Download 2.7 - San Juan College

SECTION 2.7
COMPLEX ZEROS OF A
QUADRATIC FUNCTION
SQUARE ROOTS OF
NEGATIVE NUMBERS
4
Is a value we have dealt with
up to now by simply saying
that it is not a real number.
And, up to now, we have dealt with the
following equation by simply saying it
has no solution:
x2 + 4 = 0
DEFINITION OF i
i  -1
i2 = - 1
The number i is called an imaginary
number. Imaginary numbers, along with
the real numbers, make up a set of
numbers known as the complex
numbers.
COMPLEX NUMBERS
Imaginary
i
2i
- 3i
4 + 5i
Real
5
2/3i
-7 + i
1/2 + 3/4i
-1
1/2

.7
3
COMPLEX NUMBERS
All numbers are complex and should be
thought of in the form:
a + bi
Real Part
Imaginary Part
COMPLEX NUMBERS
a + bi
Real Part
Imaginary Part
When b = 0, the number is a real
number. Otherwise, the number is
imaginary.
OPERATING ON COMPLEX
NUMBERS
Addition:
Example:
(3 + 5i) + ( - 2 + 3i)
Subtraction:
Example:
(6 + 4i) - ( 3 + 6i)
OPERATING ON COMPLEX
NUMBERS
Multiplication:
Example:
(5 + 3i) • (2 + 7i)
Example:
(3 + 4i) • ( 3 - 4i)
CONJUGATES
2 + 3i = 2 - 3i
Multiplying a complex number by its
conjugate always yields a nonnegative
real number.
THEOREM: If z = a + bi
z z = a 2 + b2
Writing the reciprocal of a complex
number in standard form.
Example:
1
3  4i
Writing the quotient of complex
numbers in standard form.
Example:
1  4i
5 - 12i
Writing the quotient of complex
numbers in standard form.
Example:
2 - 3i
4 - 3i
POWERS OF i
i1 = i
i2 = - 1
i3 = - i
i4 = 1
i5 = i
and so on
QUADRATIC EQUATIONS WITH A
NEGATIVE DISCRIMINANT
Quadratic equations with a negative
discriminant have no real solution. But,
if we extend our number system to the
complex numbers, quadratic equations
will always have solutions because we
will then be including imaginary
numbers.
EXAMPLE
1  i
 4  2i
 8  2i 2 or 2 2 i
EXAMPLE
Solve the following equations in the
complex number system:
x2 = 4
x2 = - 9
WARNING!
- 12 
-3 
36
EXAMPLE
Solve the following equation
in the complex number
system:
x2 - 4x + 8 = 0
DISCRIMINANT
If b2 - 4ac > 0 Two unequal real sol’ns
If b2 - 4ac = 0 One double real root
If b2 - 4ac < 0 Two imaginary solutions
EXAMPLE:
Without solving, determine the
character of the solution of each
equation in the complex number
system:
3x2 + 4x + 5 = 0
2x2 + 4x + 1 = 0
9x2 - 6x + 1 = 0
CONCLUSION OF SECTION 2.7