Solving Equations
Prepared by: Sa’diyya Hendrickson
Name:
Date:
Package Summary
• Square Root Equations
• Absolute Value Equations
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c Sa’diyya Hendrickson
Square Root Equations
Level: X
A square root equation is an equation with polynomial terms along with at least one
square root expression (with a polynomial under the radical).
p
Consider the following example: 2
x + 2 = 2x
Recall that the ultimate goal when solving an equation is to isolate the variable.
Answer: One of our x’s is being held “hostage” by the square root!
So, even before we begin, we know that at some point we’ll need to do
the opposite of taking a square root (i.e. squaring), in order to free x!
1 Isolate the radical before squaring both sides.
Why? Consider the equation a ±
p
b = c.
(F)
e.g. p
If a = 2, b = x + 2 and c = 2x, we can create the equation
2
x + 2 = 2x, which was given as the previous example.
Now, if we square both sides of the equation given in (F), we have:
a±
p
b=c
p 2
) (a ± b) =(c)2
) (a ±
p
b)(a ±
p
b) = c2
p
p
p
) a2 ± a b ± a b + (± b)2 = c2
p
) a2 ± 2a b + b2 = c2
DP
Unfortunately, the distributive property guarantees that the radical
will reappear in the expansion, particularly in the middle term.
On the other hand, if we isolate the radical first, we have:
p
± b=c a
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Square Root Equations
Level: X
Squaring both sides of this equation gives:
p
± b=c a
p
) (± b)2 =(c a)2
P.O.E
)
b = (c
a)2
Now it’s clear that we’ve achieved our goal: undoing the square root!
Moral of the Story Isolate the radical first!
2 If we square both sides of an equation, we are
creating a new equation that may introduce additional,
invalid (extraneous) solutions. Therefore, we must
always check our answers at the end to make sure that
L.S. = R.S. for the original equation.
Why does this happen? Suppose we are given some equation:
a=b
(E1 )
If we square both sides of E1 , we create a new equation:
a2 = b 2
(E2 )
Answer: From E2 we have:
()
P.O.E.
()
a2 = b 2
p
p
a2 = b 2
|a| = |b|
In words, E2 is satisfied whenever a and b are the same distance
from zero on the real line. This results in exactly two outcomes:
(1) a and b are the same.
(2) a and b are opposites.
i.e. a = b
i.e. a = b
Notice that these outcomes result in our need to solve two equations,
one of which is E1 and another that is not. So, solving E2 will
always produce the solutions to E1 while potentially introducing
additional solutions that are invalid.
Moral of the Story Although squaring both sides helps
us to free a variable from a square root, we must “pay the
fee” of checking our solutions at the end to ensure that
we’re only taking valid solutions for the original equation!
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Square Root Equations
Level: X
The key observations previously discussed uncover some strategies that may be helpful as
we solve square root equations. Consider the following worked examples:
p
Exercise SR1: Solve the equation 2
x + 2 = 2x.
S1 Isolate the term involving a radical!
Note: If there is more than one, isolate one radical, perform S2 and square
both sides. Then, expand, collect and repeat this sequence for the others.
2
p
p
()
x + 2 = 2x
x+2=2
2x
S2 Square both sides to undo the square root operation, and write a reminder
at the bottom of your page to “check solutions.”
p
Squaring both sides, we have:
( x + 2)2 = (2 2x)2
P.O.E
)
x+2
2x)2
= (2
S3 Identify the new equation and proceed with all relevant strategies.
This is a quadratic equation! So, first stategy: get a zero on one side!
x + 2 = (2
2x)2
)
0 = (2
)
(2
DP
4
)
)
4x2
factor
)
(x
ZPP
)
x
)
2x)(2
4x
collect
2x)2
x
2x)
2
x
4x + 4x2
9x + 2 = 0
2)(4x
2=0
x
2=0
and D = ( 9)2
4(4)(2) = 49 = 72
1) = 0
2 = 0 or 4x
1
x = 2 or x =
4
1=0
S4 Check that each value satisfies the original equation.
i.e. Substitute each value and verify whether L.S. = R.S.
p
For x = 2 :
L.S. = 2
(2) + 2
R.S. = 2(2)
p
= 2
4
=4
= 2
2
=0
Because 0 6= 4,
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L.S. 6= R.S. Thus, x = 2 is invalid/extraneous.
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Square Root Equations
Level: X
Exercise SR1 (Continued)
For x =
1
4
:
L.S.
q
= 2
q
LCD
= 2
P.O.E
=
+2
R.S. = 2
1+(2)(4)
4
=
1
4
1
2
p
p9
4
2
3
2
= 2
LCD
1
4
=
2(2) 3
2
=
4 3
2
=
1
2
) L.S. = R.S.
Therefore, x =
1
is the only solution.
4
p
Exercise SR2: Solve the equation
2x + 5
p
3x
2=1
Solution: Notice that we have two terms involving a radical. So, we’ll have to carry
out S1 followed by S2 twice!
p
p
S1 Isolating the first radical, we have: 2x + 5 = 1 + 3x 2.
S2 Squaring both sides and writing our “check” reminder, we have:
p
p
( 2x + 5)2 = (1 + 3x 2)2
P.O.E
)
2x + 5 = (1 +
DP
)
2x + 5 = 1 +
collect
2x + 5 = 1 +
)
p
3x
p
3x
|
2)(1 +
p
p
2 + 3x
{z
p
2 3x 2
3x
2)
p
2 +( 3x
}
2)2
+ 3x 2
p
S1 Isolating the next radical term and collecting gives: 2 3x 2 = 6
S2 Squaring both sides again, we have:
p
(2 3x
P.O.E
)
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4 (3x
2)2 = (6
x)2
2) = (6
x)2
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x
a quadratic equation!
c Sa’diyya Hendrickson
Square Root Equations
Level: X
Exercise SR2 (Continued)
S3 Identifying and solving the quadratic equation, we have:
4 (3x
2) = (6
x)2
x)2
)
0 = (6
)
(6
x)(6
DP
36
6x
collect
)
x2
24x + 44 = 0
factor
)
(x
2)(x
ZPP
)
)
x 2 = 0 or x 22 = 0
x = 2 or x = 22
)
x)
4 (3x
4 (3x
6x + x2
2) = 0
12x + 8 = 0
and D = ( 24)2
4(1)(44) > 0
22) = 0
S4 Checking our results, we have:
p
p
For x = 2 :
L.S. =
2(2) + 5
3(2)
p
p
= 4+5
6 2
p
p
= 9
4
= 3
2)
2
R.S. = 1
2
= 1
L.S. = R.S. So, x = 2 is a valid solution!
For x = 22 :
p
2(22) + 5
3(22)
p
p
= 44 + 5
66 2
p
p
= 49
64
L.S. =
p
= 7
=
2
R.S. = 1
8
1
Because 1 6= 1, L.S. 6= R.S. Thus, x = 22 is
invalid/extraneous.
Therefore, x = 2 is the only solution.
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Absolute Value Equations
Level: X
Consider the following examples of absolute value equations:
(a) |3x 5| 3 = 1
(b) |2x 1| = 2x + 4
(c) 3 |x| = |x 5|
Answer: x is in “jail”, behind
absolute value bars. So, to free x we must
use the definition of absolute value!
Note that there are di↵erent types of absolute value equations. Below we will discuss two,
both of which have only one term involving absolute value.
Type I One absolute value term and all other terms are constants.
Exercise AV1: Solve the equation |3x
5|
3=1
S1 Isolate the absolute value term by moving all constants/numbers
to the other side. i.e. Create an equation of the form: |a| = b.
|3x 5| 3 = 1
()
|3x 5| = 4
S2 Use the definition of absolute value to produce the relevant cases.
In general, for some nonnegative constant b, |a| = b means that
a is a distance of b units from zero. So, either a = b (Case 1)
or a = b (Case 2).
We have |3x 5| = 4, which means that a = 3x 5 is a distance
of 4 units from zero! This gives the following two cases:
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Case 1 :
3x
5=4
Case 2 :
3x
5=
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c Sa’diyya Hendrickson
Absolute Value Equations
Level: X
Exercise AV1 (Continued)
S3 Solve the equation under each case.
Case 1
3x
Case 2
5 = 4
+5
3x
+5
+5
3x
9
=
3
3
)
x=3
5 =
4
+5
3x
1
=
3
3
or
x=
1
3
Type II One absolute value term along with at least one non-constant term
(i.e. a term involving the variable to some non-zero power).
Consider the following example: |2x 1| 4 = 2x. Notice that if we isolate the absolute
value term, we have the equation |2x 1| = 2x + 4.
1 The distance from 0 is represented by “2x + 4,” which
changes as x changes.
2 Because distance is a nonnegative quantity, 2x + 4 is
restricted. Consequently, there are restrictions on x.
These di↵erences require us to proceed with caution. A popular and safe approach is to
take cases as dictated by the algebraic definition of absolute value. While
doing so, we must make note of the restrictions and check all solutions! Recall that:
8
< a
if a 0
(Case 1)
|a| =
:
a
if a < 0
(Case 2)
Exercise AV2: Solve the equation |2x
1| = 2x + 4.
S1 Use the algebraic definition of absolute value to determine the
relevant cases and restrictions for the equation.
8
<
2x 1
if 2x 1 0
(Case 1)
|2x 1| =
:
(2x 1)
if 2x 1 < 0
(Case 2)
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Absolute Value Equations
Level: X
Exercise AV2 (Continued)
S2 Solve the equation under each case. Be sure to identify and state
the restrictions before solving!
Case 1: 2x
1
0
) 2x
1
Case 2: 2x
) 2x < 1
1
2
) x
1<0
1
2
) x<
Note: A solution is valid under each case only if it lies in the
intervals stated above!
def’n
def’n
Here, |2x 1| = 2x 1
Here, |2x 1| =
) the equation becomes:
2x
) the equation becomes:
1 = 2x + 4
2x
(2x
DP
)
2x
1 = 4
2x + 1 =
2x + 4
2x
2x
)
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1) = 2x + 4
=
4x + 1 = 4
This is a contradiction!
So, in this interval, there
is no solution!
Therefore, x =
(2x 1)
1 =
1
4x
=
4
3
4
x=
3
1
<
4
2
3
is the only solution for this equation.
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