Question

evaluate the following expression: if a=9, b=6, and c=12

Answer 1

there isnt one thats all it says

Answer 2

That doesn't make sense.

Answer 3

perhaps this is giving you a polynomial

Answer 4

remember, when you plug in an expression into the quadratic formula, you use this breakdown of a polynomial:

\(\Large ax^2 +bx +c\)

Answer 5

so if you plug in the values of
a=9, b=6 and c=12, you can get yourself a polynomial

Answer 6

@DARTHVADER2900 lol...you weren't even the one who was explaining. Unless you have something to offer, leave this to me please

Answer 7

I was just asking if he/she understood what you were explaining, so please, no need to get mad. lol.

Answer 8

lol it's fine. It's just that it seems unnecessary to do that, if I'm the one guiding the user through the steps. It's just logic

Answer 9

lol

Answer 10

lol, sorry about that x'D Are you following thus far? I could help you find the values of x's, if we're just using the general form of a polynomial

Answer 11

yeah

Answer 12

alright, so if we plug the numbers that are given in this problem to the formula that is given, what would the expression be?

Answer 13

..

Answer 14

oh lol, let me clear myself up

Answer 15

if a=9, b=6, and c=12
and if you have the formula \(\large ax^2+bx+c\). then you have

\(\Large 9x^2 +6x+12\).

Make sense so far? @partyrainbow276

Answer 16

yeah

Answer 17

right then. So we can either factor this right here, or we could use the quadratic formula. I'd say that in this case, factoring would be easiest. DO you know how to factor?

Answer 18

no

Answer 19

hmm.. Then perhaps, it may be best if we use the quadratic formula. I could give it to you, if you're not familiar with it

Answer 20

oh okay

Answer 21

right. So the quadratic formula is

\(\Huge x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Answer 22

for the values of a,b, and c that you were given in this problem, just plug them in to this formula. Then, you can find your values of x

Answer 23

So can you do that, or would you like me to show you how it's done?

Answer 24

show

Answer 25

what grade is this?

Answer 26

im in 8th grade. it looks like u can just multiply the numbers then average. ya think?

Answer 27

alright.

since a=9, b=6, c=12,

\(\Huge x=\frac{-6\pm\sqrt{6^2-4(9)(12)}}{2(9)}\)

Answer 28

@tootzrll It doesn't appear to be so. as we're trying to evaluate an expression.

Answer 29

albeit, we don't know what the expression is, so we're assuming that it's polynomial

Answer 30

@partyrainbow276 what math are you taking? I find it odd how ambiguous of a question this is

Answer 31

algebra

Answer 32

huh. seems legit. sorry i couldnt help :/

Answer 33

its fine

Answer 34

if you're in algebra, it's highly likely that you will be needing to know the quadratic formula. Would you like some help in simplifying this down?

Answer 35

yes

Answer 36

So we have this:

\(\Huge x=\frac{-6\pm\sqrt{6^2-4(9)(12)}}{2(9)}\)

we're going to want to start simplifying the inside of the square root first.
\(\Huge x=\frac{-6\pm\sqrt{36-432}}{18}\)

Answer 37

you with me so far?

Answer 38

yep

Answer 39

okay. So we keep simplifying this down until we have
\(\Huge x=\frac{-6\pm\sqrt{-396}}{18}\)

This has to be left like this. Therefore, x is equal to two roots.

\(\Large x= \frac{-6+396i}{18}\) and \(\Large x= \frac{-6-396i}{18}\)

Answer 40

the reason why we have imaginary numbers is because we have the square root of a negative number. Basically, this means that the roots of this polynomial do not cross the x-axis

Answer 41

hope this has helped you! :)

Answer 42

thank youuu

Answer 43

you are most welcome