Solving Solving Inequalities Inequalities AlgebraicallyAlgebraicallySection P.6 – the last section of the chapter!!!

Solving Absolute Value InequalitiesSolving Absolute Value Inequalities

First, take a look at this graph (visually):

x

y y = |x|

a–a

a

|x| < a|x| < a |x| > a|x| > a|x| > a|x| > a

Solving Absolute Value InequalitiesSolving Absolute Value Inequalities

Let u be an algebraic expression in x and let a be a real numberwith a > 0.

1. If |u| < a, then u is in the interval (–a, a). That is,

||uu| < | < aa if and only if – if and only if –aa < < uu < < aa

2. If |u| > a, then u is in the interval (– ,–a) or (a, ). That is,

||uu| > | > aa if and only if if and only if uu < – < –aa or or uu > > aa

8 8

Guided PracticeGuided Practice

Solve |x – 4| < 8 for x. Write your answer in intervalnotation.

– – 4 < x < 124 < x < 12 (– 4, 12)(– 4, 12)

8 4 8x

Solution:Solution:

Solving Quadratic InequalitiesSolving Quadratic Inequalities

First, replace the inequality sign with an equal sign

Next, solve for the variable as with any otherquadratic equation

Finally, examine the graph to see which valuesof x satisfy the original inequality (often a visual…)

Guided PracticeGuided Practice

Solve x – x – 12 > 0 for x. Write your answer ininterval notation.

2

Now graph:2 12y x x 3,4x

For what values of x isthe graph above zero?

Solution:Solution:

Final Solution:Final Solution:

( , 3) (4, )

Guided PracticeGuided Practice

Solve x + 2x + 2 < 0 for x. Write your answer ininterval notation.

2

No Solution!!!No Solution!!!

Solve2 2 2 0x x Graph

2 2 2y x x For what values of x is this function below zero?

Nothing below zero!Nothing below zero!

Guided Practice:Guided Practice:

Solve x + 2x – 1 > 0 graphically.3 2

1.618, 1 0.618, Solution:Solution:

Guided Practice:Guided Practice:

Solve:

,

22 5x x

SolutionSolution::

How about a word problem?

Basic Projectile MotionBasic Projectile Motion(if an object is propelled, and then only

subject to the force of gravity)

If an object is launched vertically from a point s feet above theground with an initial velocity of v ft/sec, then the vertical positions (in feet) of the object t seconds after it is launched is:

0

0

s = –16t + v t + ss = –16t + v t + s2200 00

How about a word problem?

20 016s t v t s

A rugby ball is kicked straight up from a height of 3 feet withan initial velocity of 57 ft/sec.

216 57 3s t t When will the ball’s height above ground be 44 feet?

At At tt = 1 sec (on the way up) and = 1 sec (on the way up) andat at t t = 2.5625 sec (on the way down)= 2.5625 sec (on the way down)

216 57 3 44t t Solve:

How about another word problem?

20 016s t v t s

A rugby ball is kicked straight up from a height of 3 feet withan initial velocity of 57 ft/sec.

216 57 3s t t When will the ball’s height above ground be at least 44 feet?

At At tt between 1 and 2.5625 sec between 1 and 2.5625 sec[ 1 , 2.5625 ][ 1 , 2.5625 ]

…and another word problem?

20 016s t v t s

A projectile is launched straight up from ground level with aninitial velocity of 272 ft/sec.

216 272s t t 1. When will the projectile’s height above ground be 960 ft?

At At tt = 5 sec (on the way up) and = 5 sec (on the way up) andat at t t = 12 sec (on the way down)= 12 sec (on the way down)

216 272 960t t Solve:

Whiteboard Problem:Whiteboard Problem:

Solve 2x + 3x < 20 for x. Write your answer ininterval notation.

2

[ – 4 , 2.5 ][ – 4 , 2.5 ]Solution:Solution:

Whiteboard Problem:Whiteboard Problem:

Solve x – 4x + 1 > 0 for x. Write your answer ininterval notation.

2

( – , 0.268] U [ 3.732 , )( – , 0.268] U [ 3.732 , )88 88Solution:Solution:

Whiteboard problem:Whiteboard problem:

Solve :–3x > –2x – 62

1.120,1.786Solution:Solution:

Whiteboard problem:Whiteboard problem:

Solve: x + 2x – 4x + 2 < 02

, 3.365

3

Solution:Solution:

Homework: p. 59 1-29 oddHomework: p. 59 1-29 odd

Tomorrow = review dayTomorrow = review daythen Exam Chapter P (first exam!!)then Exam Chapter P (first exam!!)