4
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Small typo: $\sqrt{2x-2}^2$ should read $\sqrt{2x-3}^2$. $\endgroup$ – T. Fo Dec 29 '18 at 5:52
  • 8
    $\begingroup$ Possible duplicate of Why do extraneous solutions exist? or When do we get extraneous roots? $\endgroup$ – Carmeister Dec 29 '18 at 8:09
  • $\begingroup$ The simplest way to "debug" such things is to check the solutions at each step, then you can (usually) find out the first wrong step. $\endgroup$ – user202729 Dec 29 '18 at 9:07
  • 2
    $\begingroup$ It's not duplicate of course. $\endgroup$ – Michael Rozenberg Dec 30 '18 at 19:04
16
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Maybe it would be a little bit more correct to say that the RHS should be non-negative, since the LHS must be non-negative. There is no way to make the LHS negative. $\endgroup$ – Kai Dec 29 '18 at 22:47
10
$\begingroup$

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 $3-x \ge 0$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

To build off of the other answers provided, the equation you wish to solve is

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

To do so, you square both sides, and solve

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

which has two solutions. However this equation can also be obtained by squaring

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

The second solution is the solution to this second equation. This is easy to see with a plot:

enter image description here

$\endgroup$
1
$\begingroup$

Your reasoning is a chain of implications: if $x_0$ is a solution, then..., then $x_0$ should be $\cdot$ or $\cdot$. As you do not have equivalences instead of implications at each step, the final potential $x$s are only a superset of solutions, that should be plugged into the original problem, to see if they fit.

A change in variables can show some additional insights. To get rid of the root, you can choose a positive $y$ such that $y^2 = 2x-3$, and thus $ 2x-3=\sqrt{y}$.

You can rewrite your equation as:

$$y+(y^2+3)/2 = 3$$

or

$$2y+y^2 = 3$$

This system has at most two solutions $y_a$ and $y_b$: an obvious solution is real, satisfying $2\times 1+1^2=3$, or $y_a=1$, which yields $x=2$. Since $y_a\times y_b = -3$, the second solution would be negative, which is ruled out by hypothesis.

So $x=2$ is the only solution.

$\endgroup$
0
$\begingroup$

Squaring both sides of an equation can introduce extraneous solutions. For instance, $x=-2$ isn't a solution of $x=2$; but it is a solution of $x^2=4$.

Thus it is necessary when doing so to check your answer.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.