(by: nate epstine)
Discriminant= (b^2-4ac)- used to determine the number of REAL, UNIQUE solutions a quadratic equation has
If (b^2-4ac)>0, there are 2 distinct, real solutions
If (b^2-4ac)=0, there is 1 real solution (a double root)
If (b^2-4ac)<0, there are 0 real solutions
Example 1:
Find the Discriminant of each equation, then determine the number of real solutions for each equation
A) 2x^2+4x+1=0 B) 2x^2+4x+2=0 C) 2x^2+4x+3=0
4^2-4(2)(1) 4^2-4(2)(2) 4^2-4(2)(3)
=8 =0 =(-8)
Imaginary Number
Imaginary Number= i is defined to be the square root of -1
i^1 i^2 i^3 i^4
---------------------------
i -1 -i 1
Example 4-> Use the Quadratic Formula to solve
3x^2-7x+5=0
x=(7±square root of (-7^2-4(3)(5))/2(3)
7±square root of (-11)/6
QUIZ THURSDAY→ 5.4-5.6
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Addition to Nate's Notes:
Complex Numbers: Any number that can be written as a+bi
a - real part
b - imaginary part
Notes Continued:(by: Halina A. Wiktor)
Division: We don't really divide, we multiply by the Complex Conjugate. This is called factoring the denominator.
Complex Conjugate: The complex conjugate of a +bi is denoted a+bi, is a-bi. All you do to find the conjugate is change the operator between the real and imaginary parts.
Ex. 9: Find -2-3i -> -2+3i
Ex. 10: Simplify 2+5i/2-3i (Write your answer in standard form)
Step 1: multiply the numerator and the denominator by the complex conjugate of the denominator (2+5i)/(2-3i) (2+3i)/(2+3i)
Step 2: Foil and simplify 4+6i+10+15i^2/4+6+6i-9i^2 = -11+16i/13
Step 3: Write the answer in standard form: -11/13 + 16i/13
Graphing Complex Numbers: Complex numbers are graphed in the complex plane.
Complex Plane: The x-axis is the real axis and the y-axis is the imaginary axis.
Ex. 11: Graph 2-3 in the complex plane.
- Go over 1 on the real-axis and down 3 on the imaginary axis
- Put a dot there
- Draw a dotted line from the origin to that dot. (This is how i want complex numbers graphed)
Magnitude of a Complex Number: denoted by |a+bi|, means: find the distance the complex is from the origin. Absolute value, means distance from zero. To do this:
|a+bi| = (square root of) a^2 +b^2
Ex. 12: Evaluate |-2-3i| then sketch.
|-2-3i| = (square root of) (-2)^2 + (-3)^2 = (square root of) 4+9 = (square root of) 13
Ex. 13: Which value is greater |2+4i|or |1-5i|? Prove algebraically
|2+4i|= (square root of) 2^2+ 4^2 + (square root of) 4+16 = (square root of) 20
|1-5| = (square root of) 1+25 = (square root of) 26 -> greater
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