This post follows on from this post where the following principle was presented:

Fundamental Principle of Solving ‘Easy’ Equations

Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with thesamesolution.

There are a number of subtleties here: basically sometimes you get extra ‘solutions’ (that are not solutions at all), and sometimes you can lose solutions.

Let us write the squaring function, e.g. , by and the square-rooting function by . It appears that are an inverse pair but not quite exactly. While

and

,

check out O.K. note that

,

does not bring us back to where we started.

This problem can be fixed by restricting the allowable inputs to to positive numbers only but for the moment it is better to just treat this as a subtlety, namely while , … in fact I recommend that we remember that with an there will generally be *two *solutions.

The other thing we look out for as much as possible is that *we cannot divide by zero*.

There are other issues around such as the fact that , so that the equation has no solutions (no, is not a solution! Check.). This equation has *no *solutions.

Often, in context, these subtleties are not problematic. For example, equations with no solutions rarely arise and quantities might be positive so that if we have , only need be considered (for example, might be a length).

Highlighting these small subtleties is not therefore to scare you off but to allow you full mastery. Such as in the next example.

### Example

*Solve*

*Solution: *Take out the common factor of :

.

Assume that is non-zero (you can’t divide by zero).

Let us see what it means if this is zero:

or

Now perhaps from the context of the equation, both and are strictly positive (bigger than zero), so the second possibility can’t occur. The first possibility can happen: if we have that the original equation reads:

Therefore you can assume that because if they are equal and the equation can’t be solved for .

So assuming you do have:

Now remember because of the square there are *two* solutions (i.e. — you need plus *and * minus). Therefore the correct implication is that:

.

However, if you assume that , then it cannot be the negative square root.

In fact, under these assumptions:

- ,

we can tidy this up a little further. Multiply the number inside the square root by :

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