Quadratic Inequalities
Solving
Quadratic Inequalities
Quadratic Inequalities 27/9/2013
Inequalities Review General Inequalities
Consider: f(x) ≤ c
or f(x) ≥ c Questions to ask:
1. Is the inequality true for ALL x ?
2. Is the inequality true for NO x ?
3. Is the inequality true for SOME x but not others ?
… or f(x) < c … or f(x) > c
Quadratic Inequalities 37/9/2013
Inequalities Review General Inequalities
Questions with answers:
1. Is the inequality true for ALL x ?
Solution set: R
2. Is the inequality true for NO x ?
Solution set: { }
3. Is the inequality true for SOME x but not others ?
Solution set: { x f(x) ≤ c }
Quadratic Inequalities 47/9/2013
Examples
Example 1
Solve:
Find x such that
True for ALL x Solution Set :
y(x) = –2 ≤ 0
–2 ≤ 0
or (– , )
y
x
y(x) = –2
R
Quadratic Inequalities 57/9/2013
Examples
Example 2
Solve:
Find x such that
True for NO x Solution Set : { }
y(x) = –2 ≥ 0
y
x
y(x) = –2 –2 ≥ 0
or O
Quadratic Inequalities 67/9/2013
Example 3 Solve:
First find x such that
Examples
y = 3 – 2x ≥ 0
3 – 2x = 0
y
x
y = 3 – 2x
x
3 2 =
we have 2x > 3
No match!
32
x > 3 2
Then
For x
3 2 >
Subtracting 2x ,
0 > 3 – 2x … or 3 – 2x < 0
Quadratic Inequalities 77/9/2013
Example 3 Solve:
Examples
y = 3 – 2x ≥ 0
No match
y
x
y = 3 – 2x
32
x > 3 2
For x
3 2 >
(continued)
But for
we have 2x < 3
0 < 3 – 2x
x < 3 2
… or 3 – 2x > 0
Subtracting 2x ,
Match!
Solution Set:
x
3 2 <
=3 2 x | x { }≤ ( , ]– 3
2
Quadratic Inequalities 87/9/2013
Recall:
Basic Absolute Value Facts
x ≥ 0 for all real x
x = –x for all real x
Absolute Values in Inequalities
Quadratic Inequalities 97/9/2013
Basic Absolute Value Facts
If x < b then –b < x < b
If x > b > 0
9
x
Absolute Values in Inequalities
WHY ?
0 x b–b –x x –x –x
x
WHY ?
b0–b x x –x –x
then either x > b or x < -b
? ?
??–x
x| |
x
Quadratic Inequalities 107/9/2013
Example 4: Now where is x ?
Either x ≥ 2
Absolute Values & Quadratics
| x | ≥ 2
x2–2 x 0 x ] [
Solution set:
OR OR { x | x ≥ 2 }
{ x | x ≤ -2 } {x | x ≥ 2 }∪(- , -2] [2, )∪
So how does this connect to quadratics ?
OR x ≤ –2
Quadratic Inequalities 117/9/2013
Example 4:
Absolute Values & Quadratics
| x | ≥ 2 x2 xx 2–2 0
] [
So how does this connect to quadratics ?
If x ≤ –2 < 0 then x is negative, and x2 = x(x)
≥ x(–2) > 0
x ≤ –2 < 0 also implies= (–2)2 = 4(–2)x ≥ (–2)(–2)
Hence for x ≤ –2 ,
x2 ≥ (–2)x
WHY ?
(continued)
≥ (–2)(–2) = 4
Quadratic Inequalities 127/9/2013
Example 4:
x
Absolute Values & Quadratics
x2 xx 2–2 0
] [
Similarly, for x ≥ 2 , we have
x2 = x(x) ≥ 2x… and thus x2 ≥ 4
Hence x2 – 4 ≥ 0 if and only ifx ≥ 2 OR x ≤ –2
… that is, for | x | ≥ 2
2x ≥ 2(2) = 4and WHY ?
(continued)| x |
≥ 2
Quadratic Inequalities 137/9/2013
Quadratic Inequalities Example 5
Solve: First solve
If | x | > 2 ,
y = x2 – 4 ≤ 0
x2 – 4 = 0
y
x
x = ±
4 = ± 2
● (2, 0)(–2, 0)
y > 0y > 0
y = x2 – 4
Then x2 = 4 and thus
y = x2 – 4 > 0Either way
then x2 > 22 = 4 … or x2 > (– 2)2 = 4
No Match !
Quadratic Inequalities 147/9/2013
Quadratic Inequalities Example 5
Solve: If | x | > 2
y = x2 – 4 ≤ 0
y
x● (2, 0)(–2, 0)
y < 0
y = x2 – 4
y = x2 – 4 > 0
If | x | < 2
then –2 < x < 2 andx2 < 4 so that
Match !
x2 – 4 < 0
Including boundary points,Solution Set
{ x | –2 ≤ x ≤ 2 } = [ –2 , 2 ]
(continued)
Quadratic Inequalities 157/9/2013
Example 5 -- Revisited Find boundary points for
First solve:
Splits domain into three intervals
Test a point x in each interval
y
x
y = x2 – 4
Quadratic Inequalities
y = x2 – 4 ≤ 0
x2 – 4 = 0
x = ± 4 = ± 2 (2, 0)(–2, 0)
y > 0y > 0
y < 0
x2 = 4So
WHY ?
Quadratic Inequalities 167/9/2013
Example 5 Revisited Test a point x in each interval
y
x
y = x2 – 4
Quadratic Inequalities
(2, 0)(–2, 0)
Interval x Inequality True/ False(– , –2) –4 (–4)2 – 4 ≤ 0 False∞ (–2 , 2) 0 (0)2 – 4 ≤ 0 TRUE
( 2 , ) 4 (4)2 – 4 ≤ 0 False∞● ●
(continued)
(4, 0)(–4, 0)
Test boundary pointsx = –2 : (–2)2 – 4 ≤ 0
(2)2 – 4 ≤ 0 x = 2 : Solution set: { x │–2 ≤ x ≤ 2 } = [ –2, 2 ]
● ●●
Quadratic Inequalities 177/9/2013
Quadratic Inequalities Example 6
Solve:
Rewrite:
Find x such that
x2 – 2x + 1 ≥ 0
x2 + 4 ≥ 2x + 3
y
x
y = x2 – 2x + 1
y = x2 – 2x + 1
For x ≠ 1y = (x – 1)2 > 0
= (x – 1)2 = 0and thus x = 1
Quadratic Inequalities 187/9/2013
Quadratic Inequalities Example 6
Solve: x2 + 4 ≥ 2x + 3
y
x
y = x2 – 2x + 1
●
y > 0y > 0
1
For x ≠ 1y = (x – 1)2 > 0
Including the boundary
point Solution set: = ( , )–
R
Question: What if x2 + 4 < 2x + 3 ?
(continued)
What if x2 + 4 = 2x + 3 ?
Quadratic Inequalities 197/9/2013
Example 7 Solve:
Quadratic Inequalitiesy
x (4, –1)
y = –(x – 4)2 – 1 ≥ 0
y = –(x – 4)2 – 1 ≥ 0 Note that for any x,
–(x – 4)2 ≤ 0 Thus
So, NO (real) x satisfies< 0 –(x – 4)2 – 1 ≤ –1
Solution set: { } OR O
Question: y = –(x – 4)2 – 1 ≤ 0 ?What if
Quadratic Inequalities 207/9/2013
Example 8 Solve:
Factoring the difference of squares
Factors must have the same sign
Quadratic Inequalities
x2 – 5 ≥ 0
WHY ?
(x – 5 )(x + 5 ) ≥ 0
and
ORand
So,
(x – 5 ) ≥ 0 (x + 5 ) ≥ 0
(x – 5 ) ≤ 0 (x + 5 ) ≤ 0
Quadratic Inequalities 217/9/2013
Exercise Solve: Factor out 3 , rewrite as an
equivalent inequality
Factoring again:
There are several ways to solve
Quadratic Inequalities
3x2 + 3x – 18 > 0
x2 + x – 6 > 0
(x – 3)(x + 2) > 0
Quadratic Inequalities 227/9/2013
Exercise Solve: There are several ways to solve
1. Find boundary points and test adjacent intervals
2. Note signs of factors, solve in pairs
3. Solve graphically
4. Solve by building tables
Quadratic Inequalities
3x2 + 3x – 18 > 0
Solution Set: { x x < –2 } { x x > 3 } = (– , –2) (3, )
Quadratic Inequalities 237/9/2013
Exercise Solve: Add 1 and rewrite as an equivalent
inequality
There are several ways to solve1. Find boundary points and test
adjacent intervals
2. Note signs of factors and solve in pairs
Quadratic Inequalities
x2 – 10 < –1
x2 – 9 = x2 – 32 < 0(x – 3)(x + 3) < 0
Quadratic Inequalities 247/9/2013
Exercise Solve:
3. Solve graphically
4. Solve by building tables
Quadratic Inequalities
x2 – 10 < –1 (x – 3)(x + 3) < 0
Question: Can we use the square root property ?
Note: For x2 < 9 we have
x2 =
| x | = 3
and we know | x |< 3 means –3 < x < 3
< 9
Quadratic Inequalities 257/9/2013
Think about it !