SOLUTION: what types of roots does {{{x^2-6x+10=0}}} have? my work: a=1 b=-6 c=10 4ac=40 {{{(6+ or - sqrt(36-40))/2}}} {{{(6+ or - sqrt(-4))/2}}} {{{(6+ or - i sqrt(4)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: what types of roots does {{{x^2-6x+10=0}}} have? my work: a=1 b=-6 c=10 4ac=40 {{{(6+ or - sqrt(36-40))/2}}} {{{(6+ or - sqrt(-4))/2}}} {{{(6+ or - i sqrt(4)      Log On


   

Question 697487: what types of roots does x%5E2-6x%2B10=0 have?
my work:
a=1
b=-6
c=10
4ac=40
%286%2B+or+-+sqrt%2836-40%29%29%2F2

%286%2B+or+-+sqrt%28-4%29%29%2F2

%286%2B+or+-+i+sqrt%284%29%29%2F2
sqrt%284%29=2
%286%2B+or+-+2i%29%2F2
cancel the 2's out and simplify the 6. final answer:
3+ or - i
I know I found the roots. Because of the i does this mean the roots are irrational?
Answer by Positive_EV(69) About Me  (Show Source):
You can put this solution on YOUR website!
You can use the discriminant to determine the nature of the roots: the discriminant is the part of the quadratic formula under the square root sign; that is, b%5E2+-+4%2Aa%2Ac. If the discriminant is positive, there are two real roots. If it is zero, there is a single double root. If it is negative, there are two imaginary roots.

Here, a = 1, b = -6 c = 10, so b%5E2+-+4%2Aa%2Ac+=+%28-6%29%5E2+-+4%2A%281%29%2A%28-10%29+=+-4.

The discriminant is negative, so the equation has two imaginary roots.