Question

Felix exclaims that his quadratic with a discriminant of -1 has no real solutions. Felix then puts down his pencil and refuses to do any more work. Create an equation with a negative discriminant. Then explain to Felix, in calm and complete sentences, how to find the solutions, even though they are not real.

Answer 1

I just need help with the last part on how to find the solutions.

Answer 2

@zepdrix @mathmale @solidrebirth @superhelp101 @tejasvir

Answer 3

haha in calm sentences

Answer 4

yeah :/ it's dumb but what can you do lol

Answer 5

brb.

Answer 6

ok

Answer 7

Woah woah woah, calm down Felix.

So here's the thing...
You got a discriminant of -1, \[\Large\rm b^2-4ac=-1\]

Answer 8

The quadratic formula gives us solutions, both real and complex.\[\Large\rm x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]Plugging in our discriminant gives us:\[\Large\rm x=\frac{-b\pm\sqrt{-1}}{2a}\]From this point Felix, you need to recall that the imaginary unit is defined as: \(\Large\rm \mathcal i =\sqrt{-1}\)
So again, we'll plug this in.\[\Large\rm x=\frac{-b\pm\mathcal i}{2a}\]We'll write this as two separate fractions,\[\Large\rm x=\frac{-b}{2a}\pm\frac{1}{2a}\mathcal i\]And here we can see that we have solutions of the form: \(\Large\rm x=a+b\mathcal i\)

Answer 9

They might not be real, no, but they're solutions.
So you need to sit down and cool your jets Felix.

Answer 10

That's what I would tell him at least :d

Answer 11

Oh they want you to actually create an equation...
not leave the b and a like that.
Ugh-_- thinking..

Answer 12

I have part of this answered, I'll write what I wrote there, here! Give me one second it will probably help.

Answer 13

To find the discriminant solutions, you need to first have the problem in standard form, like so:

ax^2 + bx + c = 0

From here, you use the formula to find the discriminant, which is b^2 - 4ac.

Example problem:

-2x^2 + 2 - 4 = 0

Discriminant: 2^2 - 4(-2)(-4) = -28

Answer 14

I'm not sure if I answered the whole question, but this is what I have so far :)

Answer 15

Ok great so far.
Then to actually find the solutions, you need to use the quadratic formula.
And plug in your imaginary unit.

The example you came up with works very well for this :)

Answer 16

Okay! Can you see if I get them right, here?

Answer 17

Just make sure you understand how to deal with the sqrt(-28),\[\Large\rm \sqrt{-28}=\sqrt{28}\cdot\sqrt{-1}=\sqrt{28}~\mathcal i\]

Answer 18

Sure.

Answer 19

In your quadratic, the middle term should have an `x`.
Just making sure that was a typo.

Answer 20

The solutions are -1 and 2, right?

Answer 21

What? D:
Nooo.
We shouldn't end up with `real` solutions.
Where did those numbers come from....?

Answer 22

-_- i used the quadratic formula

Answer 23

This is what your quadratic setup should look like.
\[\Large\rm x=\frac{-2\pm\sqrt{-28}}{-4}\]

Answer 24

Yeah I just realized :) Let me see if I can get it now.

Answer 25

So the solutions are 1+ 2i sqr(7) / 4 and 1- 2i sqr(7) / 4

Answer 26

Why a 1 for the first number?

Answer 27

Wasn't that a 2 before?

Answer 28

reducing!

Answer 29

Reduced from where? :o
Nothing else changed.

Were you trying to divide it with the 4?

Answer 30

So you should get here (before simplifying anything),
\[\Large\rm x=\frac{-2\pm2\sqrt{7}~\mathcal i}{-4}\]From this step, yes you can divide a -2 out of everything.\[\Large\rm x=\frac{1\pm\sqrt{7}~\mathcal i}{2}\]

Answer 31

Which is... close to what you have but uhhh....
yah

Answer 32

so those are the "unreal" solutions, right @zepdrix ?

Answer 33

Yes, those are the complex solutions.
It contains both real and imaginary parts.
But the solution as a whole is not all real, so we call it complex.

Answer 34

Okay, awesome. Thanks for your help! :)