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avatar+1681 

\(Express\ \frac{2+i}{4+i}\ in\ the\ form\ a+bi ,\\ where\ a\ and\ b\ are\ real\ numbers.\)

 

Express (2+i ) /(4+i) in the form a+bi

where a and b are real numbers.

 

 

How do I do this, do I use formulas?

 

I thought you had to just multiply I to the numerator and

to the denominator, but you still have a fraction with I in it...

So what do you do?

 

EDIT:

What I did:

\(\ \ \ \frac{2+i}{4+i} \\=\frac{(2+i)i}{(4+i)i}\ \\=\frac{2i-1}{4i-1} \\I\ just\ made\ it\ worse.... \\please\ help!\)

 

 

-\(tommarvoloriddle \)

 Jul 13, 2019
edited by tommarvoloriddle  Jul 13, 2019
edited by tommarvoloriddle  Jul 13, 2019
 #1
avatar+792 
+4

is 1 or i in \((2+???)/(4+i)\)

.
 Jul 13, 2019
edited by travisio  Jul 13, 2019
 #2
avatar+1681 
+1

i ...

 

whoops ill fix that

tommarvoloriddle  Jul 13, 2019
 #3
avatar+792 
+4

\((2+i)\ /\ (4+i)\ =\ (9/17)\ +\ (2/17)*i\)

.
 Jul 13, 2019
edited by travisio  Jul 13, 2019
edited by travisio  Jul 13, 2019
 #4
avatar+1681 
+1

so 2i/17?

 

+9/17?

 

how did you get that?

 

 

I dont really get it.

tommarvoloriddle  Jul 13, 2019
 #5
avatar+792 
+4

no \(11/17)\ i\)

because in the equation i cancled out to be multiplied later in the equation because \(I\ /\ I\ =\ 1i\)

 

Travisio

travisio  Jul 13, 2019
 #6
avatar+1681 
+1

so the answer is:

\(\frac{11i}{17}\)

 

I still dont get it.

tommarvoloriddle  Jul 13, 2019
 #7
avatar+792 
+5

no no no the I will be multiplied to the equation

 

\(\frac{11}{17}i\)

.
 Jul 13, 2019
 #8
avatar+19787 
0

Multiply the given by  (4 - i) / (4- i)

 

= 8 -2i +4i - i^2    /   16 -4i  + 4i - i^2

= 8 + 2i +1     /    16 + 1

9 + 2i  / (17)

 

9/17 + 2i/17

 Jul 13, 2019
 #9
avatar+7761 
+3

\(\dfrac{2+i}{4+i}\\ = \dfrac{(2+i)(4-i)}{(4+i)(4-i)}\\ = \dfrac{9 + 2i}{4^2 + 1^2}\\ = \dfrac{9}{17} + \dfrac{2}{17}i\)

.
 Jul 13, 2019
 #10
avatar+105462 
+3

2 + i

_____   multiply numerator and denominator by the conjugate of 4 + i   ⇒  4 - i

4 + i

 

2 + i      (4 - i)           8 + 4i  - 2i  - i^2           8 + 2i  -  (-1)         9 + 2i

____                =    ______________ =     ___________ =   _______  =

4 + i     (4 - i)              16   - i^2                    16  -  (-1)                17

 

 

 9           2

__  +    ___ i

17        17

 

 

cool cool cool

 Jul 13, 2019
 #11
avatar+2505 
+3

This is very simple! Don't be afraid of imaginary numbers guys! If you want to simplify in the form a+bi, like \(\frac{2i+1}{i+3}\), multiply it by the conjugate of the denominator where it is also 1.

 

So for \(\frac{2i+1}{i+3}\), you multiply it by \(\frac{i-3}{i-3}\), which is the same as 1. This will simplify and won't change the fraction as you are basically multiplying it by 1. Remember to ALWAYS multiply by the conjugate.

 

 

This is your method for problems like these

 Jul 13, 2019
edited by CalculatorUser  Jul 13, 2019
edited by CalculatorUser  Jul 13, 2019
edited by CalculatorUser  Jul 13, 2019
 #14
avatar+1681 
0

What's a conjugate again?

 
tommarvoloriddle  Jul 22, 2019
 #12
avatar
+2

For these types of problems, just multiply by the conjugate like some people said, and you will get a+bi form.

 Jul 13, 2019
 #13
avatar+1681 
+6

:O WOW THANK YOU ALL OF YOU

 

Everyone was very helpful.

 Jul 14, 2019

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