How do you find the zeros, real and imaginary, of #y=x^2-5x+8# using the quadratic formula?

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Answer 1 · 2018-04-27T01:04:37

# x = 5/2+sqrt(7)/2i# and # x = 5/2-sqrt(7)/2i#

We have:

# y=x^2-5x+8 #

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

We get the the roots of the equation #y=0#:

# x = (-(-5) +- sqrt((-5)^2-4(1)(8)))/ (2(1)) #

# \ \ = (5 +- sqrt(25-32))/ (2) #

# \ \ = (5 +- sqrt(-7))/ (2) #

# \ \ = (5 +- sqrt(7)i)/ (2) #

Leading to the two conjugate roots:

# x = 5/2+sqrt(7)/2i# and # x = 5/2-sqrt(7)/2i#