Is there a pattern to extraneous solutions when an equation has a square root sign? I came across this specific question.
$$ \sqrt{8x^2+17} = 3x-2 $$
Squaring both sides and solving, two roots can be reached: x = 13 and x = -1.
Due to the square root in the original equation, x = -1 is an extraneous solution as the square root, by definition, returns the principal square root. This is found by plugging the solutions back into the original equation.
I have two questions.
First, when the equation's highest degree is 2, will there always be one true solution and one extraneous one? That is, can there be two true solutions or two extraneous solutions?
Second, is there a pattern to extraneous solutions to questions of a similar setup (those that involve a square root needing to be squared to be solved)? Of course, this second question is moot if the answer to the first question is no.
This question touches on the question tangentially: Implied plus-minus sign in radical equation?.
Thanks and apologies for the elementary-levelled question in advance.
Edit: In addition, is there a way to visualize what this extraneous solution (x = -1) is and why it is not a valid solution?
 A: If the radical equation involves some even index radicals, and no prior restrictions have been made, the existence of real extraneous roots is always possible.
Note that, if the necessary constraints have already been made, you will not get extraneous roots. For example, applying the restriction $3x-2≥0$, then, we won't get any extraneous solutions.
For instance, let $A(x),\,B(x),\,C(x)$ be some algebraic expressions, such that:
$$\sqrt {\sqrt [4]{A(x)}+B(x)}=C(x)$$
The given equation contains even index radicals. Therefore, the existence of extraneous roots is possible, if we don't apply all the necessary restrictions.
Because, the set of roots of the radical equation $$\sqrt {\sqrt [4]{A(x)}+B(x)}=C(x)$$ is a subset of the set of roots of the polynomial $$\left(\left(C(x)\right)^2-B(x)\right)^4-A(x)=0.$$ These sets may not always be equal.
Now, consider the odd index radical equation:
$$\sqrt [7]{\sqrt[5] {A(x)}+B(x)}=C(x)$$
In this case, the set of root of the equation $$\sqrt [7]{\sqrt[5] {A(x)}+B(x)}=C(x)$$ is equivalent to the set of root of the polynomial
$$\left(\left (C(x)\right)^7-B(x)\right)^5-A(x)=0.$$
In other words, all real numbers satisfying the polynomial will also be the root of the original equation. No real extraneous root can exist.
If the radical equation contains both odd and even index radicals, then it is still possible to have an real extraneous root.
The reason for this is simple. Observe that,
$$A(x)=B(x)\iff A^{2n-1}(x)=B^{2n-1}(x)$$
holds for all $n\in\mathbb N^{+}$.
But,
$$A(x)=B(x) \require {cancel} \hspace{1em}\cancel{\hspace{-1em}{\iff}\hspace{-1em}}\hspace{1em}A^{2n}(x)=B^{2n}(x)$$
doesn't hold for all $n\in\mathbb N^{+}$.
Also, it should be noted that we only work on real-valued roots here. When solving radical equations, we sometimes get extraneous roots because the set of real-valued roots of the polynomial we get as a result of raising to the power is not always equal to the set of roots of the original radical equation.
