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The problem is to find all real solutions (if any exists) for $sqrt{2x-3} +x=3$.

Now, my textbook says the answer is {2}, however, I keep getting {2, 6}. I've tried multiple approaches, but here is one of them:

I got rid of the root by squaring both sides,
$$sqrt{2x-3}^2=(3-x)^2$$
$$0=12-8x+x^2$$
Using the AC method, I got
$$(-x^2+6x)(2x-12)=0$$
$$-x(x-6)2(x-6)=0$$
$$(-x+2)(x-6)=0$$
hence, $$x=2, x=6$$

Of course, I can always just check my solutions and I'd immediately recognize 6 does not work. But that's a bit too tedious for my taste. Can anyone explain where I went wrong with my approach?

Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.

As a very simple example, notice the following two equations:

$$x = sqrt 4 iff x = +2$$

$$x^2 = 4 iff vert xvert = 2 iff x = pm 2$$

The first equation has only one solution: $+sqrt 4$. The second, however, has two solutions: $pmsqrt 4$. And you get the second equation by squaring the first one.

The exact same idea applies to your example. You have

$$sqrt{2x-3} = 3-x$$

which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving

$$0 = 12-8x+x^2 iff color{blue}{pm}sqrt{2x-3} = 3-x$$

which has a $pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:

$$2x-3 = 9-6x+x^2; quad color{blue}{x leq 3}$$

$$0 = 12-8x+x^2; quad color{blue}{x leq 3}$$

Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.

When we square both sides, we could have introduce additional solution.

An extreme example is as follows:

Solve $x=1$.

The solution is just $x=1$.

However, if we square them, $x^2=1$. Now $x=-1$ also satisfies the new equation which is no longer the original problem.

Remark: Note that as we write $$sqrt{2x-3}=3-x,$$

there is an implicit constraint that we need $3-x ge 0$.

Squaring both sides of an equation can introduce extraneous solutions. Thus it is necessary when doing so to check your answer.

Notice:$$sqrt{2cdot 6-3}+6=9neq3$$.

The initial question is actually:

If $x$ exists, then it satisfies $sqrt{2x-3}+x=3$. What is $x$?

With each algebraic step, if-then logic is used to rephrase the initial question, eventually leading to:

If $x$ exists, then it satisfies $x in {2, 6}$. What is $x$?

If all the logical steps are reversible, then we are done. We could 'let $x = 2$ or $x = 6$' and follow the logic backwards to demonstrate that x is a solution to the original equation. Unfortunately, as noted on other answers, squaring is not a reversible step; the inverse of the square root function is not the same as the square function. We can see this by noting that the square function takes positive and negative numbers and maps them to positive numbers. Meanwhile, the square root function takes positive numbers only and maps them to positive numbers only.

All this is a long way of saying that the alternative to checking answers is understanding which algebraic steps are reversible and which are not. In practice, this is much harder than just checking every answer.

We can use also the following way.

We need to solve
$$2sqrt{2x-3}+2x=6$$ or
$$2x-3+2sqrt{2x-3}+1=4$$ or
$$left(sqrt{2x-3}+1right)^2=4$$ or
$$sqrt{2x-3}+1=2$$ or
$$sqrt{2x-3}=1$$ or $$x=2.$$