Download Lecture Notes for Week 11

ZEROS OF QUADRATICS 5 FINDING ZEROS OF QUADRATICS: SUMMARY
Now we have THREE ways to find zeros of quadratic functions:
1) Use the Square Root of Both Sides Trick, if possible. If that is not possible then ...
2) Factor. If that fails, use . . .
3) The Big Enchilada! Remember, the Big Enchilada is just „CTS and solve for x‟ in one step!
ZEROS OF QUADRATICS 6 THE DISCRIMINATE
Cautionary note to myself: It is tricky to come up with functions "on the fly" to illustrate the
discriminate! So, I have prepared the following list to use during a lecture!!
f(x)
zeros
b2 - 4ac
(x-3)2 or x2 - 6x + 9
3,3 double*
0
2
2
(x-3) - 2 or x - 6x + 7
3+/- sqrt(2)
Not a perfect square => IRRational
2
2
(x-3) - 4 or x - 6x +5
5, 1 integer
Rational: b2 - 4ac is a perfect square
(2x-1)(x+1) or 2x2 + x -1
1/2, -1 Fraction Rational: b2 - 4ac is a perfect square
*
Note: Double roots => a horizontal shift only – no up/down
Give Discriminate Review Handout Sheet.
Bonus: Prove Sqrt(2) is irrational.
ZEROS OF QUADRATICS 7
ZEROS BACKWARDS:
The problem: Given two zeros, find the function!
Example Problems:
Find two quadratic functions, one that opens upward and one that opens downward, whose
graphs have the given x-intercepts. (There are many correct answers.)
5
−1 & 3; ±5; 0 & 10; 4 & 8; −3 & 0;  & 2; −2 & −2; −8 & −4; −4 & 5
2
Try This One:
Find f(x) = Ax2+Bx+C if the zeros are -2 & 5.
A=____; B=_______; C=_________.
Check to see if f(-2)=0 and f(5)=0.
Technique #1: f(x) = (x- -2)(x-5) = (x+2)(x-5)
Technique #2: x=-2 is a zero => x+2 is a factor; similarly x=5 is a zero => x-5 is a factor. So f(x)
= (x+2)(x-5). Note: This technique does not work so well when the zeros are more complex such
as 2+√3 or 2+3i.
ZEROS OF QUADRATICS 8
FRACTIONAL & IRRATIONAL ZEROS
Try These:
1) Given 2  3 are zeros, find f(x) = Ax2 + Bx + C . Answer: x2 - 4x +1
2) What is the difference between the zeros of f(x) = (x - ½)(x+1) = x2 + ½x - ½ and
g(x) = 2(x - ½)(x + 1) = (2x - 1)(x + 1) = 2x2 + x - 1?
Answer: The "c" in c▪f(x) does not affect zero location!
CHAPTER 12 COMPLEX NUMBERS
What happens if the discriminate is negative?
Use x2 + 1 = 0 & (x-5)2 + 4 = 0 to introduce complex numbers.
COMPLEX 1 COMPLEX ROOTS
Until we learn how to do complex arithmetic, use your TI calculator in „a+bi‟ mode to check your
answers and verify that the graph does not cross the x-axis!!
Notes about the TI and Complex Numbers:
1) The TI does not allow use of y(x) notation if x is complex!! I do not know why not, but you
must do it “by hand" ex. If y(x) = x2 -10x+29, the the TI will not compute y(5+2i), so you must
do the calculation on the calculator like this: Either store 5+2i ->A, then compute A2-10A+29 or
just compute (5+2i)2 -10(5+2i)+29.
2) The TI calculator yields “NonReal Ans” for √(-1) if it is in “Real” mode. The TI calculator
yields “i” for √(-1) if it is in “a+bi” mode.
Review discriminate handout. Expand to include discriminates for complex zeros.
Use f(x) = 2x2 - x + 1 to demo the QF and the Discriminate
COMPLEX 2 COMPLEX ARITHMETIC
+ - x Show Vector add, subtract
Note: Can make up and check problems on TI calculator. Must be in a+bi mode to get a complex
answer!
Find Complex Zeros Again, only this time check by hand to see that f(z)=0!!
Demo the complex number "a + bi", the "a" is the Real Part, and the "b" is the Imaginary part.
So, in 3 - 5i, a = 3 and b = -5, not -5i. Note: 3 = 3 + 0i and 5i = 0 + 5i
COMPLEX 3 A HANDLE ON COMPLEX NUMBERS
Our "handle" on Complex numbers is that they are zeros of these types of quadratics:
Try one: ex f(x) = (x-5) 2 + 7 = x2 -10x + 32. Now use the QF to find the zeros of f. Interesting!
The complex zeros are x = 5 ± sqrt(7)i wow!!!
Try This:
Given the 1 ± i are zeros, find f(x) = Ax2 + Bx + C
A = ____; B = ______; C = _____
COMPLEX 4
COMPLEX DIVIDE
COMPLEX 5 COMPLEX ZEROS COME IN CONJUGATE PAIRS
Conjugate Zeros Theorem: If a polynomial with real coefficients has a zero a+bi where a & b are
real numbers then a−bi is also a zero.
Irrational Conjugate Zeros Theorem: If a polynomial with rational coefficients has a zero a+b√c
where √c is irrational and a & b are rational, then a−b√c is also a zero.
Note: If the zero is of the form b√c then the conjugate theorem does not apply.
Example 1: f ( x)  x3  2 has one irrational zero,
2
3
2
2 , and two complex zeros 2 3  2 3 3 i .
1
3
Example 2: f ( x)  x  40 has one irrational zero, 2 5 , and two complex zeros 5  5
3
3
1
3
3i .
COMPLEX 6 POWERS OF COMPLEX NUMBERS
Try These:
i37 =
i38 =
i39 =
i40 =
COMPLEX 7 THOUGHT PROBLEMS INVOLVING COMPLEX NUMBERS
Some good problems:
1) Prove that the complex conjugate of the product of two complex numbers A+Bi and C+Di is
the product of their complex conjugates.
2) Prove that the complex conjugate of the sum of two complex numbers A+Bi and C+Di is the
sum of their complex conjugates.
3) Describe the error: 1 = √(−1▪−1) = √−1▪√−1 = i▪i = −1
T or F:
4) There is no complex number that is equal to its complex conjugate. (F)
5) −i√6 is a solution of x4 – x2 + 14 = 56 (T)
6) i44 + i150 − i74 − i109 + i61 = −1 (F)
COMPLEX 8 HOW COMPLEX NUMBERS ARE USED IN THE REAL WORLD TO SOLVE
“COMPLEX” (no pun intended) PROBLEMS IN UNIMAGINABLE (again, no pun intended!!)
WAYS!!
Example: Pick any four integers, say 5,3,7,2. Find two integers, A & B, such that:
(52 + 32)(72 + 22) = A2 + B2: i.e. A = ? and B = ?. Now find two more:
= C2 + D2 where C = ? and D = ?
Solution: Write 52+32 =(5+3i)(5−3i), etc.; Then re-associate and multiply the products & see
what happens! Note: this is an example of how complex numbers can be used to solve problems
that are tough to solve using only real numbers!!!
COMPLEX 9 COMPLEX NUMBER SUMMARY
The Key Points about Complex Numbers:
1) In the complex world know how to +, -, x, ÷
2) To divide, use the complex conjugate.
3) Note: (a+bi)(a-bi) = a2 + b2 which is easy to calculate and is a real number - wow!!
4) Every f(x) has zeros. A Real zero indicates that the graph of f(x) crosses the x-axis.
5) Complex zeros come in conjugate pairs and indicate that the graph of f(x) does not cross
the x-axis.
6) Irrational zeros come in „conjugate‟ pairs, which leads into a discussion about the type of
coefficients. i.e. If f(x) = Ax2+Bx+C, are A, B, C Rational, Irrational, or purely complex???
Answer: If f(x) = Ax2+Bx+C and the zeros of f always come in „conjugate‟ pairs, then A, B, C are
rational.
Example:
Find the zeroes of f(x) = 2x2 - x + 1 and check them on the TI84 using complex mode.
Now graph f(x).
Key point: Complex roots a ± bi => no x-intercepts, but they are still zeros, and they come
in pairs!
CHAPTER 13
HIGHER ORDER POLYNOMIALS
The two problems we will solve in this section are:
I. Given some zeros, find a function f(x) = Ax3+Bx2+Cx+D that has those zeros;
II. Given f(x) = Ax3+Bx2+Cx+D, find the zeros of f(x).
HIGHER ORDER 1 CREATING POLYNOMIALS FROM REAL & COMPLEX ZEROS
Given some zeros, find a function f(x) = Ax3+Bx2+Cx+D that has those zeros:
Examples:
Zeros −1, 2,−3 => f(x) = Ax3+Bx2+Cx+D where A=1, B=2, C=-5, D=-6; or f(x)=x3 + 2x2 - 5x -6.
Zeros 0, 4,−3 => f(x) = Ax3+Bx2+Cx+D where A=1 B=−2 C=1 D=−2; i.e f(x) = x3 − x2 − 12x.
Zeros 2, i, −i => f(x) = Ax3+Bx2+Cx+D where A=1 B=−2 C=1 D=−2; i.e f(x) = x3 − 2x2 + x – 2.
Zeros ½, 1, −2=> f(x) = Ax3+Bx2+Cx+D where A=2 B=−1 C=−5 D=2; i.e f(x) = 2x3 + x2 – 5x + 2.
Zeros −1, 0, 0 (twice) => f(x) = Ax3+Bx2+Cx+D where A=1, B=1, C=D=0; i.e. f(x)=x3 + x2.
Zeros 3, 1+i, 1−i => f(x) = Ax3+Bx2+Cx+D where A=1, B=−5, C=8, D=−6; i.e. f(x) = x3−5x2+8x–6.
Zeros 3, 1±i√5 => f(x) = Ax3+Bx2+Cx+D where A=1, B=−1, C=0, D=−18; i.e. f(x) = x3 − x2 – 18.
Zeros −2, 1±i√2 => f(x) = Ax3+Bx2+Cx+D where A=1, B=0, C=−1, D=6; i.e. f(x) = x3 – x + 6.
Zeros 1, (−1±i√3)/2 => f(x) = Ax3+Bx2+Cx+D where A=1, B=C=0, D=−6; i.e. f(x) = x3 – 1.
Zeros −3, 2±√5 => f(x) = Ax3+Bx2+Cx+D where A=1, B=−1,C=−13,D=−3; i.e. f(x) =x3−x2–13x–3.