Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 110 pps

A.3 Solving Equations 1071
Linear Equations
Definition
An equation is linear in x if it can be put into the form ax +b =0, where a and
b are constants or expressions not containing x, and where a = 0.
To determine whether an equation is linear in x, clear all x’s from the denominators; if
x is inside parentheses, remove the parentheses and multiply out.
◆
EXAMPLE A.24
(a) 3x − 2y =7 is linear in x and linear in y.
(b) 3y
2
+ 7xy = 6xz is linear in x and in z,butnot linear in y.
(c) (x + 2)(x − 3) = x is not linear in x.It’squadratic, because x is raised to the second
degree. ◆
The graph of a linear function f(x)=ax +b is a line. If a = 0, then f(x)=K (where K
is a constant) must have exactly one solution.
Solving. Suppose an equation is linear in x or can be converted to such an equation. Solve
for x as follows.
1. Clean up the equation if necessary; if x is inside parentheses, remove the parentheses.
If x is in the denominator, clear the denominator.
2. Get all terms with x’s on one side of the equation and all terms without x’s on the other
side:
x + x + x = .
3. Isolate x: Factor out the x and then divide both sides by the coefficient of x.
If x was in the denominator of the original equation, the equation is not really linear;
however, it appears linear after clearing denominators. Be sure to check that your solution
doesn’t make the denominator zero.
◆
EXAMPLE A.25 Solve the following equations for x.
(a) 2βx −5(x − β) =γ +3x ∝

2
(b)
√
yx +z
2
(x − 1) =xy
3
+2
(c)
2
x+1
=
1
x
(d)
2
x
+
m
2
+
m
x
(2 −m) = 3m
SOLUTION (a) Solve for x in terms of α, β, and γ .
2βx −5(x − β) =γ +3xα
2
2βx −5x +5β =γ +3xα
2
2βx −5x −3xα

2
=−5β+γ
x(2β −5 −3α
2
) =−5β+γ
x=
−5β+γ
2β − 5 − 3α
2
1072 APPENDIX A Algebra
(b) Solve for x in terms of y and z.
√
yx +z
2
(x − 1) =xy
3
+2
√
yx +z
2
x −z
2
=xy
3
+2
√
yx +z
2
x −y
3

x =2 +z
2
x(
√
y + z
2
− y
3
) = 2 + z
2
x =
2 +z
2
√
y + z
2
− y
3
(c) Multiply both sides of the equation by x(x +1) to clear the denominators. This is fine
as long as x = 0 and x =−1.
x(x +1)
1
·
2
x + 1
=
1
x
·
x(x +1)

1
2x = x +1
x = 1
(d) Solve for x in terms of m:
2
x
+
m
2
+
m
x
(2 −m) = 3m.
First we’ll multiply both sides by 2x to clear all denominators. This is fine as long
as x = 0.
2x

2
x
+
m
2
+
m(2 −m)
x

= 3m · 2x
4 + mx + 2m(2 − m) = 6mx
mx − 6mx =−4−2m(2 − m)
x(m −6m) =−4−2m(2 − m)

x =
−4 − 4m +2m
2
m − 6m
=
−4 − 4m +2m
2
−5m
◆
◆
EXAMPLE A.26 Solve for x:
4
x−2
+ 2 =
5x−6
x−2
.
SOLUTION Clear the denominator by multiplying both sides of the equation by (x − 2). This is fine
provided x = 2.
(x − 2)

4
x − 2
+ 2

=

5x − 6
x − 2


(x − 2)
4 + 2(x −2) = 5x − 6
4 + 2x −4 = 5x − 6
2x = 5x − 6
−3x =−6
x=2
Youmight be lulled into thinking that the solution to this equation is x = 2, but if you check
it by substituting x = 2 into the original equation you see that we have undefined terms, so
x = 2isnot a solution.
A.3 Solving Equations 1073
What happened here? When we multiplied both sides of the equation by x − 2 we made
the provision that x = 2 (otherwise we are multiplying both sides of an equation by zero).
When we come up with x = 2 we are left to conclude that there are no solutions to the
original equation.
5
◆
Quadratic Equations
Definition
An equation is quadratic in x if it can be put into the form ax
2
+ bx + c = 0,
where a and b are constants or expressions not containing x, and a = 0.
x
2
+ bx + c = a, (x + 1)(x + 2) = 3x
2
+ 5, and x
2
y
3

+ 7xy = x are all quadratic in x.
The methods that worked for solving linear equations fall apart with quadratics. For
example, suppose we want to solve x
2
=−2x−1. We can get all x’s on one side,
x
2
+ 2x =−1,
butwedon’thaveaconstant times x on the left-hand side. If we factor out at x, then
x(x + 2) =−1
x=
−1
x + 2
.
This is not solved. We’ve got x on both sides.
How many solutions do we expect for the equation ax
2
+ bx + c = 0? The graph of the
quadratic function f(x)=ax
2
+ bx + c is a parabola, opening upward if a>0,downward
if a<0;the location of the turning point [vertex] and the steepness are determined by a, b,
and c.
k
2
x
k
1
k
0

y
y = –1
y = 4
y = x
2
Figure A.2
5
Here’s another common way of understanding why there are no solutions to this equation. Getting a common denominator
on the left-hand side, we have
4+2x−4
x+2
,or
2x
x−2
. On the right-hand side, we have
5x−6
x−2
,or
2x
x−2
+
3x−6
x−2
,or
2x
x−2
+ 3
(x−2)
x−2
. The

original equation can be written
2x
x−2
=
2x
x−2
+ 3
(x−2)
(x−2)
. This equation can only hold if 3
(x−2)
(x−2)
= 0. But this is impossible; 3
(x−2)
(x−2)
,
if it is defined, must be equal to 3.
1074 APPENDIX A Algebra
We see that unlike the case of ax + b = k, the equation ax
2
+bx +c = k may have 0, 1,
or 2 solutions.
As a concrete example look at g(x) =x
2
. Solve g(x) =−1, g(x) =0, and g(x) =9.
(a) x
2
=−1. No solutions (x
2
is always nonnegative).

(b) x
2
= 0 ⇒ x = 0 . One solution.
(c) x
2
= 9 ⇒ x =±3. Two solutions.
Below is a sequence of examples leading to the quadratic formula.
6
◆
EXAMPLE A.27 Solve. (a) x
2
= 5 (b) x
2
+ 6x + 9 = 5 (c) x
2
+ 6x + 4 = 0
SOLUTION (a)
x
2
= 5
x =±
√
5
(b) Express the left-hand side as a perfect square.
x
2
+ 6x + 9 = (x + 3)
2
(x + 3)
2

= 5
x + 3 =±
√
5
x =−3±
√
5
(c)
x
2
+ 6x + 4 = 0
x
2
+ 6x =−4
This is actually equivalent to part (b), presented in a different form. We want to write
the left-hand side as a perfect square so that we can then solve for x by taking the square
root of both sides as in Examples A.27(a) and (b). In order to achieve this, we will add
an appropriate constant to both sides of the equation. This process is called completing
the square.
How can we write x
2
+ 6x +
as a perfect square? k below is a constant.
We’ll solve for k; this will help us to figure out what to add to both sides of the equation.
x
2
+ 6x +
= (x + k)
2
x

2
+6x +
= x
2
+ 2kx +k
2
So 2k must be equal to 6.
k = 3
(x + 3)
2
= x
2
+ 6x + 9
So add 9 to both sides of the equation.
x
2
+ 6x + 9 =−4+9
(x + 3)
2
= 5
x =−3±
√
5 ◆
6
Recall that
√
5 denotes, by definition, the positive square root of 5 (
√
5 =2.236 ). To denote the negative square root of
5, we must write −

√
5. A common error is to use
√
5 to denote both the positive and the negative square roots; this is not valid.
A.3 Solving Equations 1075
General Case: Deriving the Quadratic Formula Solve for x: ax
2
+bx +c =0
ax
2
+bx =−c
x
2
+
b
a
x =−
c
a
We want to write the left-hand side as a perfect square and are perfectly willing to add
the appropriate constant to both sides in order to accomplish this feat.
x
2
+
b
a
x +
= (x + k)
2
=x

2
+2kx +k
2
2k =
b
a
k =
b
2a

x +
b
2a

2
= x
2
+
b
a
x +
b
2
4a
2
.
So we must add
b
2
4a

2
to both sides of the equation.
x
2
+
b
a
x +
b
2
4a
2
=−
c
a
+
b
2
4a
2

x +
b
2a

2
=
b
2
− 4ac

4a
2
x +
b
2a
=±

b
2
− 4ac
4a
2
x +
b
2a
=±
√
b
2
− 4ac
2a
x =
−b
2a
±
√
b
2
− 4ac
2a

x =
−b ±
√
b
2
− 4ac
2a
We’ ve derived the quadratic formula by completing the square for a general quadratic
equation.
We’ ve shown that if ax
2
+bx +c =0, then
x =
−b ±
√
b
2
− 4ac
2a
If b
2
− 4ac > 0, we’ve got a positive and a negative square root so the equation has
two solutions.
If b
2
− 4ac =0, there’s one solution:

x =−
b
2a


.
1076 APPENDIX A Algebra
If b
2
− 4ac < 0, we have the square root of a negative number, so there are no real
solutions.
7
Using the Quadratic Formula to Solve Quadratics
Key step: Write the equation in the form ax
2
+bx +c =0.
a = coefficient of x
2
term
b =coefficient of x term
c =constant (i.e., any term or terms with no x’s)
Then use the quadratic formula.
◆
EXAMPLE A.28 Solve for x.
(a) x
2
+ 1 = 2x
2
− 2x (b) xy +x +3yx
2
=7y (c) (x − 5)
2
− 1 = 0
SOLUTION (a) x

2
+ 1 = 2x
2
− 2x.
x
2
+ 1 = 2x
2
− 2x
0 =x
2
− 2x − 1
Then a = 1, b =−2, and c =−1, so
x =
+2 ±
√
4 − 4(1)(−1)
2
=
2 ±
√
8
2
=
2 ±2
√
2
2
=
2(1 ±

√
2)
2
= 1 ±
√
2.
(b) Solve for x in terms of y. Treat y as a constant.
xy +x +3yx
2
=7y
3yx
2
+xy + x −7y =0
3yx
2
+(y + 1)x − 7y = 0
Then a = 3y, b =y + 1, and c =−7y,so
x=
−(y + 1) ±

(y + 1)
2
− 4 · 3y · (−7y)
2 ·3y
=
−y − 1 ±

y
2
+ 2y + 1 + 84y

2
6y
=
−y − 1 ±

85y
2
+ 2y + 1
6y
.
(c) Solve for x: (x − 5)
2
− 1 = 0.
7
The square of any real number, x
2
, is nonnegative.
A.3 Solving Equations 1077
Easier than using the quadratic formula is the following:
(x − 5)
2
= 1
x − 5 =±1
x=5±1=4or6.
Alternatively,
x
2
− 10x + 25 − 1 = 0
x
2

− 10x + 24 = 0
(x − 6)(x − 4) = 0.
So x = 6orx=4.
◆
Solving a Quadratic Equation by Factoring
Given ax
2
+ bx + c = 0, if the left-hand side factors, we can use the fact that
If A · B = 0, then A = 0 and/orB = 0.
8
◆
EXAMPLE A.29 Solve for x. (a) 3x
2
= x (b) x
2
− 5x =−4
SOLUTION (a) Note: We can’t divide by x because x could be zero.
3x
2
= x
3x
2
− x = 0
x(3x − 1) = 0
x = 0or3x=1
x=0orx=
1
3
Notice that we would have lost the solution x = 0 had we made the mistake of dividing
by x.

(b) Instead of using the quadratic formula we can factor.
x
2
− 5x + 4 = 0
(x − 4)(x − 1) = 0
x = 4orx=1.
◆
How to Solve a Quadratic Equation. If the equation is quadratic in x, we put the equation
in the form ax
2
+ bx + c = 0 and
8
Note: AB = 2 does not imply A = 2orB=2; for example, we could have
1
2
· 4 = 2or(−1)(−2) = 2. Only if the product
of numbers is zero can we draw a conclusion.
1078 APPENDIX A Algebra
1. use the quadratic formula,
x =
−b ±
√
b
2
− 4ac
2a
,or
2. factor into two linear factors (if possible). If A · B = 0, then A =0 and/or B = 0.
9
Higher Degree Equations

Higher degree equations are usually much harder to solve algebraically. There exist formulas
for solving cubic and fourth degree equations—but they are quite messy. Amazingly enough,
it can be proven that such a formula for equations of degree five does not exist.
10
If we are working with a higher order polynomial equation presented to us in the form
of a product of linear or quadratic factors, then we can use the approach that if A · B ·C = 0,
then A = 0orB=0orC=0.
◆
EXAMPLE A.30 Solve (x
6
− x
4
)(x
2
+ 2x + 1)(2x
2
− x − 1) = 0.
SOLUTION First factor completely.
(x
6
− x
4
)(x
2
+ 2x + 1)(2x
2
− x − 1) = 0
x
4
(x

2
− 1)(x + 1)
2
(2x + 1)(x − 1) =0
x
4
(x − 1)(x + 1)(x + 1)
2
(2x + 1)(x − 1) =0
x
4
(x − 1)
2
(x + 1)
3
(2x + 1) = 0
So
x
4
= 0
x = 0
x = 0
or
(x − 1)
2
= 0
x − 1 = 0
x = 1
or
2x + 1 =0

2x =−1
x=−
1
2
or
(x + 1)
3
= 0
x + 1 = 0
x =−1
◆
Suppose we are not handed our equation in factored form. Then the situation is more
difficult—but we can sometimes make progress.
◆
EXAMPLE A.31 Solve x
3
− 3x =−x+1.
SOLUTION x
3
− 2x + 1 = 0
There are at most three solutions to any cubic and at least one. Think about the graph of
y = x
3
− 2x + 1; for x large enough y is positive, and for x negative enough y is negative.
A cubic is continuous, therefore the graph must cut the x-axis at least once. On the other
hand, a third degree equation can have at most three roots. We could try some numbers in
an educated hunt-and-peck, hope-we-luck-out manner.
Let’s first look at positive x.Ifxis large enough, x
3
outweighs −2x and x

3
− 2x + 1
is positive. If x = 0, x
3
− 2x + 1 = 1, again positive. Perhaps if x is small enough, −2x
9
If the equation happens to be in the form of ()
2
=constant, you can short-cut all of this if you like.
10
This was proven by Niels Abel at age 21 in the early 1800s. A general result was proven by
´
Evariste Galois in 1833. Galois’
life was uncharacteristic of mathematicians: At the age of 21, he was killed in a duel over politics. He spent the night before the
duel (which he expected to lose, being no marksman) writing down all the mathematics he had thought of in his life, including
his method for determining whether a polynomial equation of degree n had a formula. These content packed scribblings laid the
foundations for what is now known as Galois theory.
A.3 Solving Equations 1079
can outweigh x
3
+ 1. Try x =1:
1
3
− 2(1) +1 = 1 − 2 + 1 =0.
Eureka!
11
We’ve been lucky. Now we’re inbusiness—either we can consider this graphically
to see if there are two more roots or we can use the following fact:
If P(a)=0, then (x − a) is a factor of P(x).
So x

3
− 2x + 1 =(x − 1) · q(x), where q(x) is some quadratic. We can find q(x) by long
division:
q(x) =
x
3
−2x +1
x − 1
= x
2
+ x − 1.
(See the section on long division of polynomials for this.)
So x
3
− 2x +1 = (x − 1)(x
2
+ x − 1). We must solve
x
3
− 2x +1 = 0
(x − 1)(x
2
+ x − 1) = 0
x = 1orx
2
+x−1=0
Use the quadratic formula.
x =
−1 ±
√

1 − 4(1)(−1)
2
=−
1±
√
5
2
x = 1, or x =
−1+
√
5
2
,orx=
−1−
√
5
2
.
An alternative initial attack: Given a graphing calculator, you can use that to help you.
A cubic looks like one of the figures below with the turning points possibly taken out. Using
a graphing calculator you can estimate the roots. You might even guess that one of the roots
is exactly 1; you couldn’t be sure until you tried it and saw that it actually worked exactly. As
for the other two roots—if you simply want numerical approximations, a calculator is great.
By zooming in you can get increasingly more accurate numerical estimates. Depending upon
the sophistication of your calculator, you may get exact answers. Try using your equation
solver to test this out.
◆
11
Legend has it that this phrase was coined by Archimedes of Syracuse, 287–212 b.c.e. While sitting in the bathtub, he realized
that the amount of weight lost by a floating body was precisely equal to the weight of the amount of water displaced, solving a

problem he had been given by the king. He was so excited he ran through the streets of Syracuse naked and dripping wet, shouting
“Euρ ´ηκα!” which means “I have found it!” We do not recommend such displays of exuberance.
1080 APPENDIX A Algebra
Disguised Quadratics
◆
EXAMPLE A.32 Solve for x in the following equations.
(a) x
2
(x
2
+ 3) = 4 (b) x
2
+
6
x
2
= 5
SOLUTION (a) Having the left-hand side factored when the right-hand side is not zero is not useful!
Multiply it out:
x
4
+ 3x
2
= 4.
Let u = x
2
. Then
u
2
+ 3u − 4 = 0

or equivalently, x
4
+ 3x
2
− 4 = 0.
(u + 4)(u −1) = 0
or (x
2
+ 4)(x
2
− 1) = 0. Therefore,
u =−4oru=1
x
2
=−4orx
2
=1
impossible x =±1
(b) Clear the denominator; multiply by x
2
on both sides. Make sure that x = 0 is not a
solution.
x
2
+
6
x
2
= 5
x

4
+ 6 = 5x
2
x
4
− 5x
2
+ 6 = 0
Let u = x
2
. Then
u
2
− 5u + 6 = 0
(u − 3)(u −2) = 0
or equivalently, (x
2
− 3)(x
2
− 2) =0.
u = 3or u=2
x
2
=3or x
2
=2
x=±
√
3orx=±
√

2 ◆
Equations with Radicals
The procedure for solving an equation with radicals (e.g., square roots) is to isolate and then
eliminate the radical. Consider the equation
√
x + 1 =x − 5. To eliminate the square root,
we square both sides of the equation. Then we’ll have a quadratic and can solve as in the
previous section.
◆
EXAMPLE A.33 Solve for x: x −
√
x + 1 − 5 =0.
SOLUTION
x −
√
x + 1 − 5 =0
−
√
x + 1 =−x+5
√
x + 1 = x − 5