SOLUTION: Dear Sir/Madam, Please help me with this equation: [(1)/(2z+i)]+[(1)/(2z-1)]=[(1)/(z+2i)] I know that I will be multiplying the equation by the least common denominator but how

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Dear Sir/Madam, Please help me with this equation: [(1)/(2z+i)]+[(1)/(2z-1)]=[(1)/(z+2i)] I know that I will be multiplying the equation by the least common denominator but how      Log On


   

Question 766269: Dear Sir/Madam,
Please help me with this equation: [(1)/(2z+i)]+[(1)/(2z-1)]=[(1)/(z+2i)] I know that I will be multiplying the equation by the least common denominator but how will I use the quadratic formula in arriving to the final answer? Thank you very much for your help and kindness.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me with this equation: [(1)/(2z+i)]+[(1)/(2z-1)]=[(1)/(z+2i)] I know that I will be multiplying the equation by the least common denominator but how will I use the quadratic formula in arriving to the final answer? Thank you very much for your help and kindness.
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[(1)/(2z+i)]+[(1)/(2z-1)]=[(1)/(z+2i)]
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Multiply thru by (2z+i)(2z-1)(z+2i) to get:
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(2z-10)(z+2i) + (2z+i)(z+2i) = (2z+i)(2z-1)
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2z^2 +4zi-10z-20i + 2z^2+4zi+zi-2 = 4z^2-2z+2zi-i
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4z^2 + 9zi -10z -20i -2 = 4z^2 -2z + 2zi -i
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7zi - 8z -19i -2 = 0
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z(7i-8) = 2+19i
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z = (2+19i)/(-8+7i)
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Cheers,
Stan H.