This section shows techniques to solve an equality that has a radical that can’t be simplified into a non radical form. This has potential drawbacks which is also covered in this section.

#### Dealing with radicals that can’t be simplified away.

So far, we have been working with equations where the radicals were able to be simplified away entirely, such as the previous example where we turned into thereby removing the radical entirely. Unfortunately that is rarely the case, in reality it is far more common that we cannot simplify our problem to the point of removing the radical entirely, either because we can’t factor the radicand in a type two radical, or the powers just don’t happen to work out the way we need them to in order to remove the radical entirely. In this case, we need to use the inverse function property to remove the radical. Consider the following example;

The previous example may seem a bit straight forward, but let’s look at another one to see what kind of dangers tend to lurk in this process.

However, when we plug in those values to the original equation we get (predictably) , not . Thus and are extraneous solutions.

So, in the previous example we discovered that, in fact, *neither* of our proposed solutions actually work. So what gives?
The fact that we got positive two and not negative two when we plugged the proposed solutions back in should
have been predictable, since we were essentially solving the problem from the prior example all over again. But
the starting equations weren’t the same, so why did the solutions merge and give us the same answers in both
examples? Again it comes back to the negative sign; notice that the real difference is that in one example the
square root was equal to positive two, but in the other case it was equal to negative two. However, since the first
step of our method was to square both sides, we essentially annihilated the negative sign and the two equations
became the same.

By squaring both sides of the equality, we are obliterating any potential negative multipliers, which could accidentally change
the solution set. The good news is that the process will still find any real solutions that exist, but the bad news is that it may find
a number of extra solutions as well. These ‘extra’ solutions that don’t *actually* satisfy the original equality are called *extraneous
solutions*.

An easy conceptual way to see why using a power on both sides of the equality can increase your solution set is to consider a polynomial. If you had some polynomial, say , then the number of solutions to a polynomial is equal to the degree of the polynomial (by the fundamental theorem of algebra, if we don’t worry about requiring real versus complex solutions). However, if we square both sides the polynomial then the left goes from degree 2 to degree 4, meaning that (again by the fundamental theorem of algebra) now there are four solutions instead of just two.

We conclude this section with another example of an especially difficult problem of this type. Sometimes the setup is such that you have to use powers more than once. Typically the goal is to isolate the root on one side of the equation before using a power, but if there is more than one root that might not be possible. Consider the following:

Now we can square both sides again in order to get rid of the square root, giving:

Then factoring the quadratic we get the factored form , which gives solutions of and . Plugging these back in we have:

For ;

= | ||

= |

For , on the left hand side we have: for the right hand side we have: So, since the left hand side is and the right hand side is
we have that does *not* satisfy the equality, which means it is an *extraneous* solution.

You can watch a video with more examples of solving these kinds of radical functions.