The quickest way to solve a quadratic equation with imaginary solution

Question

I had an exam question for solving an equation with imaginary roots: $$ m^2-3m+6=0 $$ Obviously this yields imaginary roots. What's the quickest way to find the solution?

Answer 1

I think the quadratic formula could really simplify things $$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ for the general quadratic equation $ax^2+bx+c=0$

Going like this the answer to your question is $$\frac{3 \pm \sqrt{-15}}{2}$$

Answer 2

As we can see its not possible to factorise given equation. So using quadratic formula one can solve easily.

$$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Also one can solve it using completing the square method.

Answer 3

We can also solve this using Completing the Square, as metioned by Kanwaljit Singh

(Side note: I think quadratic formula is faster, I just wanted to provide an alternative method. This can also be easier if we are doing this by hand and have a large constant value and a small $x$ coefficient as we only need to divide the (small) $x$ coeff. by $2$ and square it, we don't have to multiply the (large) constant by the $x^2$ coefficient)

We have $$m^2−3m+6=0$$

First we look at the coefficient of $m$, which is $-3$

We divide this by two and sqaure to get $$\left(\frac{-3}{2}\right)^2=-\frac 94$$

We can then rewrite the equation as \begin{align}m^2-3m+6&=0\\ \left(m-\frac 32\right)^2-\left(\frac32\right)^2+6&=0\tag{$\dagger$}\\ \left(m-\frac 32\right)^2-\frac94+6&=0\\ \left(m-\frac32\right)^2+\frac{15}{4}&=0 \end{align}

We can now solve this for $m$:

\begin{align}\left(m-\frac32\right)^2+\frac{15}{4}&=0\\ \left(m-\frac32\right)^2&=-\frac{15}{4}\\ m-\frac 32&=\pm \sqrt{-\frac{15}{4}}\\ m&=\frac 32 \pm \sqrt{\frac{-15}{4}}\\ m&=\frac 32 \pm \frac{\sqrt{-15}}{\sqrt{4}}\\ m&=\frac 32 \pm \frac{\sqrt{-15}}{2}\\ m&= \frac{3\pm\sqrt{-15}}{2} \end{align}

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Proof that $(\dagger)$ is correct:

\begin{align}\left(m-\frac 32\right)^2-\left(\frac32\right)^2+6&=\left(m-\frac32\right)\left(m-\frac32\right)-\left(\frac32\right)^2+6\\ &= \left(m^2-\frac32m-\frac32m+\left(\frac32\right)^2\right)-\left(\frac32\right)^2+6\\ &= \left(m^2-2\left(\frac32m\right)+\left(\frac32\right)^2\right)-\left(\frac32\right)^2+6\\ &= \left(m^2-3m+\left(\frac32\right)^2\right)-\left(\frac32\right)^2+6\\ &= m^2-3m+\left(\left(\frac32\right)^2-\left(\frac32\right)^2\right)+6\\ &= m^2-3m+6\end{align}