Question

Quadratic equations help?

Answer 1

Okay, first of all, what is it called when you use this equation? (Wait, I'll write it.)

Answer 2

x = -p/2 +- square root of (p/2)^2 -q

Answer 3

Does anyone know?

Answer 4

I don't understand what you have written, is it something like this?\[\Large x\quad=\quad \frac{-p}{2}\pm \sqrt{(p/2)^2-q}\]

Answer 5

Yes, exactly!

Answer 6

Sorry, I couldn't figure out how to write it in the equation thingy.

Answer 7

It looks like the `Quadratic Function` was used to find solutions for x.
Do we need to put it back into standard form or something?
Or you were just curious what that was called? :o

Answer 8

Quadratic Formula*
Sorry

Answer 9

I thought it was called some special thing...whatever.

Answer 10

hmm

Answer 11

Anyway, I have an equation that I did, but the answer keeps coming out wrong, so I was wondering if you could help me find out why. I'll post the equation, just a minute.

Answer 12

(x-3)^2 + (x-2)(x+3) = x^2 + x - 2

Answer 13

Okay, so that comes to x^2 - 6x + 5 = 0
Right?

Answer 14

Mmm yes! Looks good so far!

Answer 15

Okay, the way we did in class was with that equation I showed you. p would be -6, and q would be 5.

Answer 16

Then you plug it into that other equation, so that is looks like this:
x = - (-6/2) +- square root of (-6/2)^2 - 5

Answer 17

\[\Large x\quad=\quad \frac{6}{2}\pm \sqrt{(-6/2)^2-5}\]Mmmm ok good, looks like you're on the right track! :)

Answer 18

Yeah, that's exactly what I have.

Answer 19

So then under the square root sign I got 9 -5.

Answer 20

OOOH I GET IT!

Answer 21

Sorry, I just realized my mistake!! :D

Answer 22

oh cool c:

Answer 23

Yeah, you have to subtract the numbers first, before square rooting them.

Answer 24

ah yes :3

Answer 25

Okay, but still it helped to talk through it. Thanks! :D

Answer 26

yay team!

Answer 27

lol