Assignment 30
"The deep study of nature is the most fruitful source of mathematical discovery."  (Joseph Fourier, 17681830)
Michael Stifel (Germany, 14861567). One of the great algebraists of the 16th century, Stifel was a number mystic. He prophesied the end of the world on Oct. 3, 1533. When this event failed to happen, he had to take refuge in a prison to escape the wrath of peasants who gave up all of their worldly goods to follow him to Heaven. Extremely antiCatholic, he produced a "proof" that Pope Leo X was the antichrist described in the Bible. (Revelation 13:18 ).
"This calls for wisdom: let him who has understanding reckon the number of the beast, for it is a human number, its number of six hundred and sixtysix."
Stifel certainly qualifies as one of the most eccentric personalities in the history of mathematics. He was a priest in his early years, and he developed an interest in number mysticism. He studied the Bible , and interpreted certain passages in terms of number associations with key words. His proof associating Pope Leo X with the number 666 is fascinating, to say the very least.

When Herkimer was a playboy, where did he record his daily activities?
Answer: In a looselife notebook.
Herky's friends:
CARA VAN...likes to lead multicar tours.
PATTY MELT.. she loves grilled sandwiches with cheese topping. 
ASSIGNMENT #30
Reading: Section 5.6, pages 291295.
Written: Pages 295297/4761 (odds), 79. 
Mathematical word analysis: INTEGER: From the Latin word integer (whole, complete, perfect). An integer is "whole" in the sense that it is not fractional or irrational. 
The real solutions for any quadratic equation ax^{2} + bx + c = 0 can be found (if they exist) using the quadratic formula. This formula is not necessarily the most efficient way to find solutions to the equation, but it will do the job, if used properly. The formula has the form x = (b plus/minus ÷(b^{2}4ac))/(2a).
The expression D = b^{2}4ac is called the discriminant of the quadratic equation. This tells you something about the nature of the roots of a quadratic equation. Understand this! It is important!
Here are three examples that illustrate the properties of the discriminant. (These are stated on page 189).
Example 1:
Solve x^{2}  6x + 9 = 0.
In this case, D = b^{2}  4ac = (6)^{2}  4(1)(9) = 36  36 = 0. This equation has only one real root. If we substitute into the quadratic formula, we get x =( (6) plus/minus ÷(0))/(2) = (6 plus/minus 0)/2 = 3. The only solution to the equation is x = 3. We really didn't need the quadratic formula. We might have noted that x^{2}6x+9  (x3)(x3) = 0 will be true only when x = 3. If you use your TI83 and graph Y1=x^2  6x + 9, you will find that the curve (a parabola) is tangent to the xaxis at x = 3. In other words, only x = 3 will produce a y value of 0.
Example 2:
Solve x^{2} + 2x  15 = 0.
In this case, D = b^{2}4ac = 2^{2}4(1)(15) = 4 + 60 = 64. Since D is positive, the equation will have two distinct real roots. Using the quadratic formula, we get x = (2 plus/minus ÷(64))/(2) = (2 plus/minus 8)/2 ==> x = (2 + 8)/2 or x = (2  8)/2 ==> x = 3 or x = 5. If you use your TI83 and graph Y1 = x^2 + 2x  15, you will find that the graph has xintercepts at x = 3 and at x = 5. This is where y = 0.
We could have originally written x^{2} + 2x  15 = (x  3)(x + 5) = 0 when x = 3 or x = 5. Since the original quadratic could be factored, we did not need the quadratic formula. However, many quadratics cannot easily be factored, and the quadratic formula is necessary.
Example 3:
If we attempt to solve x^{2} + 2x + 9 = 0, we note D = 2^{2}  4(1)(9) = 4  36 = 32. Since the discriminant is negative, the equation has no real roots. If you use your TI83 and graph Y1 = x^{2} + 2x + 9, you will note that the graph does not intersect the xaxis. In other words, no value of x exists that makes y = 0.
The quadratic equation will always provide information about the nature of the roots (solutions) of a quadratic equation ax^{2} + bx + c = 0. Make an attempt to appreciate this neat formula.
Problem: Solve 2x^{2 } 5x  8 = 0.
Solution (with communication):
D = (5)^{2}  4(2)(8) = 25 + 64 = 89.
The roots are
x = [(5) plus/minus ÷89]/[(2)(2)]
= [5 plus/minus ÷89]/4.
Simplifying, and expressing roots to two decimal places, we have x = 3.61 or x = 1.11. 
Problem: Solve x^{2} +x +1 = 0.
Solution (with communication):
D = 1^{2}  4(1)(1) = 3.
Since D < 0, the roots are complex numbers. They are
x = [1 plus/minus i÷(3)]/2. 
