quadratic inequalities solving quadratic inequalities

Quadratic Inequalities

Solving


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


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 }


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


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


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


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


Recall:

Basic Absolute Value Facts

x ≥ 0 for all real x

x = –x for all real x

Absolute Values in Inequalities


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


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


Example 4:


| 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


Example 4:

x


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 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 !



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)


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


y = x2 – 4 ≤ 0

x2 – 4 = 0

x = ± 4 = ± 2 (2, 0)(–2, 0)

y > 0y > 0

y < 0

x2 = 4So

WHY ?


Example 5 Revisited Test a point x in each interval

y

x

y = x2 – 4


(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 ]

● ●●



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



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 ?


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


Example 8 Solve:

Factoring the difference of squares

Factors must have the same sign


x2 – 5 ≥ 0

WHY ?

(x – 5 )(x + 5 ) ≥ 0

and

ORand

So,

(x – 5 ) ≥ 0 (x + 5 ) ≥ 0

(x – 5 ) ≤ 0 (x + 5 ) ≤ 0


Exercise Solve: Factor out 3 , rewrite as an

equivalent inequality

Factoring again:

There are several ways to solve


3x2 + 3x – 18 > 0

x2 + x – 6 > 0

(x – 3)(x + 2) > 0


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


3x2 + 3x – 18 > 0

Solution Set: { x x < –2 } { x x > 3 } = (– , –2) (3, )


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


x2 – 10 < –1

x2 – 9 = x2 – 32 < 0(x – 3)(x + 3) < 0


Exercise Solve:

3. Solve graphically

4. Solve by building tables


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


Think about it !

quadratic inequalities solving quadratic inequalities

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