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In this section we want to learn how to solve equations containing radicals, like 945. = - x . In order to do this we need the following property.
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In this section we want to learn how to solve equations containing radicals, like 5 x 4 9. In
order to do this we need the following property.
n-th Power Property
If a b , then
n n a b.
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 a b then it must also be a solution to
n n a b. That doesn’t mean
that if something is a solution to
n n a b that it is a solution to a b. 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:
Solve.
a. 5 x 4 9 b. 3 2 2 3 x c. 3 1 4 x
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
(^22)
x
x
x
x
x
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
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
2
11
3 3 3
3
x
x
x
x
x
Now we check the solution.
3
3
3 2
11
11 .
c. Finally, we need to raise both sides to the 4
th power since we have a 4
th root. We get
4 4 4
4
x
x
x
x
x
Check:
4
4
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.
a. 2 x 3 2 1 b. x 1 2 x c. 2 x 5 3 x 2 1
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.
16
9 .
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
2
2
2 2
2
2 2
x x x
x x x x
x x x
x x
x x
x x x
x x x
x x x x
x x x
x x
x x
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
2
2
x x
x and x
x x
x x
x x x
Now we must check both answers and keep only those which work Check:
So, 22 does not check but 2 does. Therefore, the 22 is an extraneous solution. So our
Finally we want to see some applications of radicals. The first of these requires the Pythagorean
theorem. We state it below.
The Pythagorean Theorem
In any right triangle, if a and b are the lengths of the legs of the triangle and c is the length of
the hypotenuse, then we have the following
2 2 2 a b c or alternately
2 2 c a b
2 2 a c b
2 2 b c a
Other types of applications require us to interpret and use a given formula.
Example 3:
Find the missing side of the right triangle if c 8 ft , a 20 ft.
Solution:
First we start by drawing the picture
So we can see that we are clearly missing side b. Therefore we can use the formula
2 2 b c a. We get
2 2
b
So b 2 11 ft 6. 6 ft.
Example 4:
The equation for the time of one pendulum swing (called the period of the pendulum) is given by
pendulum of a clock that has a period of 2.3 seconds.
Solution:
a
b
c
8 ft
20 ft
Solve for the indicated variable.
g
d t
for g 56. g
d t
for d
Find the missing side of the right triangle.
when the ladder reaches a height of 8 feet?
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the TV?
the diagonal of the monitor?
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32
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h v gr
2 where v is the velocity,
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r h
h v gr
2 where v is the velocity,
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2 and the escape velocity is 1.25 km/sec? (The radius of the earth is about 6440 km)