Respuesta :

Let x = p+qi be the complex root and x = r be the the real root. The p and q are real numbers.

  • If x = p+qi, then x-p-qi = 0 which indicates (x-p-qi) is a factor.
  • If x = r, then it leads to x-r = 0. This makes (x-r) the other factor.

Multiply the two factors like so:

(x-p-qi)(x-r)

y(x-r) ..... let y = x-p-qi

xy - ry ..... distribute

x( y ) - r( y )

x( x-p-qi ) - r( x-p-qi ) ........ plug in y = x-p-qi

x^2 - px - (qi)x - rx + pr + qr*i ........ distribute

x^2 + (-px-qi*x-rx) + (pr+qr*i)

x^2 + (-p-qi*-r)x + (pr+qr*i)

x^2 + ((-p-r)-qi*)x + (pr+qr*i)

We end up with an expression in the form ax^2 + bx + c where the coefficients are:

  • a = 1 = real number
  • b = (-p-r) - qi = complex number
  • c = pr + qr*i = complex number

The coefficients b and c are complex numbers. So it is possible to have one real root and exactly one complex root; however, as shown above, it leads to some of the coefficients being complex numbers. I'm assuming your teacher wants all coefficients to be real numbers. If that's the case, then a quadratic equation cannot have one real root and one complex root. You would need to have the complex roots come in conjugate pairs. Meaning that if a quadratic equation with real coefficients has a complex root, then there's another complex root as its nearly "twin" pair so to speak.

Recall that quadratic equations have at most 2 roots according to the fundamental theorem of algebra. Having 2 complex conjugate pairs for the roots will then mean there isn't enough room for that third real root. In other words, here are all the possible scenarios:

  • A) There are two different real roots
  • B) There is exactly one real root
  • C) There are no real roots. Instead, there are two complex roots that come in conjugate pairs.

Again this only applies if all coefficients a,b,c are real numbers. For scenario B, the root repeats itself and we consider it a double root. For instance, x^2+10x+25 = 0 has the root x = -5 repeat itself. In scenarios A and B, we don't have any complex roots happen.

facundo3141592

Because always that we have a discriminant different than zero, we have two solutions.

When a quadratic equation has complex roots?

A quadratic equation has complex roots when its discriminant is negative, where for the general quadratic:

y = a*x^2 + b*x + c

The discriminant is:

D = b^2 - 4ac

And the solutions of the roots are:

[tex]x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}[/tex]

Because we have two possible signs for the square root of the discriminant, we always will have two solutions when the discriminant is different than zero.

If you want to learn more about quadratic equations, you can read:

https://brainly.com/question/1214333

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