Merced College; Don Power

INTERMEDIATE ALGEBRA - CH 8, LECTURE

8.1 Quadratic Equations: Factoring and Special Forms

Factoring -- review

Square root property

...with entire algebraic expressions: (3x + 2)² = 5, or = –3

Equations "of quadratic form"

Situation: The base appears with two exponents, one of which is twice the other.

That is, the equation contains some function of x and that same expression squared.

Ex: Exp of 4 and 2

1 and 1/2 (as in text's example x – 5*sqrt(x) + 6 = 0

2/3 and 1/3

Text's last example before Ex 4 requires grouping the first 2 terms

Substitution technique works if you don't just "see" the factorization

Technique: Let a single letter represent that entire expression:

.

If expression is (x+3) Let y = (x+3); so y² = (x+3)²

If expression is (x²) Let y = (x²); so y² = (x²)² = x⁴

If expression is sqrt(x) Let y = sqrt(x); so y² = [sqrt(x)]² = x

After solving for y, substitute back and solve for x

Test solutions: you can pick up extraneous solutions, especially for radicals, rationals.

Applications:

Surface area of a sphere is 4πr². Find r (or d) given the surface area

Compound interest: Part of the formula is used in #139, 140.

The whole formula is

8.2 Completing the Square

Square root principle: If a² = b, then a = ±sqrt(b)

a and b may be any algebraic expressions.

Ex: Solve (2x + 3)² = 13

You must be able to approximate solutions with a calculator: Check previous solution with calculator

How to solve a quadratic eqn by completing the square

1. Write equation in form ax² + bx = c. Notice: move the constant to the right.

2. If a¹1, divide through by a. The equation is now in the form x² + bx = c

3. Divide b by 2, square the result, and add to both sides.

4. The left side is now a perfect square; factor and solve by the square root principle.

Ex: Easy if a=1, b is even: Solve x² -8x + 3 = 0. After moving the 3 to the right, ...

Do it by the procedure above.

Faster (if you see it):

We know we'll get (x–4)² from the x² – 8x (take half of the –8), so write the (x–4)²

There's also a constant to manage. How?

Add 16 to both sides. Why? Expanding (x–4)² results in x² – 8x +16

Ex: Harder otherwise due to fractions: Solve 3x² + 2x - 7 = 0

Supplemental Material:

Alternate technique (especially nice in the case a¹1 or b is not even):

1. Add b²/(4a) to both sides (or add and subtract on the same side)

2. To get a perfect square, either

a. Multiply through by 4a (or just by a if b is even)-- [avoids fractions to last step], or

b. Divide through by "a" (the coefficient of x) -- [most direct solution for x], or

c. Factor out "a" (the coefficient of x) -- best for completing the square on several expressions.

3. Solve by the square root principle.

Ex: (same as above) Solve 3x² + 2x - 7 = 0 Take option of clearing fractions.

8.3 Quadratic Formula

Derivation: complete the square on ax² + bx +c = 0

Always solve by easier techniques if possible

Square root principle

Factoring

Completing the square if a=1 and b is even.

Important issue: simplifying the radical and the fraction

Ex (same as in last lesson): 3x² + 2x - 7 = 0 gives result (-2 ± sqrt(88)) / 6

Teaching note:

Any even b will result in the need to simplify.

You also need to simplify if a contains a perfect square factor and the root is a factor of b

Ex: 9x² - 3x = -7

Ex: 5x - 7 / (5x) = 1 Quadratic after clearing fractions: 25x² - 5x -7 = 0

Rational equations can result in quadratics:

Ex: 2/(x+2) - 4/(x-3) = 5 Sol: (3 ± sqrt329) / 10

Options for reducing (after simplifying radical):

Write as two separate fractions, and reduce each of them, or

Factor out the common factor from the numerator before reducing, or

Divide all three terms by their greatest common divisor

Discriminant - the radicand in the quadratic equation b²-4ac

If positive -- two real solutions

If positive and a perfect square -- solutions are rational (otherwise irrational)

(you can always solve these by factoring)

If negative -- two complex solutions

If zero, one real (rational) solution. (expression is a perfect square).

Ex: 2x²-4x-1=0: discriminant is 16+4*2=24. 2 real (irrational)

2x²-5x-3=0: discriminant is 25+4*6=49. Factorable; 2 real (rational) sol.

x²+2x+7=0: discriminant is 4-4*7 = -24. Two complex solutions

9x²-12x+4=0: discriminant is 144-4*36=0. Perfect square: one real solution.

Building equations from their solutions:

Use roots (solutions) to find factors, then use factors to build an equation (not unique unless ...)

Ex: roots/solutions t=5, t= -3, t=0

corresponding factors: t-5, t+3, t.

simplest equation: (t-5)(t+3)(t)=0, or t³-2t²-15t=0

Any multiple of this equation would also have these factors, therefore these solutions.

Ex: roots/solutions with fractions: x=2/3, x=-3/5 i.e. 3x = 2, 5x = -3

corresponding factors (x-2/3) and (x+3/5) or we can write (3x-2) and (5x+3)
equation: 15x²-x-6=0

Ex: roots/solutions t=5 (multiplicity 2), t= -3 (multiplicity 3), t=0 (multiplicity 4)

[lesson 8.1 mentions repeated solutions, which is where the "multiplicity" comes from]

corresponding factors: (t-5)², (t+3)³, t⁴

simplest equation: t⁴ (t-5)²(t+3)³=0

Note: Any multiple of this equation also has the same solutions
such as 2t⁴ (t-5)²(t+3)³=0

Supplemental Material

Finding all solutions of a cubic equation, where one solution is known:

Use root to find factor; either factor it out or divide it out; solve the resulting reduced equation.

Ex: Find all the solutions to x³+2x²+x+12 if x= -3 is one of the solutions.

Ex: t=2 is a sol. of t³=8t²-25t+26

Using graphs to find solutions (this works to find all real solutions of any equation)

Fact: the x-intercepts of the graph y=f(x) are the same as the solutions of f(x)=0

So if the graph crosses the x-axis at x=5, then

1. x=5 is a solution (or "root") of the equation f(x) = 0

2. x–5 is a factor of the polynomial f(x)

8.4 Graphs of Quadratic Functions (Parabolas)

Goal: get a sketch: exact precision is not required, but do give specific coordinates and scale.

Techniques:

1. Point-plotting

(Example: Solve x²-1=2x by algebra and by graphing)

Advantage: easy to understand

Disadvantages

There is no guarantee that you will find the most important points (vertex, intercepts)

Techniques

Substitution (standard)

Synthetic division by x gives a remainder of f(x), which is y

2. Key points technique

(Example: y=x²-5x+2: and show the calculation of y for x=4 using synthetic division)

a. vertex - and plot line of symmetry

Formula for x-coordinate: x=-b/(2a) Note: this is the Quadratic Formula without the ± term.

Calculate y by substitution or synthetic division

b. Does it open upward or downward? Leading Coefficient Test

c. y-intercept

General technique: Set x=0 and solve for y. Therefore, the y-intercept is the constant.

d. x-intercept(s):

May or may not exist, depending on whether the graph crosses the x-axis or not.

Technique: Set y=0 and solve.

Use factoring if possible.

If factoring is not possible: either

"Bracket" the roots: Calculate the points to the left and the right of the x-intercept. or

Calculate the x-intercepts by the quadratic formula or completing the square

e. additional points based on line of symmetry

f. If you don't have enough, get another point by point-plotting (and its "mate" by symmetry)

Additional examples: y = x² +3x + 3 (Imaginary roots); y = 3x² - 10x + 3 (harder vertex)

8.5 Applications

Ex 1: If Profit = Revenue - Cost, then divide by the number of vehicles to get:

Avg profit = avg selling price – avg dealer cost per car.

In this problem, the base number of vehicles is different for the sale price and the dealer cost.

Ex 2: Model: Area = L * W for a rectangle

Ex 3: Compound interest problem, where compounding is done once a year, and time is 2 years:

A = P (1+r)^t

In general: A = P (1+r/n)^nt

Ex 4: Facts: total cost of trip

cost per skier

number of skiers

Logic: Notice the word "per"

Ex 5: Right triangle sides: Use a² + b² = c²

We know a + b = 200, c = 150

Get the eqn down to one letter

Ex 6: Add rates, i.e. jobs per minute

1 / A + 1 / (A+12) = 1 / 8 In general, 1/A + 1/B = 1/total

Text uses t/A + t/B = 1

Ex 7: Height of a model rocket: Eqn of an object in free-fall, (without considering air resistance)

General eqn: h = –16t² + v₀t + h₀

v₀ and h₀ are the initial velocity and height (at time 0)

heights are in feet and times are in seconds

In this example, h = –16t² + 192t [so initial velocity is 192 ft/sec, initial height is 0]

Questions:

After how many seconds (so we solve for t) is the height 432 ft (so we substitute h = 432)?

After how many seconds (so we solve for t) does the rocket hit the ground (sub h = 0)?

[Not asked] What is the height (so we solve for h) after 5 seconds (sub t = 5)

What is the maximum height (so we need to solve for h) of the rocket?

Key: max or min values occur at the vertex -- On this problem, find h at the vertex

[Not asked] When does the rocket reach its maximum height? (so we solve for t at the vertex)

8.6 Quadratic and Rational Inequalities

Quadratic Inequalities

Example: Solve x²-x-6 > 0 (solution is developed below)

Get a zero on one side of the inequality and factor the resulting expression

For rational expressions, you may have to add or subt fractions first

Expression can only change signs when the factors of the expression are zero ("critical points")

Note: these factors could be in the numerator or the denominator

The solution will involve not only these critical points, but also the intervals in between.

Problem: We can't get the intervals directly. Ex: (x-3)(x+2)>0. Sol is not x>3, x>-2 (test x= -4)

We need a technique to count the number of positive or negative factors.

Technique: Use number line

Plot solutions to the equation (and points at which the expression is undefined)

Determine whether the ineq is true or false at each of these points: record

Solution includes the points that make the inequality true

Consider the sign of each factor on each interval; record with + or -

Test point technique

Decide on overall sign for each interval

Graphing technique:

Plot the intercepts (easy to get from the factors)

2nd degree equation gives a parabola y = f(x)

Does the parabola open up or down? (from the sign of the x² term)

Partition the x-axis into intervals.

Select the intervals for which the graph is above the x-axis (for > or ³), or

Select the intervals for which the graph is below the x-axis (for < or £)

Note whether the endpoints of the intervals are to be included.

Ex: Contrast the solution above with solution to (x-3)(x+2)£0

Rational inequalities:

Get 0 on one side of the inequality

Add/subtract fractions as necessary: you must be able to factor.

Do same kind of sign analysis as for quadratic/polynomial inequalities.

Example: , which becomes ; sol is (–2,7/2]

Contrast with use of < symbol and ≥ symbol

Example: , which becomes , sol [–5,3) U (3,–1]

Extension Example: with both a point and an interval in the sol set:

1/(2-x) £ x. Equiv to (1-x)²/(2-x) £ 0

Solution is x = 1 or x > 2.

Solving by graphing:

Look for zeros;

Also locate where (in terms of x) is the graph above or below the x-axis.

Why the previous example works the way it does: graph y = 1/(2-x)-x and view where y £ 0.

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