Question

Algebra 2 help please

Answer 1

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 2

One simple equation would be \(x^2+1=0\)

Answer 3

wouldn't have to be \[x ^{2} + bx + c\]?

Answer 4

@kirbykirby

Answer 5

as long as you have an \(x^2\) term (and that is the highest power in your polynomial), then it is a quadratic function. You have have here that \(a=1, b=0, c=1\)

Answer 6

Oh alright, so would I have to find the discriminant?

Answer 7

Notice that \(a\) can be any number except 0, because then \(x^2\) disappears

Answer 8

Yeah I understand that now

Answer 9

Well you can proceed as usual using the quadratic formula. Do you know about \(i^2=-1\)? If not you can just keep your solution with a negative square root

Answer 10

I do know what i^2 is. So if I had x^2+2x+3 the discriminant would be -8?

Answer 11

Yes indeed

Answer 12

How would I find the solutions of those?

Answer 13

Let's consider the simpler one first perhaps:
\(x^2+1=0\)
Then by the quadratic formula,
\[ s=\frac{-0\pm\sqrt{0-4(1)(1)}}{2(1)}=\pm\frac{\sqrt{-4}}{2}\]
Sine \(\sqrt{-1}=i\), then you get above that \[x=\pm\frac{2i}{2}=\pm i \]

Answer 14

Since*

Answer 15

And sorry I don't know why I wrote \(s=...\), I meant to write x

Answer 16

But isn't the discriminant formula b^2 -4ac?

Answer 17

yes, b=0, a=1, c=1

Answer 18

0^2=0

Answer 19

That's the quadratic formula..

Answer 20

the discriminant is found within the quadratic formula

Answer 21

Discriminant: \[ \Delta=b^2-4ac\]

Quadratic formula:
\[ \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{\Delta}}{2a}\]

Answer 22

oh you were dividing by 2 there which confused me a bit

Answer 23

Oh i actually found out how to do this. thanks for the help

Answer 24

:)