# Finding the other solution of a quadratic equation with complex roots

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!

• Do you know what is conjugate of a complex number ? If yes, then where is your problem exactly ? May 5 at 20:46
• In particular, what is the conjugate of $2+i$? May 5 at 21:00

$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=0$$
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
$$x^2-4x+5=0$$
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. $$2-i$$.