What are the solutions of #x^2 - 2x + 9 = 0# in simplest #a + bi# form?

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Answer 1 · 2016-01-28T03:47:05

The roots are #-1/2-sqrt8/2i and -1/2+sqrt8/2i#

For the quadratic equation, #x^2-2x+9=0#, we can use the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#= (2+-sqrt((-2)^2-4*1*9))/(2*(-2)#

#= (2+-sqrt(-32))/-4#

#= (2+-sqrt(32)*sqrt(-1))/-4#

We represent #sqrt(-1)# as #i# and simplify #sqrt32#:

#= (2+-sqrt(4)*sqrt(8)*i)/-4#

#= (2+-2*sqrt(8)*i)/-4#

#=-1/2-sqrt8/2i and -1/2+sqrt8/2i#