Solve the equations set in complex numbers.

Question

1) Solve the equation $x^2+4=0$ set in complex numbers. My solution, pretty sure this is right $$x^2+4=0$$ $$x^2=-4$$ $$x=2i$$

But I have issues with this one:

2) Solve the equation $x^2+x+1=0$ set in complex numbers.

Not quite sure what to do here, I tried calculating delta:

$$\Delta = 1-4$$ $$\Delta = -3$$ $$\sqrt{\Delta}=\sqrt{3}i$$

But delta is negative so theres no solutions.

What's the right method for this example?

Answer 1

So for 1), remember that:

$x^2 = -4 \Rightarrow x = \pm \sqrt{-4} = \pm 2i$

As for 2), you can use the quadratic equation to your advantage here, which you can look up how to derive if you need to do things from first principles.

Write our equation in the following form $ax^2 + bx + c = 0$, then:

$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} \Rightarrow x = \frac{-1\pm \sqrt{1-4\cdot1\cdot1}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm\sqrt{3}i}{2}$.

However, on a note of pedantry, I would like to remind you that there is no "right method" so to speak. Any method that doesn't contain a false piece of logic is ultimately valid, though obviously some methods are quicker and easier than others.

I hope I helped. Feel free to ask any questions, i'm more than happy to clear anything up for you.

Answer 2

$x^ 2 +x +1 = 0$

$x^2 + \frac{2x}{2} + \frac{1}{4} + \frac{3}{4} = 0 $

$(x+\frac{1}{2})^2 = -\frac{3}{4}$

$(x+\frac{1}{2}) = \pm i\frac{\sqrt {3}}{2}$

$x = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$

Answer 3

$$x^2+4=0$$ $$x^2=-4$$ $$x=+2i,-2i$$.

$$x^2+x+1=0$$ $$(x+\frac{1}{2})^2+\frac{3}{4}=0$$ $$(x+\frac{1}{2})^2=-\frac{3}{4}$$ $$(x+\frac{1}{2})=-\frac{\sqrt{3}i}{2},\frac{\sqrt{3}i}{2}$$ $$x=\frac{-1-\sqrt{3}i}{2},\frac{-1+\sqrt{3}i}{2}$$

Answer 4

1) Solve the equation $x^2+4=0$ set in complex numbers. My solution, pretty sure this is right $$x^2+4=0$$ $$x^2=-4$$ $$x=2i$$

Just as the equation $x^2-4=0$ has two solutions (namely $x=\pm 2$), you're missing the solution $x=-2i$.

But I have issues with this one:

2) Solve the equation $x^2+x+1=0$ set in complex numbers.

If the discriminant $\Delta = b^2-4ac$ of a quadratic equation $ax^2+bx+c=0$ with real coefficients ($a,b,c \in \mathbb{R}$) is negative, then the (two) complex (conjugate) solutions are given by: $$\frac{-b\pm\sqrt{-\Delta}i}{2a}$$