# Complex roots with two variables [closed]

$a + ai$ is a root of $x^2 − 6x + c = 0$, where $a, c ∈ \mathbb{R}$. Find all possible roots and all possible values of $c$.

## closed as off-topic by José Carlos Santos, zipirovich, mfl, StubbornAtom, DidAug 15 '18 at 19:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, zipirovich, mfl, StubbornAtom, Did
If this question can be reworded to fit the rules in the help center, please edit the question.

• Do you know what it means for $a+ai$ to be a root? If so, have you tried using that to learn about $a$ or $c$? – Mark S. Aug 15 '18 at 16:38
• Oh gosh, I'm so lost with this one and the comple roots of i^2 = -1 – MissPolite Aug 15 '18 at 16:54

If $r_1$ and $r_2$ are the roots of a general, real quadratic equation

$x^2 + px + q = 0, \; p, q \in \Bbb R, \tag 1$

then writing

$(x - r_1)(x - r_2) = x^2 + px + q \tag 2$

yields

$x^2 - (r_1 + r_2)x + r_1 r_2 = x^2 + px + q, \tag 3$

from which we see

$r_1 + r_2 = -p, \; r_1 r_2 = q; \tag 4$

also, if $\alpha \in \Bbb C$ is a non-real zero of (1), then so is $\bar \alpha$, since

$\alpha^2 + p \alpha + q = 0 \Longrightarrow \overline{\alpha^2 + p \alpha + q} = 0 \Longrightarrow \bar \alpha^2 + p \bar \alpha + q = 0; \tag 5$

therefore, if we have the quadratic equation

$x^2 - 6x + c = 0, \tag 6$

with root $\alpha$, so that

$\alpha = a + ai = a(1 + i), \; p = -6, \; q = c, \tag 6$

we take

$a + ai = \alpha = r_1, a - ai = \bar \alpha = r_2, \tag 7$

and then we have from (4)

$p = -(\alpha + \bar \alpha) = -(a + ai + a - ai) = -2a = - 6 \Longrightarrow a = 3; \tag 8$

$q = \alpha \bar \alpha = (a + ai)(a - ai) = a^2 + a^2 = 2a^2 = 18; \tag 9$

$x^2 - 6x + 18, \tag{10}$

and the roots are

$\alpha = 3 + 3i = 3(1 + i), \; \bar \alpha = 3 - 3i = 3(1 - i). \tag{11}$

• Thank you! really appreciated. – MissPolite Aug 15 '18 at 17:22
• My pleasure, ma'am! Cheers! – Robert Lewis Aug 15 '18 at 17:23

If $$a+ai$$ is a root then must be

$$(a+ai)^2-6(a+ai)+c=0$$ and $$a-ai$$ must be also a root. You will get $$i(2a^2-6a)+c-6a=0$$

• Sorry? Just a little lost over it at the minute, if I expand that do I use the quadratic equation to find the roots? – MissPolite Aug 15 '18 at 16:41
• You have to find $a,c$ – Dr. Sonnhard Graubner Aug 15 '18 at 16:42

we have complex roots in the quadratic equation if $D<0$ and the roots are $$\frac{-b\pm\sqrt D}{2a}$$ thus you can clearly see that if $\alpha+i\beta$ is root than the second root must be $\alpha - i\beta$

thus the second root of the given equation is $a-ai$

also, the sum of the roots = -coefficient of $x$/coefficient of $x^2$

thus $$-\frac{-6}{1}=a+ai+a-ai$$ $$2a=6$$ thus $a=3$

so the roots of the given the quadratic equation are $3+3i$ and $3-3i$.

also the multiplication of two roots = constant term/coefficient of $x^2$

thus $$(3+i3)(3-i3)=\frac{c}{1}$$ $$c=18$$

• see the edited answer to understand fully :) cheers!! – Deepesh Meena Aug 15 '18 at 17:06