can a quadratic equation has exactly one complex solution?

Answer 1

The discriminant (b^2-4ac) is either positive (two real solutions), zero (one solution only), or negative (two complex solutions). That's probably the answer that is expected.

However, it is possible to construct an equation with a complex value for b, where b is equal to the positive or negative square root of (b^2-4ac), in which case one of the roots is zero. A very simple example is:

x^2 + 3ix = 0

... which factors to:

x(x + 3i) = 0

That equation has one real root (x=0) and one complex one (x = -3i).

Answer 2

If the quadratic equation has only real coefficients, the answer is No.

Proof: Suppose the equation is:

0 = Ax^2 + Bx + C, where A, B and C are real. Then divide by A:

0= x^2 + Px + Q, where P and Q are also real.

Now suppose that this equation has two general complex roots, a+bi and c+di.

Then 0 = (x - (a+bi))(x-(c+di))
= x^2 - (a+bi + c+di)x + (a+bi)(c+di)
= x^2 - x((a+c) + i(b+d)) + ((ac-bd) + i(bc+ad))

This can only be equivalent to the equation above if:
b + d = 0 (=> d = -b)
and
bc + ad = 0 (=> 0 = bc -ba = b(c-a) )

This means either b = 0 or c = a.

But if b= 0, d = -b also = 0, and the roots are both real.
And if c = a, the two roots are a+bi and a-bi; so if b is not = 0, and both roots are equally complex.

So either both roots are real, or they are equally complex
( one is a+bi and the other is a-bi: complex conjugates).

The answer is "No, a quadratic equation cannot have exactly one complex solution."

Answer 3

i think you could argue yes since real numbers are a subset of complex numbers - any real number is also a complex number (with zero imaginary component). so if the discriminant is zero and there's one real number solution, that is also one complex number solution. probably the question means complex numbers with nonzero imaginary component though.

Answer 4

Not if the equation has real coefficients.

Complex solutions to polynomials with real coefficients always come in conjugate pairs.

Answer 5

The two roots are either both real or imaginary. You cannot have one real root and one imaginary (if this is what you mean by complex).

Answer 6

hi

Answer 7

nope

Answer 8

no it has methods four methods of solving it