I was given this equation $x^2 - 6x + 15 = 0 $
I tried to look for numbers whose sum is big and product of ac and i could not find any. I tried using the quadratic formula $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ and this is what i got
$x = \frac{6\pm\sqrt{-24}}{2}$
I just dont know what to do from here: any help would be appreacited.
$\endgroup$ 63 Answers
$\begingroup$Another approach
$$x^2-6x+15=0$$
$$x^2-6x=-15$$
$$x^2-6x+9=-6$$
$$(x-3)^2=-6$$
Take the square root of both sides
$$x-3=i\sqrt6,\;x-3=-i\sqrt6$$
$$\boxed{\color{red}{x=3\pm i\sqrt6}}$$
$\endgroup$ 2 $\begingroup$So you need to find the roots as follow: Note that:
$$ i^2 = -1$$
$$ x = \frac{6\pm \sqrt{-24}}{2}=\frac{6\pm i\sqrt{24}}{2} $$
$$ x= \frac{6\pm 2i\sqrt{6}}{2} = 3\pm i\sqrt{6}$$
Further explanation ($i^2 = -1$):
let $a = \sqrt{-24}$
$a^2 = -24= 24 \times -1= 24i^2$
$a =\pm \sqrt{24} \times \sqrt{i^2}= \pm i\sqrt{24} = \pm i \times \sqrt{4} \times \sqrt{6} = \pm2i\sqrt{6}$
$\endgroup$ 9 $\begingroup$This quadratic has two complex roots, ie no real solution. It depends now on what this equation stems from: for example if you got this equation while trying to get the intersection between a line and a parabola, the lack of real solutions represents the fact that they don't intersect.
If you're getting this equation in a complex analysis course, then you're supposed to know how to solve this and John just did that for you.
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