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Equations and Inequalities
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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
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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.
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Example
x2 – 3x + 4 = 0
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Solution
x2 – 3x + 4 = 0
• (x – 3/2) 2 = -4 + 9/4 = 5/4
• x = 3/2 ± 5 /4
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Quadratic Formula
• For ax2 + bx + c = 0, we have
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
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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)
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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
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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
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Absolute Value
• |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
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Solve
• |2x – 7| = 5
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Solution
• |2x – 7| = 5
• (2x – 7 ) = 5 or (2x – 7) = -5
• x = 6 or x = 1
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Solve
• | x + 1| = 3x - 3
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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
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More Examples
• |x – 4| - 5 = 2
• |x + 6| + ½ = 0
• |5 – 6x| = 0
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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
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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
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Equations with Radicals
• Solve: 2 + 10 − 𝑥 = x
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• Solve: 2 + 10 − 𝑥 = -x
•
10 − 𝑥 = -x- 2, square both sides
• 10 – x = x2 + 4x + 4 or x2 + 5x – 6 = 0
• This has roots (-6, 1)
• x = - 2 + 16 = 1; doesn’t work
2 + 9 = 6 works
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Example
• Solve 2𝑥 + 3 −2 𝑥 − 2= 1
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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
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Chapter 2.3
Inequalities
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Inequalities
• 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)
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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, ∞)
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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
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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
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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
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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
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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
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Example
• |x – 5| < 3
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Solution
• |x – 5| < 3
• x – 5 < 3 or -x + 5 < 3
x < 8 or x > 2;
2<x<8
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Example
• |1 – t/5| > 3
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Solution
• |1 – t/5| > 3
• 1 – t/5 > 3 or 1-t/5 < -3
• So t < -10 or t > 20
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Examples
• |x – 4| > 4
• |x – 4| < 4
•
3(𝑥 −2)
4(𝑥 −1)
+
4
3
<2
• | 3(x+2)2 – 3x2| < 1/10
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• |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
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Chapter 2.4
More on Inequalities
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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
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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
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Example
• Find the regions where x3 – 2x2 – 3x > 0
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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
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Example
• x4 < 14x3 – 48x2
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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 – 48,42
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
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Quotients
•
𝑥+3
𝑥 −4
>0
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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
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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
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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
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