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Chapter 2
Equations and Inequalities
1
Chapter 2.1
Quadratic Equations
2
Quadratic Equations
• General form: ax2 + bx + c = 0, a, b, c are real numbers, a ≠ 0
• General Solution methods:
– Factoring; relies on ab=0 means a = 0 or b = 0
– Completing the Square
– Quadratic Formula
3
Completing the Square
• Work to get the equation into the form:
(x – α)2 = β
Steps for an equation of the form ax2 + bx = c
1. Divide the equation by a
2. Find b/2, b2
3. (x – b/2) 2 = c + b2/4
4. Take the square root of both sides.
4
Example
x2 – 3x + 4 = 0
5
Solution
x2 – 3x + 4 = 0
• (x – 3/2) 2 = -4 + 9/4 = 5/4
• x = 3/2 ± 5 /4
6
Another Example
• 4x2 – 8x - 16 = 0
7
Solution
• 4x2 – 8x - 16 = 0
• x2 – 2x = 4
• (x – 1) 2 = 5
• x=1± 5
8
Quadratic Formula
• For ax2 + bx + c = 0, we have
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
9
Using the Quadratic Formula
• x2 = 5 – 3x
10
• x2 = 5 – 3x
• Rewrite: x2 + 3x - 5 = 0: a = 1, b = 3, c = 5
𝑥=
−𝑏± 𝑏2 −4𝑎𝑐
2𝑎
=
−3± 32 −20
2
= (-3 ± 29)/2
11
The Discriminant
• We call
b2
– 4ac (of
−𝑏± 𝑏2 −4𝑎𝑐
)
2𝑎
the discriminant
• It discriminates (determines) what kind of roots we have
– If the discriminant is > 0, we have two real roots;
if it is > 0 and a perfect square, we have two rational roots
– If the discriminant is < 0 we have two complex roots;
the roots will be complex conjugates
– If the discriminant = 0, we have one rational root
(sometimes called a repeated root)
12
Some Practice
Use the quadratic formula to determine the type of roots and find
the roots
• x2 – 6x – 2 = 0
• 16 s2 + 8s + 1 = 0
• (x – 5)(x + 3) = 1
• y2 + y + 1 = 0
13
Solutions
• x2 – 6x – 2 = 0; 3  3 2, two reals
• 16 s2 + 8s + 1 = 0; s= -1/4, one rational
• (x – 5)(x + 3) = 1;1  17; two reals
• y2 + y + 1 = 0; [-1  𝑖 3 ] / 2; two complex
14
Using Roots to Build Equations
• If we have the roots r and q then we have an equation
– r and q can be real or complex
– This holds for equations beyond quadratic equations
• If the x intercepts of a graph are x = 3 and x = -2, then
(x – 3) = 0 and (x + 2) = 0
so (x – 3) (x + 2) = x2 – x – 6 = 0; a quadratic
with roots 3 and - 2!
15
We can also use complex roots, radicals
• Given roots x = -2  i 5
• (x + 2 - i 5 )(x + 2 + i 5 ) = x2 + 4x +9 = 0
16
Finding Equations
Find a quadratic equation with roots as given
• 3, 11
• -4, -9
• 2 5
• 1/3[ 4  5 ]
17
Solutions
• 3, 11; x2 – 14x + 33 =0
• -4, -9; x2 +13x + 36 = 0
• 2  5; x2 – 4x + 1 = 0
• 1/3[ 4  5 ]; 9x2 – 214x + 11 = 0
18
Examples
• Show that (x – p)(x – q) = r2 p and q real and p  q
has two real roots
19
• (x – p)(x – q) = r2
• x2 – (p + q) x + pq – r2 = 0
• x = [p + q 
(𝑝 + 𝑞)2 −4𝑝𝑞 + 4 𝑟 2 ] / 2
• Examine quantity under the radical:
p2 + q2 + 2pq – 4pq + 4r2 = (p-q) 2 + 4r2 which, as the sum
of two squares is positive.
• The quantity under the radical is > 0, hence, the roots are real
20
Product and Sum of Roots
• If we have x2 + bx + c = 0, and roots r1 and r2, a and b
rational, then we have:
(r1)(r2) = c and r1 + r2 = - b
21
proof
−𝑏+ 𝑏 2 −4𝑎𝑐
• (r1)(r2) =
2𝑎
=
−𝑏− 𝑏 2 −4𝑎𝑐
2𝑎
=
𝑏 2 −(𝑏 2 −4𝑐)
4
4𝑐
=c
4
• r1 + r2 = (-2b)/2 = -b
22
Example
If we have x + 3x – 7/2 = 0 then
our two roots, (r1)(r2) = -7/2
and r1 + r2 = -3
Right now, you can use this to check roots
23
Example
• Find A and B such that the roots of x2+ Ax + B = 0 are A and B
24
Solution
• Find A and B such that the roots of x2+ Ax + B = 0 are A and B
• We need AB = B and A + B = - A
• The first equation gives A = 1, the second gives B = -2
• Check: x2+ x - 2 = 0 ;
subst x = 1; 1 + 1 – 2 = 0,
subst x = -2; 4 – 2 – 2 = 0
25
More Generally
• If you have a polynomial with rational coefficients:
xn + an-1xn1 + an-2xn-2 …..+ a1x + a0 = 0
• If we factor it as (x – r1)(x- r2)(x – r3)…..(x – rn) = 0, the
product of the roots r1 r2 r3 … rn = a0
26
Some more cool facts
If a polynomial has rational coefficients:
– If ( + i ) is a root, then so is ( - i ) [conjugate root thm]
– If a polynomial has an odd degree, than at least one
root is real [fundamental thm of algebra]
– A polynomial can be factored into terms of the form (x – a)
and (x2 + b2) (Note, (x2 + b2) factors to (a + ib)(a – ib) )
[corr. of fundamental thm of algebra]
27
Chapter 2.2
Other Kinds of Equations
28
Absolute Value (Review)
• |4| = | -4| = 4
• View absolute value as the “distance from zero”
• Solving requires breaking into two equations, one for each
branch
• BE CAREFUL: Check your results – sometimes one of the
roots (solutions) violates the absolute value principles
29
Solve
• |2x – 7| = 5
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Solution
• |2x – 7| = 5
• (2x – 7 ) = 5 or (2x – 7) = -5
• x = 6 or x = 1
31
Solve
• | x + 1| = 3x - 3
32
Solution
• | x + 1| = 3x - 3
• (x + 1) = 3x – 3; x = 2
• Or (x + 1) = 3 – 3x; x = ½; However, x = ½ does not work
since 3/2 – 3 < 0;
• The only solution is x = 2
33
More Examples
• |x – 4| - 5 = 2
• |x + 6| + ½ = 0
• |5 – 6x| = 0
34
Solutions
• |x – 4| - 5 = 2; | x – 4| = 8, (x – 4) = 8 or x = 12; (x – 4) = -8,
x = -4
• |x + 6| + ½ = 0; |x + 6| = -1/2 no solution
• |5 – 6x| = 0; 6x = 5, x = 5/6
35
Moving beyond quadratic equations
36
nth roots
• Solve (x – 2)4 = 8
37
• Solve (x – 2)4 = 8
4
• x–2= 8
38
Solve
• 3x4 = -64
39
• 3x4 = -64
• No solution; x4 must be > 0, if it is real
40
Solve
• 5x5 = -64
41
• 5x5 = -64
• x= -
5
64
5
42
Solving by Factoring
• x3 – 3x2 + 2x = 0
43
Solution
• x3 – 3x2 + 2x = 0
• x(x2 – 3x + 2) = 0 = x(x – 2)(x – 1)
• x = 1, x = 2, x = 0
44
Solve
• x4 – 3x2 + 2 = 0
45
• x4 – 3x2 + 2 = 0
• (x2 – 2)(x2 – 1) = 0
• x =  2, x =  1
46
Solve
• x4 – 3x2 + 1 = 0
47
Solution
• x4 – 3x2 + 1 = 0
• Doesn’t factor!
• Can solve for x2 then take roots
• Substitution: Let r = x
• r = (-3  9 − 4)/2 = (-3  5)/2
• x =  (−3  5)/2
48
More examples
• 6x -2 – x-1 – 2 = 0
49
Solution
• 6x -2 – x-1 – 2 = 0
• Two methods:
– Let x-1 = y and solve for y
– Multiply both sides of the equation by x2
50
BEWARE THE EXTRANEOUS SOLUTION
• If you square both sides of an equation (or any other even
power) you may introduce an incorrect (extraneous) solution
𝑥=x–2
squaring gives x = x2 – 4x + 4, or x2 – 5x + 4 = 0
factoring gives (x – 4)(x – 1), but x = 1 does not satisfy the
equation
if a2 = b2, a may not equal b
• Also, if you multiply both sides of an equation by an
expression containing the variable you may introduce an
additional solution
51
A bit on extraneous roots
• Suppose we have x = 2
When we square both sides we get x2 = 4
Solving x2 = 4 gives x = ± 2, only one of which is true in
the original equation
Squaring the equation has introduced another root
• x-1 = 5, or x = 6
Again, squaring gives x2 – 2x + 1 = 25, x2 – 2x - 24= 0
Roots are x = 8 and x = 6, but only x = 6 is true in the
original equation
Squaring the equation has introduced another root
52
Extraneous roots, continued
• Suppose we have x - 2 = 3, clearly x = 5
• Let’s multiply both sides by x
x2 – 2x = 3x, or x2 – 5x = 0
• The roots to this equation are x = 0 and x = 5
we introduced an additional root by multiplying by x
• The same could hold by multiplying both sides by any function
of x
• Always check for extra roots when squaring or
multiplying by a factor containing a variable
53
Recap
We have been working to solve (find the roots of) polynomials
• Quadratics:
– Taking the square root of both sides
– Factoring
– Completing the square
– Quadratic formula
• A quadratic equation has:
– Two real roots
– Two complex roots (complex conjugates)
– One repeated roots
The types of roots depend on the discriminant
54
Recap Continued
• In a quadratic equation, x2 + bx + c = 0, if b and c are rational,
the roots, r1 and r2:
– Have a product c: (r1)(r2) = c
– Have a sum –b: r1 + r2 = -b
• We expanded these techniques to solve equations of higher
degree, e.g., cubics and quartics:
– Taking roots of both sides
– Factoring
– Substitution to get quadratic equations
• Reminder: If we square both sides, or multiply by a variable,
we may introduce extraneous roots – check them.
55
Equations with Radicals
• Solve: 2 + 10 − 𝑥 = x
56
Solution
• Solve: 2 + 10 − 𝑥 = x
•
10 − 𝑥 = x - 2, square both sides
• 10 – x = x2 - 4x + 4 or x2 - 3x – 6 = 0
• x = [3 ± 9 + 24 ]/2 = [3± 33 ]/2
• x = [3 ± 33 ]/2; is not problem with either root for a negative
discriminant as 33 < 6
• Try in x2 - 3x – 6 = 0
x2 = [9 ± 6 33 + 33]/4 = [42 ± 6 33]/4
-3x = -[9 ± 3 33]/2 = -[18 ± 6 33]/4
• x2 - 3x – 6 =[42 -18]/4 – 6 = 24/4 – 6 = 0
57
In General
• If you have an equation with a radical, isolate the radical and
square both sides
• What if we have two radicals? Keep isolating and squaring
58
Example
• Solve 2𝑥 + 3 −2 𝑥 − 2= 1
59
Solution
•
2𝑥 + 3 −2 𝑥 − 2= 1
•
2𝑥 + 3 = 1 + 2 𝑥 − 2 or 2x + 3 = 1 + 4x – 8 + 4 𝑥 − 2
• -2x + 10 = 4 𝑥 − 2 which is –x + 5 = 2 𝑥 − 2
• Square both sides:
x2 – 10X + 25 = 4x – 8
(x – 3)(x – 11) = 0, x = 3, x = 11
only x = 3 works
60
Chapter 2.3
Inequalities
61
Inequalities (Review)
• Treat just like an equation to solve them: add the same thing
to both sides, multiply both sides by the same thing,
EXCEPT: when you multiply or divide by a negative number,
you must change the orientation of the inequality
• 2x – 3 < 5 gives 2x < 8 or x < 4
• However, -2x – 3 < 5 gives -2x < 8 or x > - 4
(You can do this without multiplying: -3 – 5 < 2x, through
addition, then x > - 4)
62
Solution Set Notation
• If the solution is t > 4, we write (4, ∞), sometimes t  (4, ∞)
If the solution is t ≥ 4, we write [4, ∞)
• Sometimes we get a solution like -1 < t < 4, which we write
as (-1, 4)
• If we have |t| > 4, we have the two sets (-∞, -4) or (4, ∞)
We can write the or as the union of the sets, (-∞, -4)  (4, ∞)
63
Compound Inequality
We have two or more inequalities we have to solve:
3 −𝑥
-½ <
<½
4
Solve them separately:
-1/2 <
3 −𝑥
4
3 −𝑥
<
4
1/2
2 > (3 – x)
(3 – x) > -2
x>1
5>x
Then combine to get the solution:
1<x<5
64
Pitfalls
When you complete both parts, you might find:
– There is no joint solution: x < 2 and x > 3 has no solution
– Only one part of the solution is relevant: x < 2 and x < 5,
leaving x < 2 as the solution
Just as with an equation, you can put your answer into the
inequality to check it
65
Examples
• 1 – 2(t + 3) – t < 1 – 2t
• (x – 1)/4 – (2x + 3) /5 < x
• -3 < 2x + 1 < 5
• 2/3 < (5 – 3t)/2 < ¾
• 9/10 < (3x – 1) / (-2) < 91/100
66
Solutions
• 1 – 2(t + 3) – t < 1 – 2t; 1 – 2t – 6 – t < 1 – 2t
-3t -5 < 1 – 2t, t > -6
• (x – 1)/4 – (2x + 3) /5 < x
5x – 5 – 8x – 12 < 20x , 23x > -17, x > -17/23
• -3 < 2x + 1 < 5
2x > -4, x > -2; 2x < 4, x < 2; -2 < x < 2
• 2/3 < (5 – 3t)/(-2) < ¾; 8 < -30 + 18t < 9
18t > 38 or t > 19/9, 18t < 39, t < 13/6, 19/9 < t < 13/6
• 9/10 < (3x – 1) / (-2) < 91/100
90< 50 – 150x < 91; -40/150 > x, -41/150 > x,
so we have -45/15 > x > -41/150
67
Inequalities with Absolute Values
• |x| < 1 becomes -1 < x < 1; the distance from x to zero is < 1
• |x| > 1 becomes x < -1 or x > 1; the distance is > 1
68
Example
• |x – 5| < 3
69
Solution
• |x – 5| < 3
• x – 5 < 3 or -x + 5 < 3
x < 8 or x > 2;
2<x<8
70
Example
• |1 – t/5| > 3
71
Solution
• |1 – t/5| > 3
• 1 – t/5 > 3 or 1-t/5 < -3
• So t < -10 or t > 20
72
Examples
• |x – 4| > 4
• |x – 4| < 4
•
3(𝑥 −2)
4(𝑥 −1)
+
4
3
<2
• | 3(x+2)2 – 3x2| < 1/10
73
• |x – 4| > 4; x – 4 > 4 or x -4 < -4, x > 8 or x < 0
• |x – 4| < 4; x – 4 < 4 and x – 4>-4, x < 8 and x > 0; 0 < x < 8
•
3(𝑥 −2)
4(𝑥 −1)
+
4
3
< 2; |9x – 18 + 16x – 16| < 24
|25x – 34| < 24, or 25x – 34 > 24 and 25x – 34 > -24
x < 58/25, x > 2/5, 2/5 < x < 58/25
• | 3(x+2)2 – 3x2| < 1/10 this is |(x + 2) 2 – x2| < 1/30
|4x + 4| < 1/30, 4x + 4 < 1/30 and 4x + 4 > -1/30
-121/120 < x < -119/120
74
Chapter 2.4
More on Inequalities
75
Key Numbers for Graphics
• Graph y = x2 – 4x + 3
• The graph has 3 regions: x < 1, 1 < x < 3, and x > 3;
at these points y changes sign
76
We can use this to solve inequalities!
• y > x2 – 4x + 3 or y < x2 – 4x + 3
• Clearly, from the graph, the first holds for x < 1 and x > 3,
the second holds for 1 < x < 3
• However, we can solve too!
• x2 – 4x + 3 = (x – 3)(x – 1)
For x < 1, both terms are < 0, so y > 0
For x > 3, both terms are >0, so y > 0
For x between 1 and 3, one is positive and one negative,
so y < 0
77
Example
• Find the regions where x3 – 2x2 – 3x > 0
78
Solution
• x3 – 2x2 – 3x > 0
• Factor: x (x – 3)(x + 1)
Roots are x = 0, x = 3, x = -1. Put in order: -1, 0, 3
•
For x < -1, all terms are negative, so the expression is negative
•
For -1<x<0, two terms are negative, so the expression is positive
•
For 0 < x < 3, two terms are positive, so the expression is negative
•
For x > 3, all terms are positive, so the expression is positive
79
Example
• x4 < 14x3 – 48x2
80
Solution
• x4 < 14x3 – 48x2
• Divide by x; we can do that unless x = 0. However, zero is
also a critical point;
• x2 < 14x – 48
• Rewrite: x2 – 14x + 48 = 0, (x – 6)(x – 8) = 0,
x = 6, x = 8, x = 0 are the critical or key points
•
Put numbers into the inequality to check regions:
For 0< x < 6: Let x = 1. 1 ? 14 – 48, clearly 1 > 14-48, so the inequality
doesn’t hold for x < 6
For x > 8: Let x = 10. 100 ? 140 – 48, clearly 100 > 140-48, so the inequality
doesn’t hold for x > 8
Try x = 7: 49 ? 98 – 48 = 50. 49 < 50, so holds for 6 < x < 8
•
We also need to check x < 0, since 0 is also a root. Use – 1, - 1 ? -14 – 48,81
Does not holds for x < 0
There aren’t always key numbers
• x2 – 4x + 5 > 0
• If we make it an equation, it has no real solutions. But, if it has
no real solutions, it doesn’t intersect the x axis and can’t
change sign.
• Therefore, it is always > 0 or < 0
• Substituted x = 0 shows that it is always positive, or always
true
82
Quotients
•
𝑥+3
𝑥 −4
>0
83
Solution
•
𝑥+3
𝑥 −4
>0
• Key numbers are -3 and 4 (and x cannot be 4!)
• -4 < -3, so try it. We get -1 / -8 which is > 0
• 0 is between -3 and r, 3/-4 < 0, so doesn’t hold
• 5 > 4, and 8/1 > 0, so holds
• Equation holds for x < -3 or x > 4
84
What if we have ≥ instead of >
• y ≥ x2 – 4x + 3
• x3 – 2x2 – 3x ≥ 0
• x2 – 4x + 5 ≥ 0
•
𝑥+3
𝑥 −4
≥0
• Still try key points
85
Solution
• y ≥ x2 – 4x + 3; points are 1 and 3, 1 works, 3 too
• x3 – 2x2 – 3x ≥ 0; points are 0, -1, 3; all solve equation
• x2 – 4x + 5 ≥ 0; no real number solves the equation, but the
inequality holds for all real numbers
•
𝑥+3
𝑥 −4
≥ 0; can’t have x = 4, but x = -3 solves
86
What you need to know
• Working with Complex Numbers
• Solving equations using substitution, factoring quadratic
formula
• Solving equations with radicals
• Checking for extraneous solutions
• Solving inequalities (review)
• Key Numbers
87