SOLUTION: We are suppose to solve the quadratic equations in the complex number system using the quadratic formula. 3x^2-2x+7=0

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Question 668244: We are suppose to solve the quadratic equations in the complex number system using the quadratic formula.
3x^2-2x+7=0
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
3x^2-2x+7=0
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a = 3 ; b = -2 ; c = 7
------
x = [-b +-sqrt(b^2-4ac)]/(2a)
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x = [2 +- sqrt(4-4*3*7)]/(6)
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x = [2 +- sqrt(80)]/6
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x = [2 +- 4sqrt(5)]/6
----
x = (1/3) +- (2/3)sqrt(5)
=============================
Cheers,
Stan H.
=============

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The complex solutions always come in pairs
which look like +r%5B1%5D+=++a+%2B+b%2Ai+ and
+r%5B2%5D+=++a+-+b%2Ai+
I can say:
+%28+x+-+r%5B1%5D+%29%2A%28+x+-+r%5B2%5D+%29+=+0+
+x%5E2+-+%28+r%5B1%5D+%2B+r%5B2%5D+%29%2Ax+%2B+r%5B1%5D%2Ar%5B2%5D+=+0+
-----------------------------------
+r%5B1%5D+%2B+r%5B2%5D+=+a+%2B+bi+%2B+a+-+bi+
+r%5B1%5D+%2B+r%5B2%5D+=+2a+
and
+r%5B1%5D%2Ar%5B2%5D+=+%28+a+%2B+bI+%29%2A%28+a+-+b%2Ai+%29+
+r%5B1%5D%2Ar%5B2%5D+=+a%5E2+%2B+b%5E2+
------------------------
+3x%5E2+-+2x+%2B+7+=+0+
Divide both sides by +3+
+x%5E2+-+%282%2F3%29%2Ax+%2B+7%2F3+=+0+
-------------------------
+-2%2F3+=+-2a+
+a+=+1%2F3+
and
+7%2F3+=+a%5E2+%2B+b%5E2%0D%0A%7B%7B%7B+7%2F3+=+1%2F9+%2B+b%5E2+
+b%5E2+=+21%2F9+-+1%2F9+
+b%5E2+=+20%2F9+
+b+=+%282%2F3%29%2Asqrt%285%29+
The solutions are:
+x+=+1%2F3+%2B+%282%2F3%29%2Asqrt%285%29%2Ai+
+x+=+1%2F3+-+%282%2F3%29%2Asqrt%285%29%2Ai+
--------------------------
check:
go back to
+x%5E2+-+%28+r%5B1%5D+%2B+r%5B2%5D+%29%2Ax+%2B+r%5B1%5D%2Ar%5B2%5D+=+0+
+x%5E2+-+%28+2%2F3+%29%2Ax+%2B+1%2F9+-+%284%2F9%29%2A5%2Ai%5E2+=+0+
+x%5E2+-+%28+2%2F3+%29%2Ax+%2B+1%2F9+%2B+20%2F9+=+0+
+x%5E2+-+%28+2%2F3+%29%2Ax+%2B+21%2F9+=+0+
+x%5E2+-+%28+2%2F3+%29%2Ax+%2B+7%2F3+=+0+
Multiply both sides by +3+
+3x%5E2+-+2x+%2B+7+=+0+
OK