How do you solve #-4n ^ { 2} + 2n - 5= 0#?

1 Answer
George C.
Aug 9, 2018

#n = 1/4+-sqrt(19)/4i#

Explanation:

We can solve this by completing the square and using the imaginary unit #i#, which satisfies #i^2=-1#. First multiply by #-4# so we can do much of the arithmetic using integers...

#0 = -4(-4n^2+2n-5)#

#color(white)(0) = 16n^2-8n+1+19#

#color(white)(0) = (4n-1)^2+(sqrt(19))^2#

#color(white)(0) = (4n-1)^2-(sqrt(19)i)^2#

#color(white)(0) = ((4n-1)-sqrt(19)i)((4n-1)+sqrt(19)i)#

#color(white)(0) = (4n-1-sqrt(19)i)(4n-1+sqrt(19)i)#

Hence:

#4n = 1+-sqrt(19)i#

So:

#n = 1/4+-sqrt(19)/4i#

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