Complex roots with two variables

Question

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

Answer 1

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 $$i(2a^2-6a)+c-6a=0$$

Answer 2

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

Answer 3

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)

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

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

the quadratic is therefore

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

and the roots are

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