True/false: A quadratic equation always has two complex roots.

Is not the statement below true?

*A quadratic equation always has two complex roots.*

According to a test I took, it is not. However I believe it is, because a real root is also a complex root since z is real if and only if b = 0 in z = a + bi. Moreover, I believe this has something to do with the fundamental theorem of algebra, but I am not sure.

Re: True/false: A quadratic equation always has two complex roots.

Quote:

Originally Posted by

**MathCrusader** Is not the statement below true?

*A quadratic equation always has two complex roots.*

According to a test I took, it is not. However I believe it is, because a real root is also a complex root since z is real if and only if b = 0 in z = a + bi. Moreover, I believe this has something to do with the fundamental theorem of algebra, but I am not sure.

This can be seen as a trick question.

The quadratic equation $\displaystyle x^2-2x+1=0$ has exactly one real solution $\displaystyle x=1$.

The quadratic equation $\displaystyle (x-1)(x+i)=0$ has exactly two solutions $\displaystyle x=1~\&~x=-i$. Note one is real and one is complex.

BUT we can say that every real number is a complex numbers.

Re: True/false: A quadratic equation always has two complex roots.

I suspect that the phrase "two square roots" is to be interpreted as "two **distinct** complex roots". Of course, it is NOT true that "the quadratic equation always has two **distinct** complex roots".

Re: True/false: A quadratic equation always has two complex roots.

the question is ill-posed (it doesn't provide enough information to answer).

here is what is true:

1) a real quadratic polynomial always has two complex roots (which may be distinct, or coincide). if the roots are distinct: they are complex-conjugates, if they coincide (a double root), it must be real. note that real numbers count as complex numbers for the purposes of THIS statement.

2) a real quadratic polynomial can have 0,1, or 2 real roots. by statement (1) above, if a real quadratic has only one root, it must be a double root (a single complex (but not real) root is not possible).

if by "two complex roots" something along the lines of statement (2) is intended (that is "complex" means "complex, but NOT real"), then the statement is FALSE. if by "two complex roots" what is meant is "two distinct complex roots" then the statement is also false. if what is meant is "two complex roots", where the roots are allowed to coincide, THEN the statement is true.

however, the above only applies for REAL quadratic polynomials. with COMPLEX quadratic polynomials, one can have one real and one complex root quite easily as indicated in Plato's post. again, these roots may be distinct, or coincide, as in:

$\displaystyle f(x) = (x - i)^2 = x^2 - (2i)x - 1$

the variety of possible answers depending on interpretation of the question speaks poorly for the designer of the test.