This section views the square root function as an inverse function of a monomial. This is used to explain the dreaded symbol and when to use (and not use) absolute values.
So, we’ve decided to sacrifice half our solutions in order to keep our tool a function. That isn’t great - and indeed, we still need a way to ensure we account for all solutions when we involve an even root. Luckily we have come up with a solution to this conundrum, although our solution introduces one of the most common sources of confusion for precalc students, the dreaded symbol.
To understand where the symbol comes from, we first want to recognize when our sign problem is actually a problem. In our example , the sign problem is obviously an issue. In contrast however, the example only has one (real) solution, . Thus in the cube root example there is no sign issue. So what’s the difference?
The key observation is that the even power obliterates any negative sign on the solution; hence both and . In contrast the odd power preserves negative signs; hence but , not . Thus this sign problem only occurs when we are dealing with even root-values.
It turns out that we often skip some technical steps when we use radicals to cancel out powers, and those
skipped steps are where the confusion comes in. Recall that, when simplifying something like we didn’t just get ,
we got . Keeping this in mind, let’s very carefully write out all the steps of solving the equality , we can more
.
| = | ||
| = | ||
| = | ||
So, really, our solution isn’t , rather we need to solve the equality . But this representation makes it more clear that we should have as our solutions. Indeed, this is really where the is coming from, from solving/simplifying a situation of the kind something (where the can be replaced with any even number), since this generates an absolute value. Notice that, for the “something” being squared inside the radical, this applies to anything that has a possible square root - meaning that even something like would be possible, since we can rewrite this as . Essentially, the coefficients won’t matter since we can always take an appropriate root to get an appropriate coefficient for inside the power, it’s really the power on the variable that we care about.
The short version of the symbol is the last paragraph of the above. Essentially the symbol comes up when you simplify an even root-value radical of something which gives you an absolute value. In practice it is rare for someone to give you a problem that isn’t simplified (at least in terms of even power vs even root). It is far more common that you have something being raised to an even power and you the solver, introduce the even root-value radical as part of your solution method (e.g. to get rid of the even power and isolate the variable). Thus there is a handy rule of thumb as follows: If you are the one to introduce the square root (or other even root-value), always put in the . If you are given a square root (or other even root-value), then it only outputs a positive.
The narrow bit of gray area is when you are given an even root of an even power (ie an unsimplified problem) because, even though the output is positive by the definition of the root function, that doesn’t mean the absolute value that you get when you simplify the even root with an even power won’t give you a style answer. So our rule of thumb listed above is merely that - a rule of thumb, not an absolute rule. In general you should always be careful when simplifying a given radical to see if both positive and negative answers would work.
Unsurprisingly, there is more we must consider. However, when it comes to simplifying radicals that is pretty much it. So let’s consider the following example in an effort to see a (full) proper solution to an equation involving a radical.
Next we check for which means we want to solve the equation which, again moving the terms to one side gives us: which has no real solutions.
Thus we have as our proposed solutions, and . If we plug in each into the original equation, we verify that each of them do, in fact, satisfy the equation and we have our solutions.
