Explain completely why a quadratic equation cannot have one real root and one complex root. Be clear, and make it evident that you understand this. (1 point)

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

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:

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:

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.

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step-by-step explanation: