Download Solving Radical Equations: Techniques and Examples and more Study notes Advanced Calculus in PDF only on Docsity!
2. 6 Solving Radical Equations
Now we will turn our attention back to solving equations. Since we
know how to operate with radicals now, we should be able to solve basic
equations involving them. We already know that a square will โundoโ a
square root and a cube will โundoโ a cube root, etc. We also know that
we can operate on any equation as long as we do the same thing to both
sides. We can use this knowledge to isolate our variable when it is inside
of a root such as this. Please note that this is only allowed with
equations, not with expressions. You canโt just randomly square or cube
an expression because this will change it. You must have two sides of
an equation in order to do anything to it (other than rewrite it in a
different form).
Examples
Solve each of the following equations for the indicated variable.
You probably already can see the answer to this one since you
know โ
81 = 9 , but most radical equations do not have obvious
solutions, so we need a technique to solve them. We will
demonstrate our technique here, where we already know what the
answer should be. In order to isolate the variable, we need to
โundoโ what has been done to it. We can square both sides to
โundoโ the square root. The basic technique involves isolating the
radical and then squaring both sides.
The radical is already isolated.
2
2
With radical equations, we must also check our answers by
plugging back into the original radical equation:
Is โ
81 = 9? Yes, it is!
2
2
Now, we must check our answer by plugging it into the original
equation: Is โ
19 โ 3 = 4? Yes.
First, we must isolate the square root. (If we try to square both
sides right now, we will not be able to get rid of the radical since
2
= (โ 5 ๐ฅ โ 9 โ 6 )(โ 5 ๐ฅ โ 9 โ 6 ) and you would
need to FOIL this out to obtain a result that still has a radical in it!
Remember that exponents do not distribute over addition or
subtraction, as tempting as it may beโฆ.)
Now square both sides and
evaluate.
Check each answer:
1 = โ 1 Not true
Therefore, the only true solution to the original equation is ๐ง = 2.
The other solution is extraneous.
2
2
2
2
Now factor and set
each factor equal to 0
to obtain two possible
solutions.
This is a quadratic, so we
need to get everything on
one side to set it equal to 0.
Now factor and set
each factor equal to 0
to obtain two
possible solutions.
Isolate the radical first by
subtracting 5 from both
sides.
Now square both sides to
eliminate the radical. You
must FOIL here.
Check each answer:
4 + 5 = 3 Not true
Therefore, the only true solution to the original equation is ๐ฅ = 8.
The other solution is extraneous.
2
2
2
2
This is a quadratic, so we
need to get everything on
one side to set it equal to 0.
Now factor and set
each factor equal to 0
to obtain two
possible solutions.
Now square both sides to
eliminate the radical. You
must FOIL here.
Isolate the radical first by
adding 2 to both sides.
which radical you choose to isolate. You will get the same answer
in the end.
2
2
โ 20
โ 10
โ 10 โ
๐ฅโ 1
โ 10
2
2
Check: ๐ฅ = 5 โ
Clean up the right side a
bit and then isolate the
remaining radical.
Now square both sides
to eliminate one of the
radicals. You must FOIL
here.
Isolate either radical first by
taking the other one to the other
side.
You can square both sides
without isolating the radical
completely if you like (since
exponents do distribute over
multiplication), but we chose
to go ahead and divide by -
10 to isolate the radical
completely.
Now square both sides again to
eliminate the remaining radical.
2
2
โ 32
โ 8
โ 8 โ
๐ก+ 15
โ 8
2
2
Check: ๐ก = 1
3
1
3
= ๐ฅ โ 1
3
1
3
)
3
3
3
3
3
2
3
3
2
3
3
2
2
Now check both answers:
3
1
3
= 2 โ 1
3
1
3
= (โ 1 ) โ 1
1
3
= 1
1
3
= โ 2
1
3
= 1
1
3
= โ 2
Both answers are solutions!
Cube both sides since 3 is
the reciprocal of
1
3
. Thus,
1
3
โ 3 = 1 and that will give
us (๐ฅ
3
โ 7 )
1
getting rid of
the rational exponent.
Donโt forget what (๐ฅ โ
1
)
3
means. You canโt
distribute the exponent
over subtraction. You
must FOIL this out,
multiplying two of the
factors together and then
your result by the third.
When factoring, donโt forget
to pull out the GCF first, and
if the leading term is
negative, pull out the
negative to make factoring
easier. Do one step at a time.
Reminder: There is no need to
set the constant factor equal to
- Only the factors with
variables should be set equal to
2
3 = 2 (๐ฅ
2
3 )
3
2
3
2
3
2
3
2 (in radical form)
Checking is obvious here if you plug it back in while it is in
the rational exponent form 2
3
2 .
โ
5
2
= 35
First, we must isolate ๐ฅ
โ
5
2
in order to โundoโ the power:
โ
5
2
= 35
โ
5
2 = 32
Now, put both sides to the reciprocal power:
โ
5
2
)
โ
2
5
โ
2
5
1
( 32
)
2
5
1
( โ
32
5
)
2
1
2
2
1
4
Here, we put both sides to the power
3
2
since it is the reciprocal of
2
3
and
thus
2
3
โ
3
2
= 1 , so we get ๐ฅ
1
( ๐๐ ๐ฅ
) .
Donโt forget that you need a ยฑ
symbol since you are square rooting
both sides with the 2 in the
denominator of the power.
Pythagorean Theorem
Given any right triangle (a triangle that contains a 90ยฐ angle), the
lengths of the three sides of the triangle are related by the following
equation: ๐
2
2
2
This theorem is fundamental in trigonometry and very useful for
some engineering applications. We are studying this theorem here
because this equation involves using radicals to solve for a given
side. You will encounter use in your future math (and probably
science) courses.
Examples
Find the length of the missing side for each of the following
triangle:
c
a
b
The side
opposite the
right angle is
called the
hypotenuse
๐ฆ
2 meters
4 meters
In this example, we are looking for the hypotenuse, y, which is
alone on one side of the equation. It is important to note whether
you are solving for the hypotenuse or for one of the legs in order to
know where your variable goes in the equation.
2
2
2
2
2
(or ๐ฆ = 4. 47 ๐๐๐ก๐๐๐ )
Notice that we are solving for one of the legs in this example, not
the hypotenuse.
2
2
2
2
2
or ๐ฅ = 3. 87 ๐๐
๐ฅ
1 cm
4 cm
Even though we
technically get two
roots to our
mathematical
equation, we choose
the positive one since
our answer represents
length, which is
always positive.
