Solving Polynomial, and Exponential Equations and InequalitiesMary Dwyer Wolfe, Ph.D.Department of Mathematics and Computer ScienceMacon State CollegeMSP with Bibb County July 2010

Math III Standard MM3A3Students will solve a variety of equations and inequalities. a. Find real and complex roots of higher degree polynomial

equations using the factor theorem, remainder theorem, rational root theorem, and fundamental theorem of algebra, incorporating complex and radical conjugates.

b. Solve polynomial, exponential, and logarithmic equations analytically, graphically, and using appropriate technology.

c. Solve polynomial, exponential, and logarithmic inequalities analytically, graphically, and using appropriate technology. Represent solution sets of inequalities using interval notation.

d. Solve a variety of types of equations by appropriate means choosing among mental calculation, pencil and paper, or appropriate technology.

Focus of this Presentationb. Solve polynomial, exponential, and logarithmic

equations analytically, graphically, and using appropriate technology.

c. Solve polynomial, exponential, and logarithmic inequalities analytically, graphically, and using appropriate technology. Represent solution sets of inequalities using interval notation.

Polynomial Equations

Polynomial Equations•Solve

•Steps in the Analytic Solution Process▫Put the equation in standard form▫ Factor the polynomial into linear and quadratic

terms▫Use the zero-product property

3 24 8 2x x x

Polynomial EquationsSolve analytically (symbolically).3 24 8 2x x x

3 22 4 8 0x x x

2

2

2 4 2 0

4 2 0

2 2 2 0

x x x

x x

x x x

Standard form

Factor by grouping

Factor difference of squares

2 0 2 0x or x zero product property

2 2x or x

Polynomial EquationsSolve graphically and numerically.3 24 8 2x x x

3 22 4 8 0x x x Standard form

3 2( ) 2 4 8f x x x x graph Solve by finding x so that f(x)=0

Calculator Tutorial

Polynomial EquationsSolve graphically and numerically.3 24 8 2x x x

Polynomial EquationsSolve analytically (symbolically).3 24 8 2x x x

3 22 4 8 0x x x

2

2

2 4 2 0

4 2 0

x x x

x x

Standard form

Factor by grouping

2

2

4 0 2 0

4 2

2 2

x or x

x or x

x i or x

zero product property

So if we can factor into linear and quadratic factors, we can find the exact values of all real and complex roots.

Polynomial EquationsSolve graphically and numerically.3 24 8 2x x x

3 22 4 8 0x x x Standard form

3 2( ) 2 4 8f x x x x graph Solve by finding x so that f(x)=0

The calculator/graphing method can only find real roots.

Polynomial EquationsSolve graphically and numerically.3 24 8 2x x x

Polynomial Equations•How do we find analytic solutions when the

polynomial in standard form doesn't factor easily?

•Use the following:▫Rational Root Theorem▫ Factor Theorem▫Remainder Theorem▫-----and lots of trial and error to maybe factor the

beast!▫See activity p. 15: Link to Activity

•BUT – graphing can help conjecture rational roots along with the Rational Root Theorem!

Polynomial Equations• Real roots of polynomial equations can be approximated

using numerical methods on the TI83/84 calculator. (The same numerical methods that I had to learn to do by HAND in the '60's!)

• Solve 312 4 0

2x x

There is one REAL root. (Rational Root Theorem does not apply because the coefficients are not integer.)

Polynomial Equations•Solve

•Graphical solution:

312 4 0

2x x

…

…

2.649x

Polynomial EquationsSolve by Intersection of Graphs Method.

3 24 8 2x x x

Method steps:1.Enter left side of equation in Y1 and right side in Y2.2.Graph in a window where the intersection of the two functions is visible.3.Find the intersection point(s). The x-coordinates of these point(s) are the solutions.

Polynomial Inequalities

Polynomial Inequalities•Solve analytically:•Solution:

▫Put in standard form:▫Replace the > with an = and solve

▫Note that these solutions to the equation are not solutions to the inequality

3 24 8 2x x x

3 22 4 8 0x x x

2

2

2 4 2 0

4 2 0

2 2 2 0

2 2

x x x

x x

x x x

x or x

Polynomial Inequalities•Solve analytically:•Solution:

▫Put in standard form:▫Put the solutions (x= -2 or x= 2) to the equation on

a number line.

▫Pick a test point in each interval formed and determine the sign of the inequality

f(-4) = -24 f(0) = -8 f(3) = 25

3 24 8 2x x x

3 22 4 8 0x x x

Polynomial Inequalities▫Put in standard form:

3 22 4 8 0x x x

f(x) < 0 (negative) | f(x) < 0 (negative) | f(x) > 0 (positive)

We are looking for where f(x) > 0 (is positive)

The solution is x > 2 written in interval notation is 2,

Polynomial Inequalities3 22 4 8 0x x x

So you say that wasn't analytic enough, eh?

( 2)( 2)( 2) 0x x x

2 0 2 0x and x or 2 0 2 0x and x

But other cubics could have up to 8 combinations of 3 linear factors. This is an easy one!

Polynomial Inequalities▫Put in standard form:

3 22 4 8 0x x x

The solution is x > 2, or in interval notation:

We can get to the same conclusion using the TI83/84 calculator.Method:1.Put the polynomial equation in standard form2.Enter the left side as a function3.Find the zeroes4.Examine the graph to determine position or negative f(x) values

2,

Polynomial Inequalities▫Put in standard form:

3 22 4 8 0x x x Find the zeroes: x=-2 or x = 2

Examine the graph to see that the positive value occur when x > 2

This is verified in the Table.

2,

Polynomial Inequalities•Try this one: Solve•Solution:

3 26 3 10x x x

The solution set is , 1 2,5

Polynomial Inequality ApplicationThe Chamber of Commerce in River City plans to put on a 4th of July fireworks display. City regulations require that the fireworks at public gatherings explode higher than 800 feet from the ground. The mayor particularly wants to include the Freedom Starburst model, which is launched from the ground. Its height after t seconds is given by

h = 256t – 16t2

When should the Starburst explode in order to satisfy the safety regulations?

how many seconds to reach Height > 800 feet?256t – 16t2 > 800

Polynomial Inequality Application

how many seconds to reach Height > 800 feet?256t – 16t2 > 800

Polynomial Inequality Application

how many seconds to reach Height > 800 feet?256t – 16t2 > 800

4.3 11.7

The fireworks will be at a height of 800 feet or more between 4.3 and 11.7 seconds after being launched.

Exponential Equations

Exponential Equations•Analytic (Symbolic) Method

▫Solve for the exponential term and factor▫Take the log of both sides▫Use log of a power rule to get the variable out of

the exponent▫Solve the resulting equation

Exponential EquationsSolve 2x = 8 (Head problem!)

x = 3

Solve 2x = 7

log 2x= log 7x (log 2) = log 7x = (log 7)/(log 2) x ≈ 2.807

Since the solution is irrational, the best we get, even using symbolic methods, is a decimal approximation.

Exponential EquationsSolve 5(1.2)3x – 2 + 95 = 100 5(1.2)3x – 2 = 5

(1.2)3x – 2 = 5/5 log (1.2)3x – 2 = log (1) (3x – 2) log (1.2) = log (1)

log(1)3 2

log(1.2)x

03 2 0

log(1.2)x

3 2

2

3

x

x

Exponential Equations•An Application: Newton's Law of Cooling

The temperature, T, of an object after time t is modeled by

T(t) = T0 + Dat

where 0 < a < 1 and D is the initial temperature difference between the object and the room. T0 is the initial temperature of the environment.

Exponential Equations•An Application: Newton's Law of Cooling

Modeling Coffee Cooling:

A pot of coffee with temperature of 100°C is placed in a room with a temperature of 20°C. It takes one hour for the coffee to cool to 60°C.(a)Find the values of T0, D, and a for the formula

T(t) = T0 + Dat

(b) Find the temperature of the coffee after half and hour.(c) How long did it take for the coffee to reach 50°C?

Exponential EquationsA pot of coffee with temperature of 100°C is placed in a room with a temperature of 20°C. It takes one hour for the coffee to cool to 60°C.(a)Find the values of T0, D, and a for the formula

T(t) = T0 + Dat

T0 is the initial temperature of the room, so T0 = 20D is the initial temperature difference : D = 100 – 20 = 80

So far we have T(t) = 20 + 80at

The last sentence tells us that T(1) = 60So 60 = 20 + 80a1, so we can solve for a. 40 = 80a a = 0.5

so T(t) = 20 + 80(0.5)t

Exponential EquationsA pot of coffee with temperature of 100°C is placed in a room with a temperature of 20°C. It takes one hour for the coffee to cool to 60°C.(a) T(t) = 20 + 80(0.5)t

(b) Find the temperature of the coffee after half and hour.

That means find T(0.5) = 20 + 80(0.5)0.5 ≈ 76.6So after half an hour the temperature of the coffee is about 76.6°C.

Exponential EquationsA pot of coffee with temperature of 100°C is placed in a room with a temperature of 20°C. It takes one hour for the coffee to cool to 60°C.(a) T(t) = 20 + 80(0.5)t

(c)How long did it take for the coffee to reach 50°C?

Solve 50 = 20 + 80(0.5)t

Exponential EquationsSolve 50 = 20 + 80(0.5)t

50 20 80(0.5)

30 80(0.5)

30(0.5)

803(0.5)

8

t

t

t

t

3log log(0.5)83

log log(0.5)83

log8

log(0.5)

t

t

t

t ≈ 1.415

Note that in this case, we went to a LOT of work symbolically to arrive at only an approximate result because the solution is an ugly irrational number!

Exponential EquationsSolve by Intersection of Graphs Method 50 = 20 + 80(0.5)t

t ≈ 1.415

Returning to our problem, it will take approximately 1.415 hours for the coffee to cool to 50°C.

Exponential Inequality

Exponential Inequality• I have $10,000 to invest. The current interest

rate at my bank is 3.5% compounded daily. I want to leave the money in the account until it grows to somewhere between $20,000 and $30,000. How long must the money be left in the bank to grow to this range of amounts?

•So we must solve: 360

.03520,000 10,000 1 30,000

360

t

Exponential Inequality•So we must solve:

360.035

20,000 10,000 1 30,000360

t

Rounded to the nearest year it will take between 20 and 32 years to have an amount between $20,000 and $30,000.

Logarithmic Equations

Logarithmic Equations•Analytic solution method:

▫Rewrite equation as a single logarithmic statement

▫Translate to an exponential statement▫Solve the resulting equation

•Solve ln 4x = 1.5 e1.5 = 4x

1.5

1.1204

ex

Logarithmic Equations•Solve log (x + 1) + log (x – 1) = log 3

log (x + 1)(x – 1) = log 3 log (x + 1)(x – 1) - log 3 = 0 ( 1)( 1)

log 03

x x

0 ( 1)( 1)10

3

x x

( 1)( 1)1

3

x x

( 1)( 1) 3x x 2

2

1 3

4 0

( 2)( 2) 0

2 2

x

x

x x

x or x

x = -2 is extraneous, so the solution is x = 2

Logarithmic Equations• Solve log (x + 1) + log (x – 1) = log 3 Graphically

Pick a better window!

Logarithmic Equations•What about other log bases? Solve:

log2 x = 3

•Where is the log2 key or menu item????

Logarithmic Equations•What about other log bases as in Solve:

log2 x = 3

•Use the change of base formula: log

logloga

bb

a

2

loglog

log 2

xx

Logarithmic Equations•What about other log bases as in Solve:

log2 x = 3

•Use the change of base formula:

2

loglog

log 2

xx

x = 8

Logarithmic Equations•What about other log bases as in Solve:

log2 x = 3

•Use the change of base formula:

2

loglog

log 2

xx

x = 8

Note one problem with certain windows and log functions:

Press arrows to find a point on the first curve!

Logarithmic Equations•Try this one using the calculator method. Solve: log3(x + 24) – log3(x + 2) = 2

x = 0.75