Why do I keep getting this incorrect solution when trying to find all the real solutions for $\sqrt{2x-3}\ +x=3$.

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:

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

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

The first equation has only one solution: $+\sqrt 4$. The second, however, has two solutions: $\pm\sqrt 4$. 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 $3-x \ge 0$.


The initial question is actually:

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

With each logically sound algebraic step, the initial question is rephrased, eventually leading to:

If $x$ exists, then it satisfies $x = 2\text{ or } x= 6$. What is $x$?

Unfortunately, we still have done nothing to prove x exists. If all the logical steps are if and only if , or 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 square root function is not the same as the inverse of 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, its easier to just check your answers every time.

Tags:

Algebra Precalculus

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