Question

Question number 2, @satellite73

Answer 1

can't wait

Answer 2

\[9x^2-16x+60=0\]\[ax^2+bx+c=0\] the distriminant is
\[b^2-4ac\]which is our cases is
\[(-16)^2-4\times 9\times 60\]

Answer 3

if my arithmetic is right, this is
\[(-16)^2-4\times 9\times 60=-1904\]
does that look right?

Answer 4

From what I know which is not much lol it looks pretty right to me (:

Answer 5

k good ( i used a calculator)
there are 3 possibilities for the discriminant
a) is it positive meaning there will be two real solutions to the equation
b) it is zero, which means there will be one real solution
c) it is negative (our case) meaning there will be NO real solutions
that is the answer h ere

Answer 6

so when its negative there are no real solutions?

Answer 7

right

Answer 8

how about
\[4x^2+8x-5=0\]
my suggestion is to factor it if you can

Answer 9

Okay yeah i can do that but i have to explain as to why i chose to factor it could you tell me why you suggest it? To me it seems like the easiest method idk

Answer 10

you want a hint? t is \[(2 x-1) (2 x+5) = 0\]
factor because what you said, it is the easiest method if you can do it

Answer 11

and also in this case you don't have too many choices because \(-5=-1\times 5\) or \(5\times (-1)\) so it is not hard to come up with \[(2 x-1) (2 x+5) = 0\]

Answer 12

Okay thank you (:

Answer 13

you can solve that, right?

Answer 14

\[2x-1=0\\
2x=1\\x=\frac{1}{2}\] for the first one

Answer 15

yep i can (: and okay thanks

Answer 16

good

Answer 17

wait this is going to come out dumb i suppose but are we still on part b or did we go to C? sorry little tiny bit confused

Answer 18

we did part B solved
\[4x^2+8x-5=0\]

Answer 19

\[(2 x-1) (2 x+5) = 0\]
\[2x-1=0\\
2x=1\\
x=\frac{1}{2}\] or
\[2x+5=0\\
2x=-5\\
x=-\frac{5}{2}\] are the two solutions

Answer 20

okay yes got that, thanks for clearing up my confusion

Answer 21

k now how about the last one
i guess you can't factor it

Answer 22

yeah and factoring is like the easiest method i know ):

Answer 23

\[2x^2-12x+5=0\] we can complete the square

Answer 24

subtract 5 from both sides to get
\[2x^2-12x=-5\] then divide by 2 and ge t
\[x^2-6x=\frac{5}{2}\] are the first two steps
write them down, let me know when you want to continue

Answer 25

okay done writing that down (:

Answer 26

ok now we take half of 6, which is 3 and square it to get 9, and write
\[(x-3)^2=-\frac{5}{2}+9\] or
\[(x-3)^2=\frac{13}{2}\]

Answer 27

maybe it would have been easier to use the quadratic formula, i don't know, but we are almost done

Answer 28

okay i wrote both of them down

Answer 29

you get
\[x-3=\pm\frac{\sqrt{13}}{\sqrt2}\] so
\[x=3\pm\frac{\sqrt{13}}{\sqrt2}\]

Answer 30

next job i get will be doing math at mystic falls

Answer 31

okay thankss (: and lol well that would be a little complicated cause mystic falls is actually a fictional town from the vampire diaires xD i actually go to FLVS

Answer 32

yeah i know
i googled it

Answer 33

well, i didn't know about FLVS and don't even know what that is

Answer 34

florida virtual school (: its home school but on a computer lol

Answer 35

wow
must be hard to learn math that way
at least you can get help here
do you have to turn this in? like written out on paper? or through a computer

Answer 36

through the computer, we could mail our work to the teachers but it would take a while for it to get there so our teachers dont want us to.

Answer 37

ok i assume you want to be done right? because we could redo the last one using the quadratic formula if you think it would look better to the teacher

Answer 38

Oh no its fine, plus i kinda need to finish up cause its 11pm and i have to go to bed soon lol so can we go onto my last question?

Answer 39

you got one more?

Answer 40

cause that was the last one, the one where we got \[x=3\pm\frac{\sqrt{13}}{\sqrt2}\]

Answer 41

oh yeah i know lol but i have one last question i need to work on if you cant help me now thats fine we can finish up tomorrow (:

Answer 42

go ahead an post, maybe we can do it quickly