The solutions to a quadratic equation show up as the xintercepts of the corresponding quadratic function. The parabola \begin{align*}y=x^22x+3\end{align*}
What does this graph tell you about the solutions to \begin{align*}x^22x+3=0\end{align*}
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Guidance
When humans first created the concept of numbers, they only had the counting numbers (whole numbers) \begin{align*}\{1,2,3,...\}\end{align*}
\begin{align*}\sqrt{1}=i\end{align*}
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 \begin{align*}i\end{align*}

\begin{align*}\sqrt{4}=\sqrt{4\cdot 1}=\sqrt{4}\sqrt{1}=2i\end{align*}
−4−−−√=4⋅−1−−−−−√=4√−1−−−√=2i 
\begin{align*}\sqrt{5}=\sqrt{5\cdot 1}=\sqrt{5}\sqrt{1}=\sqrt{5}i\end{align*}
−5−−−√=5⋅−1−−−−−√=5√−1−−−√=5√i 
\begin{align*}\sqrt{16}=\sqrt{16\cdot 1}=\sqrt{16}\sqrt{1}=4i\end{align*}
−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 \begin{align*}i^2=1\end{align*}

\begin{align*}2i+13i+4\end{align*}
2i+1−3i+4 simplifies to \begin{align*}i+5\end{align*}−i+5 or \begin{align*}5i\end{align*}5−i 
\begin{align*}3i\cdot 2i\end{align*}
3i⋅2i can be multiplied to \begin{align*}6i^2=6(1)=6\end{align*}6i2=6(−1)=−6
Numbers that are a combination of imaginary and real numbers, such as \begin{align*}5i\end{align*}
Example A
Express \begin{align*}\sqrt{49}\end{align*}
Solution: \begin{align*}\sqrt{49}=\sqrt{49\cdot 1}=\sqrt{49}\sqrt{1}=7i\end{align*}
Example B
Express \begin{align*}\sqrt{40}\end{align*}
Solution: \begin{align*}\sqrt{40}=\sqrt{4\cdot 10\cdot 1}=\sqrt{4}\sqrt{10}\sqrt{1}=2\sqrt{10}i\end{align*}
Example C
Simplify the following expression: \begin{align*}(4+3i)+(65i)\end{align*}
Solution: Simplify by combining the real numbers with the real numbers and the imaginary numbers with the imaginary numbers:
\begin{align*}(4+3i)+(65i)=102i\end{align*}
Concept Problem Revisited
The parabola \begin{align*}y=x^22x+3\end{align*}
This means that the solutions to the equation \begin{align*}x^22x+3=0\end{align*} are not real numbers. The solutions are complex numbers. You can still find the solutions using the quadratic formula, but your result will be 2 complex number solutions.
Vocabulary
 Complex Number
 A complex number is a number in the form \begin{align*}a + bi\end{align*} where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are real numbers and \begin{align*}i^2=1\end{align*}.
 Imaginary Number
 An imaginary number is a number such that its square is a negative number.
 \begin{align*}\sqrt{16}\end{align*} is an imaginary number because its square is –16.
 \begin{align*}\sqrt{16}=i\sqrt{16} = 4i\end{align*}
Guided Practice
Simplify each of the following:
1. \begin{align*}(53i)(2+4i)\end{align*}
2. \begin{align*}3i(4i^25i+3)\end{align*}
3. \begin{align*}(7+2i)(3i)\end{align*}
4. \begin{align*}\sqrt{12}\end{align*}
Answers:
1. \begin{align*}(53i)(2+4i)=53i24i=37i\end{align*}
2.
\begin{align*}3i(4i^25i+3)&=3i(4\cdot 15i+3)\\ &=3i(45i+3)\\ &=3i(15i)\\ &=3i15i^2\\ &=3i15(1)\\ &=3i+15\end{align*}
3.
\begin{align*}(7+2i)(3i)=& 217i+6i2i^2 \\ &= 21i2i^2 \\ & =21i2({\color{red}1}) \\ & =21i{\color{red}+2} \\ & =23i\end{align*}
4. \begin{align*}\sqrt{12}=\sqrt{4\cdot 3\cdot 1}=2i\sqrt{3}\end{align*}
Practice
Express each as a simplified imaginary number.
 \begin{align*}\sqrt{300}\end{align*}
 \begin{align*}\sqrt{32}\end{align*}
 \begin{align*}4 \sqrt{18}\end{align*}
 \begin{align*}\sqrt{75}\end{align*}
 \begin{align*}\sqrt{98}\end{align*}
Simplify each of the following:
 \begin{align*}(8+5i)  (12+8i)\end{align*}
 \begin{align*}(7+3i)(45i)\end{align*}
 \begin{align*}(2+i)(4i)\end{align*}
 \begin{align*}3(5i4)  2(6i7)\end{align*}
 \begin{align*}5i(3i2i^2+4)\end{align*}
 \begin{align*}(3+4i) + (11+6i)\end{align*}
 \begin{align*}(5+2i)(15i)\end{align*}
 \begin{align*}(1+i)(1i)\end{align*}
 \begin{align*}2(6i3)  4(2i+6)\end{align*}
 \begin{align*}i^3\end{align*}
 \begin{align*}i^4\end{align*}
 \begin{align*}i^6\end{align*}
 \begin{align*}5i(3i2i^2+4)\end{align*}