sqrt(2x-3)$ , $B = \sqrt(5x-1)$ , and $C = \sqrt(3x-2)$

First, square both sides of the equation. To square the lefthand side, see below. Do not simply remove the radicals from each term.

$(A+B)(A+B) = A^2 + 2AB + B^2$

$\begin{cases} A^2 & = 2x+3 \\ B^2 & = 5x-1 \\ AB & = 10x^2-2x+15x-3 = 10x^2+13x-3 \\ C^2 & = 3x-2 \end{cases}$

So, squaring both sides of the equation gives:

$(2x + 3) + 2\sqrt(10x^2-2x+15x-3 = 10x^2+13x-3) + (5x - 1) = 3x - 2$

Simplified:

$-2x-2 = \sqrt(10x^2+13x-3 = 10x^2+13x-3)$

$0 = 6x^2+5x-7$

This does not give an integer value for x, so the equation does not have a rational solution (input "0").