SOLUTION: In a complex number system, any quadratic equation has either one or two solutions. (That is the T/F question) I believe this is false because I thought they will always have 2

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: In a complex number system, any quadratic equation has either one or two solutions. (That is the T/F question) I believe this is false because I thought they will always have 2       Log On


   

Question 151469: In a complex number system, any quadratic equation has either one or two solutions. (That is the T/F question)
I believe this is false because I thought they will always have 2 solutions but can someone explain if I am wrong?

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
It is TRUE.
.
Consider the quadratic equation:
+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F2a+
.
Notice that we get our TWO solutions of a quadratic from the +- portion.
.
Now, let's focus our attention on (the term after the +-):
+sqrt%28b%5E2+-+4ac%29+
.
Notice that if:
b^2 is EQUAL to 4ac we will be taking the square root of zero. In this case,
we will get only one solution:
+%28-b+%2B-+sqrt%280%29%29%2F2a+
or
+-b%2F2a+
.
As an example, consider this:
x^2+2x+1 = 0
factoring we get:
(x+1)(x+1) = 0
Notice that we ONLY have a single solution of:
x = -1