Find the quadratic equation with real coefficients and solutions $x_1,x_2$ if you know that $\Delta =b^2-4ac=-36$ and $x_1+x_2=6m$

Question

What I've done so far is: We know that $\Delta \lt 0$, which means that $x_1$ and $x_2$ are two conjugate complex numbers with the form: $x_1=r+n\cdot i \space\text{ and }\space x_2=r-n\cdot i.$

We now need to find the sum $S=x_1+x_2$ and the product $P=x_1\cdot x_2$ to form an equation $x^2-Sx+P=0$ which has the solutions $x_1$ and $x_2$.

\begin{align*} sum = x_1+x_2&=6m \tag{1}\\ \iff(r+n\cdot i)+(r-n\cdot i)=2r&=6m \tag{2}\\ \implies r&=3m \tag{3} \end{align*}

The $product = x_1\cdot x_2=(r+n\cdot i)(r-n\cdot i)=r^2+n^2 = (\cdots).\quad $ From here I don't know what relation to find between $\Delta$ and $x_1\cdot x_2.$

I tried: $x_2=\dfrac{-b\pm\sqrt{-36}}{2a} =\dfrac{-b}{2a}\pm\dfrac{6i}{2a} =3m\pm\dfrac{3i}{a}\space \text{ so }\space x_1\cdot x_2 =\dfrac{9m^2+9}{a^2} (\cdots)$

Answer 1

Alternative approach:

Without loss of generality, the equation is
$x^2 + Bx + C = 0.$

Since $x_1 + x_2 = 6m$,
and since you must have that
$(x - x_1) \times (x - x_2) = x^2 + Bx + C$,
you must have that
$B = -6m.$

---

Here, there is some ambiguity involved. Taking the constraint of (in effect)
$B^2 - 4C = -36$
at face value, you have that

$\displaystyle B^2 + 36 = 4C \implies C = \frac{36m^2 + 36}{4} = 9m^2 + 9.$

Therefore, the quadratic equation is

$\displaystyle x^2 + [-6m]x + \left[9m^2 + 9\right] = 0.$

---

I mentioned a possible ambiguity.

Generally, for an equation of the form $Ax^2 + Bx + C = 0$, whose solutions are given by

$\displaystyle \frac{1}{2A} \left[-B \pm \sqrt{B^2 - 4AC}\right]$

it is unclear whether the discriminant should be considered to be

$$ B^2 - 4AC ~~~\text{or}~~~ \frac{B^2 - 4AC}{4A^2}. \tag1 $$

In (1) above, I went with the LHS, on a guess.

Answer 2

\begin{align*} b^2-4ac=-36\implies 4ac\ge36\implies ac&\ge9\tag{1}\\ \\ x_1+x_2= \bigg(\dfrac{-b+\sqrt{-36}}{2a}\bigg)+ \bigg(\dfrac{-b-\sqrt{-36}}{2a}\bigg)= \dfrac{-b}{a}&=6m\tag{2}\\ \\ x_1\cdot x_2= \bigg(\dfrac{-b+\sqrt{-36}}{2a}\bigg) \bigg(\dfrac{-b-\sqrt{-36}}{2a}\bigg) = \dfrac{b^2+36}{4 a^2}& \tag{3} \end{align*} From $(1)$ we could say $\space x^2+36=0\space$ but $(2)\longrightarrow\dfrac{-0}{a}=6m\implies m=0.$

The first "candidates" for $\space (a, b)\space$ that will yield $\space -36\space$ are $a=1, b=2$

\begin{align*} \dfrac{-b}{a}=\dfrac{-2}{1} \implies a=1, b=2, 4ac=40\\ \implies 2^2-4(1)(10)=-36\\ \implies c=\dfrac{2^2+36}{4 (1^2)}=10\\ \\ \implies x^2+2x+10=0 \\ \implies x_1= -1+3i\quad x_2=-2-3i\\ \\ x_1\cdot x_2=( -1+3i)(-1-3i)=10 \\ \dfrac{-2}{1}=6m\implies m=\dfrac{-1}{3} \end{align*}

These last equations are in no particular order but I think we have the quadratic equation we seek, i.e. $\quad x^2+2x+10=0$

Answer 3

You already concluded from the negative discriminant that the zeroes of the quadratic polynomial must be a "conjugate pair" $ \ \alpha \ + \ \beta·i \ \ . $ The difference of the zeroes is $$ x_1 \ - \ x_2 \ \ = \ \ \frac{\sqrt{\Delta}}{a} \ \ \Rightarrow \ \ 2·\beta·i \ \ = \ \ \frac{\sqrt{-36}}{a} \ \ = \ \ \frac{6i}{a} \ \ \Rightarrow \ \ \beta \ = \ \frac{3}{a} \ \ . $$ The sum of the zeroes is given as $$ x_1 \ + \ x_2 \ \ = \ \ 2·\alpha \ \ = \ \ 6m \ \ \Rightarrow \ \ \alpha \ = \ 3m \ \ $$ (as you observed).

If we choose to construct the corresponding monic polynomial, we would set $ \ a = 1 \ \ , $ producing the conjugate pair of zeroes $ \ x_{1,2} \ = \ 3m \ \pm \ 3i \ \ . $ Multiplying out the factors $ \ (x \ - \ 3m \ - \ 3i) \ · \ (x \ - \ 3m \ + \ 3i) \ \ $ gives us the polynomial shown in user2661923's answer.