SOLUTION: solve radical equation √(2x+3i)+x=2

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Question 1086727: solve radical equation √(2x+3i)+x=2
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x%2B3i%29%2Bx=2
Isolate the square root by subtracting x fom both sides:

sqrt%282x%2B3i%29=2-x

Square both sides:

2x%2B3i=4-4x%2Bx%5E2

Get 0 on one side:

-x%5E2%2B6x-4%2B3i=0

Multiply through by -1 to get the x2 term positive:

x%5E2-6x%2B4-3i=0

Use the quadratic formula with
a=1, b=-6 and c=4-3i

 x+=+%28-%28-6%29+%2B-+sqrt%28%28-6%29%5E2-4%281%29%284-3i%29%29%29%2F%282%281%29%29+ 

 x+=+%286+%2B-+sqrt%2836-4%284-3i%29%29%29%2F2+

 x+=+%286+%2B-+sqrt%2836-16%2B12i%29%29%2F2+

 x+=+%286+%2B-+sqrt%2820%2B12i%29%29%2F2+

 x+=+%286+%2B-+sqrt%284%285%2B3i%29%29%29%2F2+

 x+=+%286+%2B-+2sqrt%285%2B3i%29%29%2F2+

 x+=+%282%283+%2B-+sqrt%285%2B3i%29%29%29%2F2+

 x+=+%28cross%282%29%283+%2B-+sqrt%285%2B3i%29%29%29%2Fcross%282%29+

 x+=+3+%2B-+sqrt%285%2B3i%29+

Checking with a TI-84 calculator in a+bi MODE,

Only  x+=+3+-+sqrt%285%2B3i%29+ checks

Edwin