Find all real values of a such that $x^2+(a+i)x-5i=0$ has at least one real solution

Question

Find all real values of a such that $x^2+(a+i)x-5i=0$ has at least one real solution.

$$x^2+(a+i)x-5i=0$$

I have tried two ways of solving this and cannot seem to find a real solution.

First if I just solve for $a$, I get $$a=-x+i\frac{5-x}{x}$$ Which is a complex solution, not a real solution...

Then I tried using the fact that $x^2+(a+i)x-5i=0$ is in quadratic form of $x^2+px+q=0$ with $p=(a+i)$ and $q=5i$

So I transform $$x^2+(a+i)x-5i=0$$ to $$(x+\frac{a+i}{2})^2=(\frac{a+i}{2})^2+5i$$

Now it is in the form that one side is the square of the other but I don't know how to find the roots since I'm not sure if I'm supposed to convert $(\frac{a+i}{2})^2+5i$ to polar form since I can't take the modulus of $(\frac{a+i}{2})^2+5i$ (or at least I don't know how).

At thins point I feel like I'm just using the wrong method if anyone could guide me in the right direction I would very much appreciate it. Thank you.

Answer 1

If $p$ and $q$ are the roots of $x^2+(a+i)x-5i=0$, then $p+q=-(a+i)$ and $pq=-5i$. If $p$ is real, then $q=-5i/p$ is imaginary. If $a$ is real, then we must have $q=-i$ and $p=-a$. We find $p=-5i/q=5$, hence $a=-5$ is the only real number for which $x^2+(a+i)x-5i=0$ has a real root.

The equations $p+q=-(a+i)$ and $pq=-5i$ come from expanding $(x-p)(x-q)=x^2-(p+q)x+pq$ and equating coefficients with those of $x^2+(a+i)x-5i$.

Answer 2

I think you were almost there when you said $a=-x+i\dfrac{5-x}{x}$.

(By the way, I think you should have mentioned $x\ne0$ before getting to that point.)

If $a=-x+i\dfrac{5-x}{x}$ and $a,x\in\mathbb R$, then $5-x=0$, so $x=?$, so $a=?$.

Extra credit question: what is the other solution of the quadratic equation?