Equations in One Variable I
An equation in one variable is an equation of the form f (x) = g(x) where
f (x) and g(x) are functions.
Examples.
• x2 − 3x + 2 = loge (x) is an equation in one variable. Both x2 − 3x + 2
and loge (x) are functions.
• x − 3 = 2 is an equation in one variable. Both x − 3 and 2 are
functions. The latter is a constant function.
√
• x = 15x − 4 is an equation in one variable.
•
x2 +7
x−1
+ 27x − 3 = ex
3
−18
is an equation in one variable.
All of the examples above are equations, because they are mathematical
formulas that use an equal sign. The are in one variable, because there is
a single variable used in each equation. In all of the examples above, x is
the only variable. We’ll continue to use x as the only variable for a while,
but keep in mind that it’s just a variable, and that we could use any variable
instead. For example, there really isn’t any difference between the equation
x − 3 = 2 and the equation w − 3 = 2. The only difference is the name that
we gave the variable, x or w, but otherwise the equations are the same.
Domain of an equation
If f (x) = g(x) is an equation in one variable, then the implied domain of
the equation is the set of all real numbers that are in both the implied domain
of f and the implied domain of g.
Examples.
• The implied domain of the equation x3 − 2x + 1 = 3x − 9 is R. That’s
because x3 − 2x + 1 and 3x − 9 are each polynomials, so they each have R as
their implied domains.
√
•√
The equation x = 4 has an implied domain of [0, ∞). That’s
because x has an implied domain of [0, ∞), while the constant
√ function 4
has an implied domain of R. Therefore, the implied domain of x = 4 is the
51
set of all real numbers that are√in the set [0, ∞) and in the set R, or in other
words, the implied domain of x = 4 is the set [0, ∞).
• The implied domain of the equation loge (x) + loge (x − 2) = 1 is
(2, ∞). The implied domain of the constant function 1 is R. The implied
domain of loge (x) + loge (x − 2) is a little more tricky. We can only take the
logarithm of a positive number. That means that we want x > 0 and that
we want x − 2 > 0, or equivalently, that x > 2. Of course asking that x > 0
and x > 2 is the same as just asking that x > 2, so the implied domain for
the function loge (x) + loge (x − 2) is (2, ∞).
√
3x
is [0, ∞) − {5}. We
• The implied domain of the equation x = x−5
can only take the square-root of a number if it is greater than or equal to 0,
so x ≥ 0. We can never divide by 0, so x − 5 6= 0, or equivalently x 6= 5. The
set of all real numbers x with x ≥ 0 and x 6= 5 is the set [0, ∞) − {5}.
√
• [3, 5) is the implied domain of the equation loge (5 − x) = x − 3.
We can only take the logarithm of a number if it is positive, so 5 − x > 0,
or equivalently, 5 > x. We can only take the square-root of a number if it is
greater than or equal to 0, so x − 3 ≥ 0, or equivalently, x ≥ 3. The set of
all real numbers x with 5 > x and x ≥ 3 is exactly [3, 5).
Sometimes, you will be given an explicit domain for an equation. For
example, you might be asked to examine the equation x2 − 3x + 1 = ex for
x ∈ [0, 10]. This means that the domain of the equation is the restricted
domain [0, 10]. The domain is restricted, because the implied domain of the
equation x2 −3x+1 = ex would have been R if we hadn’t been told otherwise,
but we have been told otherwise. We are told here that the domain of the
equation is just the set [0, 10].
Solutions of equations
If someone writes an equation such as 3x − 2 = 4, they are not saying that
3x − 2 is the same function as the constant function 4. They are clearly not
the same function. One is a constant function and the other one isn’t. Rather,
they are asking you to find the set of all numbers that can be substituted
in for x to make the equation true. For example, we can substitute 2 in the
place of x and we’d be left with the equation 3(2) − 2 = 4, which is true, as
you can check. The equation usually won’t be true for every number that you
substitute for x as we can see by substituting the number 0 in the place of x.
52
That would leave us with 3(0) 2 = 4, which is clearly false. The numbers
Thatcan
would
leave us with
2 = 4,anwhich
is clearly
false.
Thesolutions
numbers
that
be substituted
for3(0)
x to−make
equation
true are
called
that
be substituted for x to make an equation true are called solutions
of
thecan
equation.
ofIfthe
equation.
f(x)
= g(x) is an equation in one variable, and if the equation has the
If
f
(x)
= domain,
g(x) is an
equation
variable,
and =
if g(x)
the equation
set D as its
then
the set inofone
solutions
of f(x)
is the sethas the
set D as its domain, then the set of solutions of f (x) = g(x) is the set
—
S = { α ∈ D | f (α) = g(α) }
Returning to the example above, if S is the set of solutions of the equation
to the2 example
above,S.if SWe’d
is thespeak
set ofthis
solutions
equation
3xReturning
2 = 4, then
solution
as 2 isofa the
of
E S, but 0
3x
−
2
=
4,
then
2
∈
S,
but
0
∈
/
S.
We’d
speak
this
as
2
is
a
solution
of
3x 2 = 4, but that 0 is not a solution of 3x 2 = 4.
3x − 2 = 4, but that 0 is not a solution of 3x − 2 = 4.
Examples.
Examples.
Suppose that S is the set of solutions of the equation √ = 4. We
•
Suppose
that
S is the
solutions
thesoequation x = 4. We
saw above
that this
equation
hasseta of
domain
of [0,ofoc),
saw above that this equation has a domain of [0, ∞), so
√
S={ce[0,oo) I=4}
S = { α ∈ [0, ∞) | α = 4 }
√ = 4. However, —10 S since
We see that 16 E S, since 16 e [0, oc) and /i
We
see
that
16
∈
S,
since
16
∈
[0,
∞)
and
16 = 4. However, −10 ∈
/ S since
√ 4.
—10 0 [0,oo), and 25 5, since /E
4
7
−10 ∈
/ [0, ∞), and 25 ∈
/ S, since 25 6= 4.
—
—
—
0
. Now suppose that S is the set of the solutions for the equation x
2 2=
that S is the
the solutions
the equation
x = x.
Both x
2•2Now
and xsuppose
are polynomials,
so set
theofdomain
of the for
equation
is R and
Both x and x are polynomials, so the domain of the equation is R and
S = {c ER a
2
S = { α ∈ R | α2== α }
Then 0, 1 E S because 0 and 1 are both real numbers and 02 2= 0 and 12 2= 1.
Then 0, 1 3∈ SSbecause
and 1 are both real numbers and 0 = 0 and 1 = 1.
However,
because0 32
0
However, 3 ∈
/ S because 32 6= 3.
51
53
x
x
1-”-••,
1-”-••,
Do
Do
. Let’s look at
•
look
at
.
Let’s
look
equationLet’s
2
—x
=
+at9 an
an equation with a restricted domain. Let’s look at the
an
equation
aWe
restricted
domain.
Let’s
at the
equation
a restricted
domain.
look
at the
0 for
are told
x E with
oo).
(0,with
here Let’s
that
the look
domain
is
2
equation
equation
−x
=for
0that
for
∈are
(0,
We
the
domain
is
2
=9 0
are are
x Exwe(0,
oo).∞).
Weinterested
toldtold
here
thatthat
the
domain
+ 9+means
(0, oo),—x
which
only
in here
finding
solutions
tois the
(0, (0,
oo),
∞),
which
means
only
interested
in
finding
to
which
thatthat
we we
are
interested
only
in finding
to the
equation
if means
solutions
those
areare
positive
numbers.
That
is,solutions
ifsolutions
S is the
setthe
of
equation
if
equation
if
those
solutions
are
positive
numbers.
That
is,
if
S
is
the
set
of
solutions
those
are
positive
numbers.
That
is,
is
the
if
S
set
of
solutions, then
solutions,
solutions,
then
then
5= {
(0, oc)
+9= 0}
2
∈ positive
(0,
+ 90}
=that
0 } are also roots of the
5=Sset
(0,
oc) ∞) | −α
+9=
{= of{ αall
Notice that S is the
numbers
Notice
that
Notice
that
is the
of+ all
numbers
are
also
roots
of the
isS the
set set
of all
positive
Spolynomial
numbers
thatthat
also
roots
of the
quadratic
2
—x
know
9 =positive
0. We
how
toare
find
roots
of
quadratic
2
quadratic
quadratic
polynomial
−x
0.+9We
how
toOf
find
roots
ofroots,
quadratic
polynomial
2
—x
know
=9—x
0.= We
how
to 3.find
roots
of
quadratic
+ 9+of
polynomials,
and the
roots
2
and
areknow
—3
these
two
only
2
polynomials,
polynomials,
and
the
roots
of
−x
+
9
are
−3
and
3.
Of
these
two
roots,
only
and
the
roots
of
2
—x
and
Of
are
—3
3.
these
two
roots,
+9
only
3 is in the domain of the equation. That is, only 3 is positive. Therefore,
in
3 is3S={3}.
domain
is the
in the
domain
of the
equation.
That
is, only
is positive.
Therefore,
of the
is, only
equation.
That
3 is3positive.
Therefore,
S={3}.
S = {3}.
54
The equation √ ==x x++5 5has
has an implieddomain
domain of [0, oc),since
since we
•••The
Theequation
equation x = x + 5 hasananimplied
implied domainofof[0,[0,oc),
∞), sincewewe
can’t takethe
the square-rootofofa anegative
negativenumber.
number.We
Wecan
cansee
seefrom
fromthe
the graphs
can’t
can’ttake
take√thesquare-root
square-root of a negative
number. We
can see
below graphs
that the
below,that
thatthe
thegraphs
graphsofof and
andx x++5 5dodonot
Thatmeans
notintersect.
intersect.That
meansthere
there
below,
graphs
of x and x + 5 do not intersect. That means there isn’t a number α
√
isn’ta anumber
numbera awith
with%/
Thatis istotosay,
therearen’t
aren’tany
solutions
say,there
+5.That
anysolutions
%/==a a+5.
isn’t
α
=
α
+
5.
That
is
to
say,
there
aren’t
any
solutions
of
the
equation
with
theequation
ofthe
equation ==x x++5.5.The
Thesolutions
solutionsset
setis is0.0.
of√
x = x + 5. The set of solutions is ∅.
//
Dorn
Dorn
An importantgeneral
general exampleis isif ifp(x)
p(x) is a polynomialand
and our
• ••An
Animportant
important generalexample
example is if p(x)is isa apolynomial
polynomial andour
our
equationisisp(x)
p(x)==0.0. Then
Thenthe
thesetsetofofsolutions
solutionsforforthis
thisequation
equationis isexactly
exactly
equation
equation is p(x) = 0. Then the set of solutions for this equation is exactly
the setofofroots
roots of thepolynomial
polynomial p(x). We
how to findroots
roots of any
We knowhow
the
theset
set of rootsofofthe
the polynomialp(x).
p(x). Weknow
know howtotofind
find rootsofofany
any
linearand
andquadratic
quadraticpolynomial,
polynomial,but
butit itis isextremely
extremelydifficult
difficulttotofind
findroots
roots
linear
linear and quadratic polynomial, but it is extremely difficult to find roots
ofmost
most polynomials. AApolynomial
polynomial equationp(x)
p(x) = 0 is aboutasasnice
niceanan
ofof
mostpolynomials.
polynomials. A polynomialequation
equation p(x)= =0 0is isabout
about as nice
of
equationasaswewecan
canhope
hopetotohave,
have,and
andeven
evenfinding
findingitsitssolutions
solutionscan
canbebea a
equation
an equation as we can hope to have, and even finding its solutions can be a
nearly impossibletask.
task.
nearly
nearlyimpossible
impossible task.
QQ
55
*
*
*
*
*
*
*
*
*
*
*
*
*
In the remainder of this chapter, we’ll cover a few basic rules of algebra that
can be used to help us find solutions of equations. We’ll learn when equations
are equivalent — either by addition, multiplication, or invertible functions —
and we’ll see that equivalent equations have the same solutions. That will
allow us to find solutions of equations by creating a sequence of equivalent
equations that terminates in an equation that’s easy for us to solve.
Equations equivalent by addition
In the definition below, D is a set of real numbers. That is, D ⊆ R.
The equation
f (x) + h(x) = g(x) with domain D
is equivalent to the equation
f (x) = g(x) − h(x) with domain D
Examples.
• The equation x2 + 3x = ex with domain R is equivalent to the
equation x2 = ex − 3x with domain R. Here we are just using the rule above
with f (x) = x2 , h(x) = 3x, and g(x) = ex .
• The domains of two equivalent equations must be the same. The
equation x2 + 3x = ex with the restricted domain (0, ∞) is equivalent to the
equation x2 = ex − 3x with the restricted domain (0, ∞).
• To repeat, the domains of two equivalent equations are equal. The
equation x2 + 3x = ex with domain [−3, 10] is equivalent to the equation
x2 = ex − 3x with domain [−3, 10]. Compare this example to the previous
two examples. Equivalent equations have the same domain.
• x2 −loge (x) = 0 is equivalent to x2 = loge (x). To see this, just subtract
− loge (x) (or add loge (x)) to both sides of the equation x2 −loge (x) = 0. Both
56
equations in this example have an implied domain of (0, ∞), because we can
only take the logarithm of a positive number.
• Whenever you add or subtract a function from both sides of an
equation, you’ll obtain an equivalent equation. Thus, 7x3 − 5 = x2 + 2x is
equivalent to 7x3 − x2 − 2x − 5 = 0. Just subtract x2 + 2x from both sides
of 7x3 − 5 = x2 + 2x.
√
√
• 3x = 2x2 is √
equivalent to 3x + x − 2 = 2x2 + x − 2. Just add
x − 2 to both sides of 3x = 2x2 .
The importance of equivalent equations is that they have the same sets of
solutions.
Equivalent equations
have the same sets of solutions.
Examples.
• The equations x2 = −x + 5 and x2 + x − 5 = 0 are equivalent. To
get the second equation from the first, just add the function x − 5.
Because these two equations are equivalent, they have the same set of
solutions. We know how to find the solutions of x2 +x−5 = 0. These
√ are just
1
2
the roots
21) and
√ of the quadratic polynomial x + x − 5, which are 2 (−1 −
1
2
words, the set
of solutions of the equation x +x−5 = 0
2 (−1+ 21). In other
√
√
is the set { 12 (−1 − 21), 12 (−1
+ 21)}. Therefore,
the set of solutions of
√
√
x2 = −x + 5 is also { 12 (−1 − 21), 12 (−1 + 21)}.
• To find the set of solutions of x2 + x − 4 = x2 , just subtract x2
from both sides of the equation. We’ll be left with the equivalent equation
x − 4 = 0. The set of solutions of this latter equation is {4}. Because the two
equations in this example are equivalent, the set of solutions of x2 +x−4 = x2
is also {4}.
√
• To find the solutions of x − x = 5, we can add the function x
to
√ both sides of the equation and we’d be left with
√ the equivalent equation
x = x + 5.√We saw earlier in this chapter that x = x + 5 has no solutions.
Therefore, x − x = 5 has no solutions.
57
*
*
*
*
*
*
*
*
*
*
*
*
*
Equations equivalent by multiplication
We say that the number α ∈ R is a zero of the function h(x) if h(α) = 0.
Examples.
• 2 is a zero of h(x) = x2 − 4 because h(2) = 22 − 4 = 0. The number
−2 is also a zero of h(x) = x2 − 4.
• 3 is a zero of h(x) = x − 3 since h(3) = 3 − 3 = 0.
• In the previous two examples, h(x) was a polynomial. If h(x) is a
polynomial then we usually call a zero of h(x) a root of h(x) instead.
It’s just a different name for the same thing.
√
√
√
• 5 is a zero of x − 5 because 5 − 5 = 0 = 0.
• 1 is a zero of loge (x).
• 4 is not a zero of 2x − 5 because 2(4) − 5 = 8 − 5 = 3 6= 0.
We say that h(x) has no zeros in a set D ⊆ R if none of the numbers in D
is a zero for h(x).
Examples.
• The linear polynomial h(x) = x − 3 has no zeros in [5, ∞). That’s
because if α ∈ [5, ∞), then α ≥ 5. Therefore, h(α) = α − 3 ≥ 5 − 3 = 2, and
if h(α) ≥ 2, then h(α) 6= 0. That is, α is not a zero of h(x) if α ∈ [5, ∞).
The function h(x) = x − 3 does have a zero. It’s 3. But h(x) has no zeros
in [5, ∞).
(co
1/”
7
58
• The function h(x) = x2 − 4 has two zeros, 2 and −2. Neither of these
zeros are in the set [−1, 1], so h(x) = x2 − 4 has no zeros in [−1, 1].
(co
(co
1/”
1/”
7 7
/
1/”
• 4 − 2x has exactly one zero, the number 2. Because 2 is the only
zero, 4 − 2x has no zeros in R − {2}.
(co
(co
2
7
1/”
7
//
(oo’o)
• ex has no zeros. It has no zeros in R.
/
22/
(oo’o)
(oo’o)
• x is a function. It’s the identity function. It only has one zero, the
number 0 itself. Thus, x has a zero in R, but it has no zeros in R − {0}.
Neither does it have zeros in (0, ∞), for example.
2
(oo’o)
59
In the definition below, D is a subset of real numbers.
The equation
h(x)f (x) = g(x) with domain D
is equivalent to the equation
f (x) =
g(x)
h(x)
with domain D
as long as h(x) has no zeros in D.
The definition above tells us that we can multiply or divide both sides of an
equation by a function—as long as the function has no zeros in the domain—
and obtain an equivalent equation.
Examples.
• If h(x) is a constant function, and not the constant 0, then it has no
zeros. Therefore, we can always divide both sides of an equation by h(x) to
x4 +1
4
obtain an equivalent
equation.
Thus,
3x
=
x
+
1
is
equivalent
to
x
=
3 .
√
√
The equation 2 x = 8 is equivalent to x = 4.
• If we multiply both sides of an equation by a constant that’s not zero
then we’ll also obtain an equivalent equation. x4 = log5 (x) is equivalent to
√
√
x = 4 log5 (x). 13 x = 4 is equivalent to x = 12.
• The equation x loge (x) = x has a domain of (0, ∞). Because the
function x has no zeros in (0, ∞), we can divide both sides of the equation
x loge (x) = x by x to obtain the equivalent equation loge (x) = 1.
• This is an example of something that can’t be done. If you have the
equation x2 = x, then you might be inclined to divide by x, similar to the
previous example. We can’t do that here though. The implied domain of the
equation x2 = x is R, and x has a zero in R. It has the number 0 as its zero.
The two equations x2 = x and x = 1 (which is what we would get by
dividing x2 = x by x) are very different equations. The former has two
solutions, 0 and 1, while the latter has a single solution, 1.
60
Equivalent equations
have the same solutions.
Examples.
√
√
• The equation 2 x = 8 √
is equivalent by multiplication
to
x = 4.
√
We know the set of solutions of x = 4 is {2}, so 2 x = 8 also has 2 as its
only solution.
• The equation x loge (x) = x is equivalent to loge (x) = 1. The only
solution of loge (x) = 1 is the number e. Therefore, e is the only solution of
the equation x loge (x) = x.
*
*
*
*
*
*
*
*
*
*
*
*
*
Equations equivalent by invertible function
Once again, in the definition below D ⊆ R.
The equation
h(f (x)) = g(x) with domain D
is equivalent to the equation
f (x) = h−1 (g(x)) with domain D
The above definition says that we can erase a function, h, on one side of an
equation if we apply its inverse, h−1 to the other side of the equation. Both
equations have to have the same domain. Equivalent equations always have
the same domain.
61
Examples.
• loge (x) = 5 has an implied domain of (0, ∞), because we can only
take the logarithm of a positive number. We could erase logarithm base e
on the left side of loge (x) = 5 by applying its inverse, exponential base e to
the right side. We’d be left with the equation x = e5 with the same domain
that we started with, the set (0, ∞). These two equations, loge (x) = 5 and
x = e5 , with the same domain, (0, ∞), are equivalent.
Equivalent equations always have the same domain.
√
√
• The functions 3 x and x3 are inverse functions. Therefore, 3 x − 2 = 7
is equivalent to x − 2 = 73 . (Both equations have a domain of R because we
can cube
or cube-root any number.) We erased the cube root from the left
√
3
side of x − 2 = 7 by applying its inverse to the right side.
Again, what’s important about equivalent equations is the following fact:
Equivalent equations
have the same solutions
Example.
• We’ll continue from the previous example. If we have the equation
x − 2 = 7, then we know it’s equivalent to the equation x − 2 = 73 . Now
73 = 343, so the previous equation is x − 2 = 343. The only solution of this
equation is√x = 343 + 2 = 345.
Because 3 x − 2 = 7 and x − 2 = 343 are equivalent,
they have the same
√
3
solutions. Therefore, 345 is the only solution of x − 2 = 7.
√
3
*
*
*
*
*
*
*
*
*
*
*
*
*
Finding solutions of equations
To find the solutions of an equation f (x) = g(x), first write down its
domain D. Next, find a sequence of equivalent equations, either equivalent
by addition, or by multiplication, or by invertible function. The goal is to
have the last equation in this sequence have a set of solutions that is easy to
62
find. It might be really easy, such as in the equation x = 2, or it might be
somewhat easy to find, as in the equation x2 +2x−5 = 0, where you can apply
the quadratic formula to the quadratic polynomial x2 +2x−5. Whatever easy
equation you are left with, find its set of solutions S, but remember to only
include in the set S those solutions that are in the domain D of your original
equation. Because equivalent equations have the same set of solutions, S will
be the set of solutions of your original equation f (x) = g(x).
Examples.
2
• Let’s find the solutions of the equation ex +x−6 = 1 where the domain
of the equation is the set D = (0, ∞). That means we are only interested in
2
those solution of ex +x−6 = 1 that are positive.
2
Using loge , the inverse function of exponential base e, the equation ex +x−6 =
1 is equivalent to x2 +x−6 = loge (1), which is the equation x2 +x−6 = 0. To
find the solutions of this quadratic equation, we use the quadratic formula
to find the roots of the quadratic polynomial x2 + x − 6. In doing so, we
would find that the solutions of x2 + x − 6 = 0 are −3 and 2. However,
the domain of our original equation is (0, ∞), and −3 ∈
/ (0, ∞). Therefore,
the only relevant solution for us is 2, because 2 ∈ (0, ∞). That means that
2
S = {2} is the set of solutions for our original equation ex +x−6 = 1 with a
2 + xof (0,
x
6 =
0 are —3 and 2. However, the domain of our original equation
domain
∞).
is (0, cc), and —3 (0, cc). Therefore, the only relevant solution for us is 2,
because 2 e (0, cc). That means that S = {2} is the set of solutions for our
_6 = 1 with a domain of (0, cc).
2
original equation ex
—
d\
e
C—
-
1
-3
2.
Let’s find the solutions of x 1 = 1 If we aren’t told what the
domain of the equation is, then we 63have to find its implied domain. For this
equation, the implied domain is R {0}, because we can’t divide by 0. That
means that we should only search for solutions of the equation that aren’t
zero.
We can multiply by the function x to obtain the equivalent equation x
—x =
2
.
—
—
11
-
-3
-3
2.2.
1
• Let’s
find
thethe
solutions
of of
x xx 1 1=
wewe
aren’t
=
=
1If IfIf
find
.. Let’s
aren’t
solutions
told
what
the
x1 .1
find
the
Let’s
we
aren’ttold
solutions
of
1
toldwhat
whatthe
.
If
we
aren’t
told
what
thethe
•
Let’s
find
the
solutions
of
x
−
1
=
x
domain
of
the
equation
is,
then
we
have
to
find
its
implied
domain.
For
this
domain
the equation
is, then
of
implied
we have
to
find
its
domain.
For
this
domain
equation
then
of
have
implied
to
find
its
domain.For
Forthis
this
domain
ofthe
thethe
equation
is, is,
then
wewe
have
to
find
itswe
implied
domain.
equation,
implied
domain
is
R
{0},
because
can’t
divide
by
0.
That
equation,
the
implied
domain
isis RR {0},
because
we
can’t
divide
by 0.0. That
equation,
the
implied
domain
because
we
can’t
divide
{0},
That
equation,
thewe
implied
domain
is R
−for
{0},
because
wethe
can’t
divide
byby
0.aren’t
That
means
that
should
only
search
solutions
of
equation
that
that
we
means
should
only
search
for
the
equation
solutions
of
that
aren’t
that
means
should
only
search
equationthat
solutions
thataren’t
aren’t
means
that
wewe
should
only
search
forfor
solutions
of of
thetheequation
zero.
zero.
zero.
zero.
WeWe
can
multiply
byby
the
function
x to
obtain
the
equivalent
equation
x22x
xxx
===
can
multiply
the
function
equivalent
xxto
obtain
the
equation
—
2
We
can
multiply
by
the
function
equivalent
to
obtain
the
equation
—
2
x
2
We
can
multiply
by
the
function
x
to
obtain
the
equivalent
equation
x
−x
=
1. 1.1.WeWe
can
add
1—1
toto
obtain
the
equivalent
equation
x 2x
x xx 1 1=
0. 0.The
can
add
obtain
the
equivalent
equation
2
The
=
We
can
add
—1
to
obtain
the
equivalent
equation
2
x
1
The
= 0.
1. We canequation
add −1 tells
to obtain
thethe
equivalent
equation
x − x −equation
1 = 0. The
quadratic
us that
solutions
toto
this
quadratic
quadratic
equation-tells-ks
that
the
solutions
this
quadratic
equation
are
p
p
quadratic
equation-tells-ks
that
the
solutions
this
quadratic
to
equationare
are
1
1
quadratic
formula
tells
us
that
the
solutions
to
this
quadratic
equation
are
(1
5)
and
(1
+
5).
Neither
of
these
numbers
are
0,
they
are
both
in
√
√
and
1
Neither
of
/‘g)
numbers
are
these
0,
they
are
both
inin
21
Neither
of
/‘g)andand21 (1 1
numbers
are
these
0,so they
are
both
(1
−
5)
+
5).
Neither
of
these
numbers
are
0,
they
are
both
2 the
2our
the
domain
of of
our
original
equation,
soso
they
are
both
solutions
ofof
our
original
domain
g(xial
equation,
they
are
solutions
original
both
our
the
ofof
our
domain
g(xial
equation,
so
they
are
solutions
original
both
of
our
1
1
in
the
domain
our
original
equation,
and
they
are
both
solutions
of
our
equation,
x
1
=
.
They
are
also
the
only
solutions
of
x
1
=
.
equation,
also
xx 11 =x= They
the
of
only
solutions
xx 11 ==
x
equation,
They
are
also
the
of
only
solutions
1
1 are
original equation, x − 1 = . They are the only solutions of x − 1 = .
—
—
—
—
—
—
—
—
—
—
—
.
.
—
.
.
—
—
—
—
x
x
/x-1
/x-1
p
3
•• The
The equation
equation 5x
5x 22 == 55has
hasan
animplied
implieddomain
domainofofR.R. We
Wecan
can
=
53
√
3
obtain
•the
The
equation
domain
equivalent
2by=
equation
invertible
implied5x5x
R.
WeNow
—2 2 = of
5 61has anfunction
can
by
35x
obtain•the
equivalent
equation
invertible
function
5 =R.
=125.
125.
Now
The
equation
5x
−
2
=
5
has
an
implied
domain
We
can
= of
53
127
obtain
we
the
equivalent
can
22totoobtain
equation
invertible
add
5x
—2x x== = .125. Now
5x
by and
divide
by
5 5totohave
we
canthe
addequivalent
obtain
5x==127,
127,
and
divide
by
have
3 function
obtain
equation
5x
−
2
=
5
=
125
by
erasing
the5127cube-root
we
can
= 127,ofand
add 2127toisisobtain
5x
divide
by function,
to have since
xsince
5function,
Of
course
ininthe
domain
original
our
is a areal
=
Of
course
the
domain
of
our
original
real
from the left and
Now we can add 25 toisobtain
5 applying the cube to the right.
p
3
Of
course
is
in
the
domain
original
function,
127
of our127
since
a real
number.
Therefore, the
set of solutions
is the
of /5x
set
number.
solutions
set { is5 }.
5x
= 127, Therefore,
and divide the
by 5settoofhave
x = 5of. 5x 2 2==5 5 isthe
number. Therefore,
the set of solutions of /5x 2 = 5 is the set127
Of course 127
original function, since 5 is a real
5 is in the domain of our √
number. Therefore, the only solution of 3 5x − 2 = 5 is 127
5 .
—
—
—
—
• The equation 5x 2 = 5 has an implied domain of R. We can
obtain the equivalent equation by invertible function 5x —2 = 53 = 125. Now
we can add 2 to obtain 5x = 127, and divide by 5 to have x =
Of course
is in the domain of our original function, since
is a real
1I1in;
(wH1
number. Therefore, the set of solutions of /5x 2 = 5 is the set
127
1I1in;
(wH1
Dornon
127
Dornon
—
—
64
••The
has
Theequation
equation3x
3x 22==0,0,
hasaasolution.
solution.It’s
It’s 23 .However,
However,ififweweare
are
1I1in;
(wH1
told
Thethe
equation
solutions
3xofof3x
23x=
solution.
It’sofof(—oo,
if we
0,2=has
awith
—2127
are
there
a adomain
toldtoto•find
find
the
solutions
=0 0with
domain
(However,
1,0),0),then
then
there
Dornon
—
—
.
.
127
Dornon
• The equation 3x − 2 = 0, has a solution. It’s 23 . However, if we are
told to find the solutions of 3x − 2 = 0 with a domain of (−∞, 0), then there
2
are no• solutions.
The3x
set of2 solutions
is ∅. That’s because
3 isn’t a
The equation
However,
if negative
= 0, has a solution. It’s
we are
number.
told to find the solutions of 3x —2 = 0 with a domain of (—oo, 0), then there
.
—
are no solutions. The set of solutions is 0. That’s because
number.
Doa’tr
1
62
65
isn’t a negative
Exercises
For #1-13, match the numbered equations on the left to their lettered
implied domains on the right.
1.) loge (x) − loge (x) = 4
A.) R
2.) x2 − 45 = x7 − 3x − 6
B.) (0, ∞)
2
3.) e−7x
√
√x
x
4.)
−3x
= x3
C.) [0, ∞)
= 3x − 2
D.) [2, ∞)
5.) loge (x − 5) = 23x9
√
171
6.)
3x − 5 = 2x + 3
2x2 −7x+3
x−5
7.)
√
3
G.) R − {5}
√
x−2
x − 7 = 3x + 4
10.) x5 +
11.)
F.) (5, ∞)
= 3x + 4
8.) loge (x + 7) =
9.)
E.) [0, ∞) − {2}
√
20
7x+3
x2 −4
=
√
x
x − 2 = 3x + 6
12.) 3x − 4x6 = x3 − 13x + 56
13.)
√
x = 3x2 − 5
66
For #14-19, match the numbered equations on the left to their lettered sets
of solutions on the right.
3
14.) 4x − 3 = 0
A.)
15.) 2x − 6 = 0
√
√
B.) { −1 − 2 2 , −1 + 2 2 }
16.) x + 4 = 0
C.) ∅
17.) x2 − 2x + 3 = 0
D.) {−4}
18.) 2x2 − 12x + 18 = 0
E.) {3}
4
19.) −x2 − 2x + 7 = 0
For #20-24, match the numbered equations on the left to the lettered equations on the right that they are equivalent to by addition.
√
20.) x2 + x = 3
A.) 5 = 3x +
21.) loge (x) + 2x − 3 = 0
B.) 2x2 − 7x + 3 = 0
22.) 5 −
√
x+2
C.) x2 + x − 3 = 0
x = 3x + 2
23.) 3x2 − 4x + 5 = x2 + 3x + 2
D.) −x2 − 3x + 3 = 0
24.) −x2 − 4 = 3x − 7
E.) 2x − 3 = − loge (x)
67
If you ever have an equation p(x) = q(x) where both p(x) and q(x) are
polynomials, it’s almost always best to begin by writing the equivalent equation p(x) − q(x) = 0, and then to search for solutions of p(x) − q(x) = 0. Use
this method to find the solutions of the equations in problems #25-28.
25.) x2 − 3x + 2 = 3x2 + x − 4
26.) x3 + 3x − 2 = x3 + x2 + x + 1
27.) x2 − 2x + 3 = x2 − x + 4
28.) x − 2 = 3x2 − 5x + 1
For #29-35, match the numbered functions on the left with their lettered
set of zeros on the right.
29.) x − 7
A.) {−1, 1}
30.) x + 5
B.) {0}
31.) x2 − 1
C.) {−5}
32.) loge (x)
D.) {1}
33.)
√
E.) ∅
x
34.) x2 + 1
F.) {7}
35.) x
68
For each equation given in #36-40, decide whether dividing by the function
x on both sides of the equation would result in an equivalent equation. For
each of these, you’ll want to check whether x has zeros in the implied domain
of the given equation.
36.) 3x2 − 5 = 2x + 8
37.) loge (x) = 3x − 5
38.)
39.)
1
x−7
√
= 3x
−x = 4x3 − 5x
40.) x2 (x − 7) = x(x − 7)2
For each equation given in #41-45, decide whether dividing by the function
x − 7 on both sides of the equation would result in an equivalent equation.
For each of these, you’ll want to check whether x − 7 has zeros in the implied
domain of the given equation.
41.) 3x2 − 5 = 2x + 8
42.) loge (x) = 3x − 5
43.)
44.)
1
x−7
√
= 3x
−x = 4x3 − 5x
45.) x2 (x − 7) = x(x − 7)2
For #46-48, match the numbered equations on the left to the lettered equations on the right that they are equivalent to by multiplication.
46.) x + x = 3
A.)
√
5− x
x+2
47.) loge (x) = 2
B.)
1
x
2
48.) 5 −
√
=
3x+2
x+2
loge (x) =
2
x
C.) ex (x2 + x) = 3ex
x = 3x + 2
69
For #49-52, match the numbered equations on the left to the lettered equations on the right that they are equivalent to by invertible function.
2
49.) e3x+5 = e2x−3
A.) 3x + 5 = x 3
50.) loge (x2 + 1) = 2
B.) x2 − 3x + 1 = 8x3
51.) (3x + 5)3 = x2
C.) 3x + 5 = 2x − 3
52.)
√
3
D.) x2 + 1 = e2
x2 − 3x + 1 = 2x
Find the set of solutions of each of the equations given in #53-61.
53.) x2 − 3 = x2 + x
54.) loge (x2 ) = 2 where x ∈ [0, ∞)
55.) 4x2 + 2x + 3 = 3x2 + 2x + 2
56.)
√
3
4x + 2 = 4
57.) 3x2 − 6 = x2 − 4x with x > 10
58.) e3x+3 = 1
59.) x5 (x2 + 3x − 2) = 0 where x < 0
60.) (log2 (2x + 1))3 = 8
61.)
455
x2 −9
= 5 where x < 5
70