Download Radical Equations

Equations with Variables in Radicals
Introduction
Remember that when working math exercises your goal is not to “get the answer”.
a) Your primary goal is to determine if you understand the concepts, which you have
been studying, well enough to use them to solve problems.
b) A secondary goal is to further learn and understand the theory and process used to
answer the question.
c) A third goal is to communicate to yourself and your instructor that you understand the
theory and processes.
d) Proper communication is a by-product of the first three goals.
The following two excepts from “Make It Stick” are relevant to the current learning task.
Putting new knowledge into a larger context helps learning.1
Prior knowledge is a prerequisite for making sense of new learning, and forming those
connections is an important task of consolidation.2
That prerequisite knowledge will be reviewed in summary form before beginning the discussion
of equations containing variables inside radicals. This review will make learning the new
material easier and more useful.
Definition: An equation is a mathematical statement which contains an = symbol.
Definition: A number (or numbers) that makes an equation true when substituted for the
variable (or variables) is called a solution of the equation.
Definition: The collection of all solutions of an equation is called the solution set of the
equation.
Definition: The graph of an equation consists of all the points, and only those points,
whose coordinates are solutions of the equation.
Definition: The process of finding all the solutions (the solution set) of an equation is
called solving the equation.
Definition: Two equations are equivalent if they have the same solution sets.
Definition: A simplest equation is an equation which has a single variable on one side
of the equal sign and a single number on the other side.
Properties of Equations:
(1) : If any expression is added to both sides of an equation the resulting equation
is equivalent to the original equation.
(2) : If both sides of an equation are multiplied by the same non-zero real number,
the resulting equation is equivalent to the original equation.
1
2
Brown, Peter C. (2014-04-14). Make It Stick (p. 6). Harvard University Press. Kindle Edition.
Brown, Peter C. (2014-04-14). Make It Stick (p. 73). Harvard University Press. Kindle Edition.
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Definition: A linear equation in one variable is an equation that can be written in the
form ax + b = 0 where a and b are real numbers with not both a and b equal to zero.
Process to solve a linear equation: The process to solve a linear equation in one
variable is to use nothing but the first two properties of equations to generate a sequence
of equations each equivalent to the previous equation until a simplest equation is
obtained.
Equations to be studied in this essay are called radical equations. An equation is a radical
equation if the equation contains a radical which contains a variable.
these equations cannot be solve as simply as linear equations.
A common misconception is that performing the same operation on both sides of an equation
will produce correct results. This idea is absolutely false. Multiplying both sides of an equation
by an expression will frequently produce so-called extraneous solutions as will squaring both
sides of an equation.
We would like to use the simple process of solving linear equations to solve all equations, but the
process breaks down for all equations other than linear. The remedy is to insert a “remedy” into
the process and then continue.
The remedy for rational and radical equations is identical and will be imitated for some future
equations so we examine the process for both types of equations. Read, study, compare, and
contrast the statements in the table below.
Rational Equations
Radical Equations
IMPORTANT FACT A: When both sides of
an equation are multiplied by an expression
containing a variable there is no assurance that
the resulting equation will be equivalent to the
original. Consequently it must be assumed that
the resulting equation is not equivalent to the
original equation.
IMPORTANT FACT B: When both sides of
an equation are multiplied by an expression
containing a variable the solution set of the
resulting equation contains the solution set of
the original equation.
Remedy: Test each solution in the solution set
of the second equation to determine which
ones are solutions to the first equation.
IMPORTANT FACT A: When both sides of
an equation are squared there is no assurance
that the resulting equation will be equivalent to
the original equation. Consequently it must be
assumed that the resulting equation is not
equivalent to the original equation.
IMPORTANT FACT B: When both sides of
an equation are squared the solution set of the
resulting equation contains the solution set of
the original equation.
Remedy: Test each solution in the solution set
of the second equation to determine which
ones are solutions to the first equation.
As soon as we realize that what we are doing will produce a list of possible solutions which
contains all the solutions, then we know we can determine the desired solution set simply by
testing each of the possible solutions to see if they are solutions. Notice how important it is to
know that no solutions are lost in the process. That would be an difficulty which could not be
overcome simply by testing.
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So here is the process as applied to radical equations
When both sides of an equation are squared the solution set of the resulting equation contains the
solution set of the original equation.
1. This means that if both sides of an equation are squared and the resulting equation is
solved, those solutions are the only candidates as solutions to the original equation.
2. It also means that if both sides of an equation are squared and the resulting equation is
solved, some of those solutions may not be solutions to the original equation.
Therefore when squaring both sides of an equation is part of the solution process, testing all the
possible solutions in the original equation must also be a part of the solution process.
In the following examples I have sometimes presented two styles distinguished by their titles.
a) Solution with Reasoning
b) Solution
In those cases where Solution with Reasoning is presented, I write most of the important thought
process that must be part of the solution process. When you are learning a new process, it is a
good idea to write out all of the reasoning as I have done when presenting Solution with
Reasoning. By writing these particular details, you force yourself to concentrate on the critical
mathematics facts which you can carry forward to other situations. After you become more
practiced at a particular process, you may revert to writing a more abbreviated form as presented
in the more sparse Solution presentations used in the following examples. However, even when
writing the simpler form, your mind should concentrate on the underlying mathematics details
and that should guide you through the process. The process should always be based on
mathematics not on a memorization of steps as presented by Teacher Fritz.
The red flag is a mnemonic device to remind everyone the equation might not be equivalent to
the preceding equation.
Exercises with Solutions
Problem 1: Solve the equation 2x  9  x  3
Solution with Reasoning: Square both sides of the original equation to obtain the equation
2x + 9 = x2 + 6x + 9
Note: This equation might not be equivalent to the original equation.
However, the solution set of this equation contains the solution set of the first equation.
Add –2x – 9 to both sides of the second equation to obtain x2 + 4x = 0.
Factorization yields x (x + 4) = 0 and then The Zero Factor Property yields the two equations
x = 0 OR x + 4 = 0. The solution sets to these two equations are {0} OR {–4}.
The solution set of the second equation 2x + 9 = x2 + 6x + 9 is therefore {0} {4}  {0, 4} .
Every solution of the original equation is contained in the set {0, –4}, but some of the elements
of the set {0, –4} might not be solutions of the original equation. Therefore we must test each of
these possible solutions.
Test 0: 2(0)  9  0  3 is TRUE.
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Therefore 0 is a solution of the original equation.
Test –4: 2(4)  9  4  3
1  1 is FALSE
Therefore –4 is not a solution of the original equation.
Therefore the solution set for the original equation is the set {0}.
Observe that {0}  {0, 4} .
An Alternate Way to Present the Solution Process
square both sides
2x  9  x  3
2
2x + 9 = x + 6x + 9
add –2x – 9 to both sides
2
x + 4x = 0
factor
x (x + 4) = 0
By The Zero Factor Property
x = 0 OR x + 4 = 0
x = 0 OR x = –4
The solution set for the equation 2x + 9 = x2 + 6x + 9 is {0, –4}
The solution set for the original equation is a subset of {0, –4}
Test 0:
Test –4:
2(0)  9  0  3 is TRUE.
2( 4)  9  4  3
Therefore 0 is a solution of the original
 1  1 is FALSE
equation.
Therefore –4 is not a solution of the original equation.
It follows that the solution set for the original equation is the set {0}.
Problem 2: Solve the equation 10x  5  1  2x
Solution:
10x  5  1  2x
add 1 to both sides
10x  5  2x  1
square both sides
10x + 5 = 4x2 + 4x + 1
4x2 – 6x – 4 = 0
2x2 – 3x – 2 = 0
(2x + 1)(x – 2) = 0
By The Zero Factor Property
2x + 1 = 0 OR x – 2 = 0
x   1 OR x  2
2
add –10x – 5 to both sides
multiply both sides by 1
2
factor
 
The solution set for 10x + 5 = 4x2 + 4x + 1 is  1 , 2
2
 
The solution set for the original equation is a subset of  1 , 2
2
 
 
Test 2: 10  2   5  1  2  2  is TRUE
Therefore 2 is a solution of the original
equation.
Test - 1 : 10 1  5  1  2  1 is TRUE
2
2
2
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Therefore  1 is a solution of the original equation.
2
The solution set for the original equation is  1 , 2
2
 
Problem 3: Solve the equation 5x  1  3x  1
Solution:
5x  1  3x  1
add 3x to both sides
5x  1  3x  1
square both sides
5x  1  3x  2 3x  1
add – 3x – 1 to both sides
2x  2 3x
multiply both sides by 1
2
x  3x
square both sides
2
x = 3x
add – 3x to both sides
x2 – 3x = 0
factor
x(x – 3) = 0
By The Zero Factor Property
x = 0 OR x – 3 = 0
x = 0 OR x = 3
The solution set for x2 = 3x is {0, 3}.
The solution set for the original equation is a subset of {0, 3}.
Test 0: 5(0)  1  3(0)  1 is TRUE
Test 3: 5(3)  1  3(3)  1 is TRUE
Therefore 0 is a solution of the original equation.
Therefore 3 is a solution of the original
equation.
The solution set for the original equation is {0, 3}.
Problem 4: Solve the equation 3x  1  1  x  4
Solution:
3x  1  1  x  4
square both sides
3x + 1 = 1 + 2 x  4 + x + 4 add –x – 5 to both sides
2x – 4 = 2 x  4
multiply both sides by 1
2
x – 2 = x4
square both sides
2
x – 4x + 4 = x + 4
add – x – 4 to both sides
x2 – 5x = 0
factor
x(x – 5) = 0
By The Zero Factor Property
x = 0 OR x – 5 = 0
x = 0 OR x = 5
The solution set for x2 = 3x is {0, 5}.
The solution set for the original equation is a subset of {0, 5}.
Test 0: 3(0)  1  1  0  4 is FALSE
Test 5: 3(5)  1  1  5  4 is TRUE
Therefore 0 is not a solution of the original equation. Therefore 5 is a solution of the original
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equation.
The solution set for the original equation is {5}.
Problem 5: Solve the equation 2x  5  x  1  x  2
Solution:
2x  5  x  1  x  2
square both sides
(2x  5)  2 2x  5 x  1  (x  1)  x  2 add – 2x – 5 to both sides
2 2x  5 x  1  (x  1)   x  3
2 2x  5 x  1  2x  2
add – x + 1 to both sides
multiply both sides by  1
2
square both sides
multiply
add – (x2 + 2x + 1) to both sides
factor
2x  5 x  1  x  1
(2x + 5)(x – 1) = x2 + 2x + 1
2x2 + 3x – 5 = x2 + 2x + 1
x2 + x – 6 = 0
(x + 3)(x – 2) = 0
By The Zero Factor Property
x + 3 = 0 OR x – 2 = 0
x = –3 OR x = 2
The solution set for (2x + 5)(x – 1) = x2 + 2x + 1 is {–3, 2}.
The solution set for the original equation is a subset of {–3, 2}.
2( 3)  5  3  1  3  2
2(2)  5  2  1  2  2
Test –3:
Test 2:
1  4  1
9 1 4
The last line shows that –3 is not in the domain of the is TRUE
original equation. (Because these square roots are not Therefore 2 is a solution of the original
real numbers.)
equation.
Therefore –3 is not a solution of the original equation.
The solution set for the original equation is {2}.
Problem 6: Solve the equation x  5x  6
Solution with Reasoning: Square both sides of the original equation to obtain the equation
x2 = –5x – 6.
Note: This equation might not be equivalent to the original equation.
However, the solution set of this equation contains the solution set of the first equation.
Add 5x + 6 to both sides of the second equation to obtain x2 + 5x + 6 = 0.
Factorization yields (x + 2)(x + 3) = 0 and then The Zero Factor Property yields the two
equations x + 2 = 0 OR x + 3 = 0.
The solution sets to these two equations are {-2} OR {-3}.
The solution set of the second equation x2 = –5x – 6 is therefore {2} {3}  {2, 3} .
Every solution of the original equation is contained in the set {–2, –3}, but some of the elements
of the set {–2, –3} might not be solutions of the original equation.
Therefore we must test each of these possible solutions.
Test –2: 2  5(2)  6  4  2 is FALSE
Therefore –2 is not a solution of the original equation.
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Test –3: 3  5(3)  6  9  3 is FALSE
Therefore –3 is not a solution of the original equation.
Neither of the two possible solutions are solutions of the original equation.
Therefore the solution set for the original equation is the empty set 
Observe that   {2, 3}
An Alternate Way to Present the Solution Process
square both sides
x  5x  6
2
x = –5x – 6
add 5x + 6 to both sides
x2 + 5x + 6 = 0 = 0
factor
(x + 3)(x + 2) = 0
By The Zero Factor Property
x + 3 = 0 OR x + 2 = 0
x = – 3 OR x = – 2
The solution set for the equation x2 = –5x – 6 is {– 3, – 2}
The solution set for the original equation is a subset of {– 3, – 2}
Test – 2:
Test – 3:
2  5(2)  6  4  2 is False
3  5(3)  6  9  3 is FALSE
Therefore – 2 is not a solution of the
Therefore – 3 is not a solution of the original equation.
original equation.
It follows that the solution set for the original equation is the empty set  .
Problem 7: Solve the equation x  5x  6
Solution with Reasoning: Square both sides of the original equation to obtain the equation
x2 = 5x – 6.
Note: This equation might not be equivalent to the original equation.
However, the solution set of this equation contains the solution set of the first equation.
Add –5x + 6 to both sides of the second equation to obtain x2 – 5x + 6 = 0.
Factorization yields (x – 2)(x – 3) = 0 and then The Zero Factor Property yields the two
equations x – 2 = 0 OR x – 3 = 0.
The solution sets to these two equations are {2} OR {3} respectively.
The solution set of the second equation x2 – 5x + 6 = 0 is therefore {2} {3}  {2,3} .
Every solution of the original equation is contained in the set {2, 3}, but some of the elements of
the set {2, 3} might not be solutions of the original equation. Therefore we must test each of
these possible solutions.
Test 2: 2  5(2)  6 is TRUE
Therefore 2 is a solution of the original equation.
Test 3: 3  5(3)  6 is TRUE
Therefore 3 is a solution of the original equation.
We conclude the solution set for the original equation is the set {2, 3}.
Observe that {2,3}  {2,3} .
An Alternate Way to Present the Solution Process
x  5x  6
square both sides
x2 = 5x – 6
add –5x + 6 to both sides
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x2 – 5x + 6 = 0 = 0
factor
(x – 3)(x – 2) = 0
By The Zero Factor Property
x – 3 = 0 OR x – 2 = 0
x = 3 OR x = 2
The solution set for the equation x2 = 5x – 6 is {3, 2}
The solution set for the original equation is a subset of {3, 2}
Test 2:
Test 3:
2  5(2)  6 is TRUE
3  5(3)  6 is TRUE
Therefore 2 is a solution of the original
Therefore 3 is a solution of the original equation.
equation.
It follows that the solution set for the original equation is the set {2, 3}.
Problem 8: Solve the equation
x5  x8  3
Solution:
x5  x8  3
(x  5)  2 x  5 x  8  (x  8)  9
2x  13  2 x  5 x  8  9
2 x  5 x  8  2x  22
x  5 x  8  x  11
(x  5)(x  8)  (x  11)2
x 2  13x  40  x 2  22x  121
9x  81
x  9 There is no gaurantee that 9 is a solution to the original
equation. However, it is assured that there is no other possible
solution. That's why we perform the following test.
Test 9:
3  9  5  9  8  4  1  2  1  1 which is false.
Therefore 9 is not a solution of the original equation and since 9 is
the only possibility, we conclude:
The solution set for the original equation is the empty set .
4x  x2
Problem 9: Solve
3 is a solution 0 is not
Problem 10: How can you tell without any computations that the
solution set for 2x  3  4 has no real solutions.
Problem 11. What is wrong with the following “solution” of
2x  5  4  x  8 ?
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2x  5  4  x  8
(2x  5)  (4  x)  64
x  9  64
x  55
This particular environment – solving equations involving radicals – demonstrates some very
important basic ideas. These ideas have far reaching consequences.
a) Solving an equation means to find all numbers which make the equation true.
b) The process used must find solutions.
c) The process used must insure that all solutions have been discovered.
d) The process must provide a means of recovery when it has a flaw.
e) The process might yield a list of possible solutions which must be tested.
f) Every step in a process must be justified by some mathematics principle.
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