Extraneous Solutions: Which step did it come from?

[yucheng]

Homework Statement

Solve the following:

Relevant Equations

$x^2-3|x|-2=0$

I used the identity $\sqrt{x^2}=|x|$ and completed the square as follows:

\begin{align*}
x^2-3|x|-2&=0 \tag1\\
\sqrt{x^4}-3\sqrt{x^2}-2&=0 \tag2\\
(\sqrt{x^2}-\frac{3}{2})^2-\frac{9}{4}-2&=0 \tag3\\
(\sqrt{x^2}-\frac{3}{2})^2&=\frac{17}{4} \tag4\\
\sqrt{x^2}-\frac{3}{2}&=\pm\frac{\sqrt{17}}{2} \tag5\\
\sqrt{x^2}&=\frac{3 \pm\sqrt{17}}{2} \tag6\\
x&= \pm\frac{3 \pm\sqrt{17}}{2} \tag7
\end{align*}

After substituting the solutions back into the equations, the solutions are:$x=±\frac{3+\sqrt{17}}{2}$

Which step did the extraneous solution $x=±\frac{3−\sqrt{17}}{2}$ originate? The method I used to search for this step is by checking from which step on did the extraneous solution satisfy the equation. I tried (5) but it was not. Or, was it because I used the identity $\sqrt{x^2}=|x|$? It seems unlikely...

 

[mfb]

From 4 to 5. Taking the square root and using $\pm$ can come with extra solutions.
Oh, and you didn't take the square root on the right side in (5).

I would use $x^2$ = $|x|^2$ and solve the quadratic equation in $|x|$, that's a bit less awkward to write and it's clearer that it needs to be positive.

 

[yucheng]

From 4 to 5. Taking the square root and using $\pm$ can come with extra solutions.

Does this mean that in general, taking square roots will come up with extra solutions, as in completing the square on any other polynomial? From what I know, completing the square does not lead to extraneous solutions.

Example:
\begin{align*}
x^2+2x+3&=0 \\
(x+1)^2-1+3&=0 \\
(x+1)^2&=2 \\
(x+1)&=\pm\sqrt{2} \\
x&=1 \pm\sqrt{2}
\end{align*}

Oh, and you didn't take the square root on the right side in (5).

Uh oh, I made a mistake. Step (5) is correct, but I made a jump at step (4). I have corrected it.

Let me try your $x^2=|x|^2$... Indeed it's neater.

 

[PeroK]

Homework Statement:: Solve the following:
Relevant Equations:: $x^2-3|x|-2=0$

My approach would be to look for solutions for $x > 0$ and $x < 0$ separately. Then nothing can go wrong!

 

[FactChecker]

My approach would be to look for solutions for $x > 0$ and $x < 0$ separately. Then nothing can go wrong!

At least nothing can go wrong as long as you remember to compare your two solutions in each case with your initial assumption about x. As usual, when the quadratic formula is used, it is necessary to check the validity of the two solutions when other constraints are considered.

 

[PeroK]

At least nothing can go wrong as long as you remember to compare your two solutions in each case with your initial assumption about x. As usual, when the quadratic formula is used, it is necessary to check the validity of the two solutions when other constraints are considered.

That's true.

 

[mfb]

The original formula is symmetric around x=0, so all solutions will be in pairs anyway.

Completing the square isn't changing anything so this doesn't add extra solutions.
With the fixed (4) it's now easy to see why taking the square root adds extra solutions. $\sqrt{x^2}\geq 0$, therefore $\sqrt{x^2}-\frac 3 2 \geq -\frac 3 2$. It can't be $\frac {-\sqrt{17}}{2} < -2$, only the positive option is meaningful.

 

[haruspex]

Which step did the extraneous solution x=±3−172 originate?

The final step. Equation (6) still doesn't allow a solution in which the RHS is negative, but equation (7) does.

Does this mean that in general, taking square roots will come up with extra solutions

Not in general. It is usually squaring that introduces extra solutions.
Here, it is that you turned $\sqrt {x^2}$ into $x$ instead of $|x|$,