Finding the other solution of a quadratic equation with complex roots

Question

I am having trouble with the following problem:

The equation $ax^2 + bx + c = 0$ has real coefficients and one solution $x_1 = 2 + i$. What can be said about the other solution $x_2$?

(a) $x_2 = x_1$

(b) $x_2$ is a real number

(c) $x_2 = 2-i$

(d) none of the above

I have tried using the fact that complex roots of a quadratic equation always come in conjugate pairs, but I am not sure how to proceed from there. I am wondering if there is a specific formula or method that can be used to find the other solution.

Any help or guidance would be greatly appreciated. Thank you in advance!

Answer 1

Yes exactly we have that solutions come in conjugate pairs, indeed

$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=0$$

which is with real coefficients if and only if $x_2=\overline{x}_1$, from which we deduce that the other root is necessarly $x_2=2-i$ which leads to

$$x^2-4x+5=0$$

Answer 2

We can use the product of roots i.e. $x_1 \cdot x_2=\frac c a$. From the question, we know that $c$ and $a$ are real, therefore $\frac c a$ must also be real provided that $a \neq 0$. We know that $x_1$ is a complex number i.e. $2+i$, the product of roots states that the other root when multiplied with the first root gives some real value and we know that a complex number when multiplied by its conjugate, give real value. Hence we get to the conclusion that the other root must be the conjugate of $2+i$ i.e. $2-i$.