Let's review some factoring from the beginning algebra section.

You can use the distributive law to see that

3(4n + 5) = 3*4n + 3*5 = 12n + 15,

and you can use FOIL to see that (n - 3)(n - 5) = n^2 - 8n + 15.

But how can you start with the answer and find the factors?

It's like Jeopardy: I say, "n^2 - 8n + 15," and you respond (in the form of a question),

"What is (n - 3)(n - 5), Alex, uh, I mean Dan?".

 

Here's another way to look at the first problem:

12n + 15 = 3(4n) + 3(5) = 3(4n + 5) .

You go backwards, and this is called factoring out a common factor. Here's another:

10 h^2 + 15 hk = 5h * 2h + 5h * 3k = 5h (2h + 3k)

100 = (100)(1); 100 + 1 = 101
100 = (50)(2) ; 50 + 2 = 52
100 = (25)(4) ; 25 + 4 = 29
100 = (20)(5) ; 20 + 5 = 25 
100 = (10)(10); 10 + 10 = 20.

Solving Quadratic Equations [ top of page ] (5/98)

Recall a linear equation is one that looks like ax + b = cx + d, and our strategy was to get all x terms on the left, all constants on the right, then divide by the coefficient on x to solve.

A quadratic equation has an x^2 (x-squared) term; "quadrat" is Latin for square. This web site is being written on a Macintosh Quadra; the word is for the "040" processor chip in the computer.

The general quadratic equation looks like

a x^2 + b x + c = 0 , . . . . where a /= 0.

If we want to find the x or x's that work, we might guess and substitute and hope we get lucky, or we might try one of these three methods:

  • Factoring
  • Completing the Square
  • The Quadratic Formula
    • Hey, let's solve the same equation three times, using each method!
    Example (Factoring): When the left hand side factors, we can use the "zero product
    property" which says if a product is zero, one or more of the factors hasta be zero:
      x^2 - 8x + 15 = 0 . . . . . . . . .our given equation
      (x - 3)(x - 5) = 0 . . . . . . . . . factor the left side
      x - 3 = 0 . . or . . x - 5 = 0 . . the "zero product property"
      x = 3 . . or . . x = 5 . . . . . . . solving the two "little equations"
       
    [ intermed algebra | top of page ]

    Example (Completing the square):
    The idea is that: "stuff squared equals a number" is easy to solve, using square roots.
    I'll use -/(n) for square root of n, meaning the positive number whose square is n.

    For example, -/49 = 7 because 7^2 = 49.

    Also, -/2 is about 1.414, because 1.414^2 = 1.999396, which is close to 2. But you'll never hit -/2 exactly by squaring a fraction (or terminating decimal), the square root of 2 is an "irrational number", meaning its decimal equivalent goes on forever:

    -/2 = 1.41421356237309504880168872420969807856967187537695 . . .

    Now let's solve x^2 - 8x + 15 = 0 using the square root method. It will take four levels of trickiness to get there:

    Level 1: Solve for x in the equation . . x^2 = 49.

    Level 2: Solve for x in the equation . . x^2 = 17.

    Level 3: Solve for x in the equation . . (x + 6)^2 = 5.

    Level 4: Solve for x in the equation . . x^2 - 8x + 15 = 0.

    So we ended up with the same solutions as before. Whew!

    [ intermed algebra | top of page ]

    Example (The Quadratic Formula): There's a formula for solving a general quadratic
    equation like a x^2 + b x + c = 0 ; it's called the quadratic formula.
    It gives the possible values for x in terms of a, b, and c.

    You can derive it by completeing the square; the result is this: . . . [ top of page ]

    If . . a x^2 + b x + c = 0 . . (*) . then either
    x = [-b +(b^2 - 4 a c)] / (2 a) . . or
    x = [-b -(b^2 - 4 a c)] / (2 a)
    The quantity D = (b^2 - 4 a c) is called the discriminant of the quadratic equation (*) . .
    So x = [- b (+ or -) D] / (2 a) .
      @ If D < 0 the equation (*) has no real solutions, because we can't do a real square root
      of a negative number. However, there are two complex solutions.
       

     

    @ If D = 0 the equation has one real (rational) solution x = - b / (2 a) because adding
    or subtracting 0 has no effect.
     
    @ If D > 0 then the equation has two real solutions. @ If D is a perfect square (of a
    whole number or fraction) then there are two rational solutions for x. This is when
    factoring will work. Trust me.
      Now back to our original equation : x^2 - 8x + 15 = 0 . In the above notation,
      We have a = 1, b = -8, and c = 15. Using the quadratic formula gives us
      x = (- b +/-(b^2 - 4 a c)) / (2 a)
      . = (-(-8) +/-((-8)^2 - 4(1)(15))) / (2(1))
      . = (8 +/-(64 - 60)) / 2 = (8 +/- 4) / 2
      . = (8 + 2) / 2 or (8 - 2) / 2 ; so again x = 5 or x = 3. . . . <;-}
       
      But the power of the quadratic formula is its ability to solve equations that don't factor.
      Example: The Golden Ratio: phi = 1.618... is one of the roots of x^2 - x - 1 = 0 ;
      the roots are x = [1 +/-5] / 2 . This is closely related to the Fibonacci Sequence.

     

    Graphing parabolas (See also functions and graphs.) (5/99)

    How can we draw a picture of y = x^2 ? As with any graph, we put a buncha dots
    at (x, x^2), and notice the pattern: We can make a table and go from there.

    x

    x^2

    -2

    4

    -1

    1

    0

    0

    1

    1

    2

    4

    3

    9

    Functions and Graphs [ top of page ]

    What's a function? [ top of page | funcs ] (5/99)

    Ok, I'm going out on a limb here, but I'd say the concept of "function" is THE most important in mathematics. Any arguments from math teachers or anyone?

    A function is a way of dealing with an "input", applying some "rule" (the function), and then getting an "output".

    What's the catch? There can be at most one output for every input. The inputs that make "sense" form the domain of the function, and the answers or outputs form the range.

    Function notation: We can call the input x , the rule f , and then the output is f(x) .

    This DOES NOT mean "f times x" , it's just a notation device to record the input and output. For example, if f(x) = x^2, then f(3) = 3^2 = 9, not f times 3 (meaningless).

    Think of f(x) = x^2 as f( ) = ( )^2 ; that way you can safely plug in negative numbers or even other expressions: f(-5) = (-5)^2 = 25 ; f(x+h) = (x+h)^2.

    The graph of a function [ top of page | beg alg | funcs ] (11/99)

    Graphs are a visual way of displaying numerical information.

    The graph of a function y = f(x) is the set of all points (x, y) with y = f(x),

    so this means put a dot at (x, f(x)) for every x that makes sense (the domain).

    		 
    Here are a few functions and their graphs (plotted in Mathematica):
           

    y = 2 x

    y = x^2

    y = 2^x

    y = 2

     f(x) = 2 x

     f(x) = x^2

    f(x) = 2^x

    f(x) = 2

     (linear)

    (quadratic; power)

    (exponential)

    (constant)

    Moving graphs around [ top of page | funcs ] (11/99)

    If you have the graph of y = f(x) you can move it to the right c units
    by graphing y = f(x - c).
     
    For example, to move the parabola y = x^2 to the right by 3 units,
    you'd graph y = f(x - 3) or y = (x - 3)^2 .
     
    To move a graph up or down, you'd change the y-coordinates,
    so graphing y = x^2 + k puts the "vertex" at (0, k).
    In general do y = f(x) + k .
     
    By doing a combination of both, we can move the graph anywhere
    in the plane. . . y = f(x - c) + k or (x - c)^2 + k .
     
    My "danimation" shows both effects -->
    		 
     
    Click the picture for a 72K
    animation in a new window
    (graphs done in Mathematica)

    Four-way functions [ top of page | funcs ] (11/99)

    Part of the "reform" movement in precalculus and calculus is to look at functions four ways:

    (1) Verbally, (2) Numerically, (3) Symbolically, (4) Graphically.

    (1) "The squaring function" would describe what's being done to the input to get the output.

    (2) Make a table of values for the inputs and outputs:

    x

    x^2

    1.0

    1.00

    1.1

    1.21

    1.2

    1.44

    1.3

    1.69

    1.4

    1.96

    1.5

    2.25

    (3) Describe the output algebraically in terms of the input: f(x) = x^2
    where f(x) is the function notation described above. I like f( ) = ( )^2.

    (4) Draw a graph of y = f(x) on a pair of coordinate axes; here's y = x^2.

    Exponential and Logarithm Functions [ top of page | funcs ] (11/99)

    If you get a job in June that pays $1 the first day, $2 the second day, $3 the third day, etc, earning one more cent per day each day, you'd have a total for the month of
    1 + 2 + 3 + . . . + 30 = 30(30+1)/2 = $465. This daily salary grows linearly, and the running total (465 is a "triangular number") grows quadratically. (See above red graphs)
     
    But what if your salary doubled every day for a month? Think about it first.
    Ok, you'd have 1 + 2 + 4 + 8 + . . . + 2^29 = 2^30 - 1 = $1,073,741,823 (a billion bucks)
    (The 30th day is 2^29 because you start with 1 = 2^0)
     

    This is called exponential growth, and you can see its rapid rise just for three days in the graph. Some more realistic examples are compound interest, bacteria or population growth, and spread of disease or rumors in a large group.

    (The graph of y = 2^x appears at the right-->)

    		 
     
    Simple and Compound Interest [ top of page | funcs ] (8/00)
     
    Say you deposit some money, initially P dollars (or your favorite unit of currency), into a bank.
    The bank pays you "interest" to keep it there; it can pay either simple or compound interest.
     
    Simple interest is simple to figure out; hence the name!
    If you make, say, 8% interest for 5 years, that makes 40%.
    The total amount of money A that you'll have after t years at interest rate r is given by
     
    A = P (1 + r t)
     
    So if you put in $2000 for 5 years at 8% interest, you get
    A = 2000(1 + (.08)(5)) = 2000(1+.40) = 2000(1.40) = $2,800.
    Notice the factor of 1.40 means you've made a profit of 40%.
     
    A graph of your money A as a function of t would look like a straight line; you always make
    8% of just $2000, or $160, each year, forever. The amount A is a linear function of t.
     
    Compound interest is more comp-licated. The difference is that you make interest on the interest.
    After 1 year your $2000 would grow by 8%, so multiply by 1.08 to include the original deposit:
    A(1) = 2000 (1.08) = $2,160. But the second year you'd want to make interest on the $2160, not
    just on the $2000. So the second year balance should be 2160(1.08) = $2332.80. But this equals
    2000(1.08)(1.08) = 2000 (1.08)^2. After 5 years you'd have a total of:
    A(5) = 2000 (1.08)^5 = 2000 (1.4693280768) = $2938.65615 ~=~ $2,938.65 (the bank rounds down)
    Note the factor of 1.469... means your effective profit is now 46.9% (compared to 40% for simple interest.)
     
    The general formula (if interest is "compounded yearly" like this) for your balance at time t yrs is:
     
    A = P (1 + r)^t . . . (where the ^ means "to the power of")
     
    Since the variable t is in the exponent, this time A is an exponential function of t., and your balance
    would grow at an ever-increasing rate (dollars per year), but a constant percentage rate.
    The graph would be an exponential curve like the y = 2^x example above.
     
    If the interest is compounded n times per year (n=4 for quarterly, for example), then this
    means you get paid that fraction (1/n) of the interest n times per year.
    With your 8% annual rate, each quarter you'd gain 8%/4 = 2% of your money; so you multiply
    your balance by 1.02, four times a year.
    In t years this makes 4t quarters (3-month periods); after 5 years you'd have 20 quarters, and so: . .
    A(5) = 2000 (1.02)^20 = 2000 (1.485947396) ~=~ $2,971.89. This represents a 48.6% profit.
     
    The general formula (if interest is "compounded yearly" like this) for your balance at time t yrs is:
     
    A = P (1 + r/n)^(nt). . . (remember to do the r/n first!)
     
    It seems to be better when the interest is compounded more often! Let's make a table of how your
    final balance after 5 years depends on n, the number of compounding periods per year.
     

    n

    (1 + r/n)^(nt)

    =

    = growth factor

    A = P*(growth)

    1 (yearly)

    (1 + .08/1)^(1*5)

    1.08^5

    1.46933

    2938.65

    2 (semiann)

    (1 + .08/2)^(2*5)

    1.04^10

    1.48024

    2960.48

    4 (quarterly)

    (1 + .08/4)^(4*5)

    1.02^20

    1.48595

    2971.89

    12 (monthly)

    (1 + .08/12)^(12*5)

    1.006667^60

    1.46933

    2938.65

    52 (weekly)

    (1 + .08/52)^(52*5)

    1..001538^260

    1.49137

    2982.73

    365 (daily)

    (1 + .08/365)^(365*5)

    1..000219^1825

    1.49176

    2983.51
     
    Apparently the total grows, but not very much, as n --> infinity. In the limit, we say the
    interest is compounded continuously, and the amount is given by the easy 'pert formula':
     
    A = P e^(rt) ....where e = base of natural logs; e ~~ 2.71828...
     
    Going back one last time to our example, we just put in r = .08 and t = 5:
    A = 2000 e^(.08 * 5) = 2000 e^0.4 = 2000 (1.49182) = $2,983.65.
     
    This is the best we can do in 5 years with the given interest rate; you get a 49.18% profit;
    it's not much stronger than the weekly compounding (hehe).
     
     
    Logarithms [ top of page | funcs ] (11/99)
     
    What are logarithms? And why are they misunderstood? Here's an easy way to deal:
    LOGS ARE EXPONENTS. (I have a t-shirt with this slogan.)
     
    Example: You know 2^3 = 8 because 2^3 means 2 * 2 * 2. (3 factors.)
    What's the base? 2. What's the exponent? 3. What's the answer? 8.
     
    So we record the exponent, 3, that you need to put on a basem of 2 to get an answer of 8.
    Write this as "log, base 2, of 8, equals 3" because the exponent (the log) is 3.
    This is written as log2(8) = 3.
     
    I'm too cheap to put a separate graph for y = log2(x), so just flip the above y = 2^x graph
    across the line y = x to get the log graph; this is because they're inverse functions.
     
     
    That's it for now. Check back often for new stuff!
    Click below for other topics, or visit the ask dan page!
     
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