Complex Number Simplification

J

Jason76

Senior Member
Joined
Oct 19, 2012
Messages
1,180
Post Edited :!:

\(\displaystyle x = \dfrac{4 \pm \sqrt{-80}}{8}\)

\(\displaystyle x = \dfrac{4 \pm i\sqrt{80}}{8}\)

:confused: What happens right in here?

Answer: \(\displaystyle x = \dfrac{1}{2} + \dfrac{i\sqrt{5}}{2} \) OR \(\displaystyle x = \dfrac{1}{2} - \dfrac{i\sqrt{5}}{2}\)
 
Last edited:
Jason76 said:
\(\displaystyle x = \dfrac{4 \pm \sqrt{80}}{8}\)................(1)

:confused: What happens right in here?

Answer: \(\displaystyle x = \dfrac{1}{2} + \dfrac{i\sqrt{5}}{2} \) OR \(\displaystyle x = \dfrac{1}{2} - \dfrac{i\sqrt{5}}{2}\).......................(2)
Click to expand...

(2) does NOT follow from (1). - unless you had \(\displaystyle \sqrt{-80}\) in (1)
 
Jason76 said:
\(\displaystyle x = \dfrac{4 \pm \sqrt{-80}}{8}\).........................(1)

\(\displaystyle x = \dfrac{4 \pm i\sqrt{80}}{8}\)...............................(2)

:confused: What happens right in here?

Answer: \(\displaystyle x = \dfrac{1}{2} + \dfrac{i\sqrt{5}}{2} \) OR \(\displaystyle x = \dfrac{1}{2} - \dfrac{i\sqrt{5}}{2}\)........................(3)
Click to expand...

Which part is giving you trouble?

(1) to (2)

or

(2) to (3)
 
Jason76 said:
\(\displaystyle x = \dfrac{4 \pm \sqrt{-80}}{8}\)

\(\displaystyle x = \dfrac{4 \pm i\sqrt{80}}{8}\)

:confused: What happens right in here?

Answer: \(\displaystyle x = \dfrac{1}{2} + \dfrac{i\sqrt{5}}{2} \) OR \(\displaystyle x = \dfrac{1}{2} - \dfrac{i\sqrt{5}}{2}\)
Click to expand...
\(\displaystyle x = \dfrac{4 \pm \sqrt{-80}}{8} = \dfrac{4}{8} \pm \dfrac{\sqrt{-80}}{8} = \dfrac{1}{2} \pm \dfrac{\sqrt{16 * 5 * (- 1)}}{8} = \dfrac{1}{2} \pm \dfrac{\sqrt{16} * \sqrt{5} * \sqrt{-1}}{8}\)

Now what?
 
JeffM said:
\(\displaystyle x = \dfrac{4 \pm \sqrt{-80}}{8} = \dfrac{4}{8} \pm \dfrac{\sqrt{-80}}{8} = \dfrac{1}{2} \pm \dfrac{\sqrt{16 * 5 * (- 1)}}{8} = \dfrac{1}{2} \pm \dfrac{\sqrt{16} * \sqrt{5} * \sqrt{-1}}{8}\)

Now what?
Click to expand...

Wouldn't you leave the \(\displaystyle i\) outside the square root?
 
Last edited:
JeffM said:
\(\displaystyle x = \dfrac{4 \pm \sqrt{-80}}{8} = \dfrac{4}{8} \pm \dfrac{\sqrt{-80}}{8} = \dfrac{1}{2} \pm \dfrac{\sqrt{16 * 5 * (- 1)}}{8} = \dfrac{1}{2} \pm \dfrac{\sqrt{16} * \sqrt{5} * \sqrt{-1}}{8}\)

Now what?
Click to expand...
\(\displaystyle \sqrt{16}= 4\), \(\displaystyle \sqrt{-1}= i\). Does that clarify it?
 
The main difficulty for some students might be "not understanding perfect squares and their relation to square root simplification". One big rule is that you have to look for a pair of numbers, one being a perfect square root. The perfect square roots are: \(\displaystyle 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144\) etc.. If you don't know these, then your going to be slow at simplifying.
 
Last edited:
Jason76 said:
The main difficulty for some students might be "not understanding perfect squares and their relation to square root simplification". One big rule is that you have to look for a pair of numbers, one being a perfect square root. The perfect square > > roots < < are: \(\displaystyle 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144\) etc.. If you don't know these, then you[']r[e] going to be slow at simplifying.
Click to expand...
Jason76, they are called "perfect squares" (in the sense of integers). They are not called "perfect square roots." You left out 0. It is the smallest perfect square.
 
The main strategy for square root simplification (or sometimes complex number simplification) is first see if the number under the radical is a perfect square. If it's not, then see if it can be factored as a product of a perfect square and some number. To accomplish this, you need to see any perfect square divides evenly (no remainder) into the number under the square root.

So you just test out the perfect square numbers:

\(\displaystyle 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144\)

Of course, you should have those numbers memorized.
 
Last edited:
Jason76 said:
The main strategy for square root simplification (or sometimes complex number simplification) is first see if the number under the radical is a perfect square. If it's not, then see if it can be factored as a product of a perfect square and some number. To accomplish this, you need to see any perfect square divides evenly (no remainder) into the number under the square root.

So you just test out the perfect square numbers:

\(\displaystyle 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144\)

Of course, you should have those numbers memorized.
Click to expand...
In practice, it helps to memorize the smallest perfect squares, but if the number under the radical is an integer, you need not have memorized anything. If the integer is greater than 1, factor it into its prime factors. If it is less than - 1, first factor it into - 1 and a positive integer, and then factor the latter into its prime factors. Memorizing perfect squares is a short cut.
 
I memorized the multiplication table in elementary school, but I did not learn about prime factorizations until the 8th grade. :D
 
You must log in or register to reply here.
Top