Question 65972: Hello! I'm having trouble with this equation because I don't know how to solve for "k". Here is the equation.
If 1+2i is a root of x^2-kx+6=0, find k.
Can you help me out?
Sincerely,
Leanne Spence
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! Hello! I'm having trouble with this equation
because I don't know how to solve for "k".
Here is the equation.
If 1+2i is a root of x^2-kx+6=0, find k.
Can you help me out?
Sincerely,
Leanne Spence
Substitute 1+2i for x and solve for k:
x² - kx + 6 = 0
(1+2i)² - k(1+2i) + 6 = 0
(1+2i)(1+2i) - k - 2ik + 6 = 0
1 + 2i + 2i + 4i² - k - 2ik + 6 = 0
7 + 4i + 4i² - k - 2ik = 0
Since i² = -1 replace i² by -1
7 + 4i + 4(-1) - k - 2ik = 0
7 + 4i - 4 - k - 2ik = 0
3 + 4i - k - 2ik = 0
Get all the k terms on the right
3 + 4i = k + 2ik
Factor k out on the right
3 + 4i = k(1 + 2i)
Divide both sides by 1 + 2i
3 + 4i
-------- = k
1 + 2i
To divide out the two complex numbers,
multiply top and bottom by conjugate
of bottom, 1 - 2i
(3 + 4i)(1 - 2i)
------------------ = k
(1 + 2i)(1 - 2i)
3 - 6i + 4i - 8i²
-------------------- = k
1 - 2i + 2i - 4i²
3 - 2i - 8i²
-------------- = k
1 - 4i²
Replace i² by -1
3 - 2i - 8(-1)
---------------- = k
1 - 4(-1)
3 - 2i + 8
------------ = k
1 + 4
11 - 2i
--------- = k
5
or in A + Bi form in fractions
k = (11/5) - (2/5)i
or in A + Bi form in decimals
k = 2.2 - 0.4i
Edwin
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