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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.
The video!
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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 which can be
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 into
something we can simplify, 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;
Determine which values of satisfy In this case we cannot factor the radicand on the left, and it certainly isn’t a perfect
square. Thus, the only way to remove the square root function is to use its inverse function; the square. Specifically
we need to square both sides of the equality, which gives the ‘equivalent’ function to solve; . Subtracting from
both sides to get our quadratic equal to zero, and then factoring yields , and so we have as potential solutions
and . Checking both these values by plugging into the original function we get that both the values satisfy the
equation.
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.
Determine which values of satisfy We can manipulate this radical by squaring both sides and factoring, which gets: . This
factors as the last example did into yielding the same solution set of and .
However, when we plug in those values to the original equation we get (predictably) , not . Thus and are not valid
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. (In fact, an especially clever student might have skipped the entire solution
process and determined there were no real solutions to the second example, purely because the square root was
equal to a negative number, and remember square roots can only yield non-negative numbers (ie zero or positive
numbers).)
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.
Extraneous Solution An extraneous solution is a ‘solution’ that is acquired as a result of some solution method, but the
solution doesn’t actually satisfy the original problem. See Example Ex1 in this topic for an example of 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 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:
Find all that satisfy the equation Typically we would try to isolate each root. If we didn’t have the “” part above, and only
had then squaring both sides would lift both square roots simultaneously. Unfortunately, that isn’t the case, so first we must
square both sides, then isolate any remaining roots before doing it again.
So squaring both sides gives:
Now we can square both sides again in order to get rid of the square root on the right, giving:
Then factoring the quadratic we get the factored form , which gives solutions of and . Plugging these back in we
have:
For :
=
=
So works.
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.