The solutions to a quadratic equation show up as the xintercepts of the corresponding quadratic function. The parabola
What does this graph tell you about the solutions to
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Guidance
When humans first created the concept of numbers, they only had the counting numbers (whole numbers)
Imaginary numbers were not commonly accepted in mathematics until the 1700s, but since then they have become an important part of our number system and are especially important in physics. They are called imaginary because they cannot be found on a traditional number line of real numbers. The square root of any negative number can be written in terms of the imaginary number

−4−−−√=4⋅−1−−−−−√=4√−1−−−√=2i 
−5−−−√=5⋅−1−−−−−√=5√−1−−−√=5√i 
−16−−−−√=16⋅−1−−−−−−√=16−−√−1−−−√=4i
You can perform addition, subtraction, and multiplication with imaginary numbers just like regular numbers (you can also do division, but that is a bit more complicated and won't be considered here). When performing multiplication, remember that

2i+1−3i+4 simplifies to−i+5 or5−i 
3i⋅2i can be multiplied to6i2=6(−1)=−6
Numbers that are a combination of imaginary and real numbers, such as
Example A
Express
Solution:
Example B
Express
Solution:
Example C
Simplify the following expression:
Solution: Simplify by combining the real numbers with the real numbers and the imaginary numbers with the imaginary numbers:
Concept Problem Revisited
The parabola
This means that the solutions to the equation
Vocabulary
 Complex Number

A complex number is a number in the form
a+bi wherea andb are real numbers andi2=−1 .
 Imaginary Number
 An imaginary number is a number such that its square is a negative number.

−16−−−−√ is an imaginary number because its square is –16. 
−16−−−−√=i16−−√=4i
Guided Practice
Simplify each of the following:
1.
2.
3.
4.
Answers:
1.
2.
3.
4.
Practice
Express each as a simplified imaginary number.

−300−−−−√ 
−32−−−−√ 
4−18−−−−√ 
−75−−−−√ 
−98−−−−√
Simplify each of the following:

(8+5i)−(12+8i) 
(7+3i)(4−5i) 
(2+i)(4−i) 
3(5i−4)−2(6i−7) 
5i(3i−2i2+4) 
(3+4i)+(11+6i) 
(5+2i)(1−5i) 
(1+i)(1−i) 
2(6i−3)−4(2i+6) 
i3 
i4 
i6 
5i(3i−2i2+4)