Solving Radical Equations

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Solving Radical Equations

In this section we want to learn how to solve equations containing radicals, like

. In
order to do this we need the following property.

Basically, this property tells us we can raise both sides of any equation to any power we would
like. However, we must be careful. There are several places where we can make serious
mistakes.

First, we need to make sure that we really are raising the entire side to the nth power and not just
each term individually. Also, when using the nth power property there is the possibility that we
end up with solutions that don’t check, called extraneous solutions. The reason we get these on
occasion has to do with the logical construction of the property. The property tells us that if
something is a solution to

then it must also be a solution to

. That doesn’t mean
that if something is a solution to

that it is a solution to

. To rectify this, we simply
check our answers and throw away the ones that do not work.

So, always remember, be careful to raise the entire side to the n-th power, and we must always
check our answers.

Example 1:

Solution:

a. We can use the n-th power property to get rid of the radical. Since we have a square
root, we should use the second power, i.e. we should square both sides of the equation.
Then we can solve like usual. We get

Now we must check our answer. We do this in the original equation. If the value does
not check, then we eliminate it and would have no solution. We get

Since the value checks, our solution is set {17}.

b. Again, we need to raise both sides of the equation to a suitable power to get rid of the
radical. This time we will use the 3^rd power since we have a cube root. We get

Since the value checks the solution set is

.

c. Finally, we need to raise both sides to the 4^th power since we have a 4^th root. We get

Since it does not check, 2 can not be a solution to the equation. Therefore, we have the
equation has no solution.

Now, sometimes the equation is a bit more complicated. What we need to know is that before
raising both sides to the n-th power, we must always isolate the radical expression first. That way
we are assured that it will cancel when we use the exponent. Also, sometimes, after raising both
sides to the n-th power, we end up with a radical still in the equation. In this situation, we need to
again isolate the radical and again use the exponent. We keep repeating this process until all the
radicals are gone. Then we solve as usual.

Example 2:

Solve.

Solution:

a. First thing we need to do is isolate the radical. Once we have done that we can square
both sides of the equation as we did in example1. Then, of course, we finish solving and
check.

The value checks. Therefore the solution set is {6}.

b. Notice this time that one of the radicals is already isolated. Therefore we can go ahead
and square both sides. Must, however be very careful with squaring the right side since it
has two terms. It generally makes it easier to square properly by writing the entire side
out twice and then multiplying as we learned in section 8.4. We get

Notice that we are left with another equation that has a radical. Therefore, we need to
isolate again and square again. In this case however, it is easier to simply isolate the
term containing the radical and the carefully square both sides. We can then continue
solving as usual.

Since the solution checks, the solution set is

.

c. Finally, we must start by isolating one of the two radicals in this equation. We will choose
to isolate the positve one. Generally we isolate the more complicated radical in order to
eliminate it first. However, either could be isolated and still produce the correct solutions.
Once we have isolated the radical then we can square both sides very carefully and then
continue as usual. We get

Recall, to solve an equation that has a squared term we must solve by factoring. That is,
we get all terms on one side, factor and set each factor to zero. This gives

So, 22 does not check but 2 does. Therefore, the 22 is an extraneous solution. So our
solution set is {2 }.

Finally we want to see some applications of radicals. The first of these requires the Pythagorean
theorem. We state it below.