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 $x = 5/2-sqrt(7)/2i$

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