Question

@johnweldon1993 Sorry :3

Answer 1

Solve this quadratic equation using the quadratic formula.
I NEED HELP WIHT THIS TOO

\[2x^2 - 10x + 7 = 0\]

Answer 2

@johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993 @johnweldon1993

Answer 3

;)

Answer 4

Haha i get it i get it XD

You tell me! we just did an extremely similar one...follow the same steps :)

What are your 'a' 'b' and 'c'?

Answer 5

POh shoot, yea yeaaaa I'll try

Answer 6

I''ll be here to check :D

Answer 7

Okay be here

Answer 8

XD I wont leave! sheesh! :P

Answer 9

Okay so i did a,b,c
now i'm doing \[\sqrt{6^2 +4ac}\]

Answer 10

O.o are we doing the same problem?

Answer 11

\[\large 2x^2 - 10x + 7=0\]

What are your 'a' 'b' and 'c' values? list em out :)

Answer 12

LOL ..duhh..i jsut wrote a formula >.>
here\[\sqrt{6x^2+4(10)(7)}\]

Answer 13

AAA DAMN
okay a=
b=10
c=7

Answer 14

Oh shoot a is 2

Answer 15

a = 2 right
b = check this one again
c = 7 yes

Answer 16

Wait wouldn't it be \[\sqrt{(10)^2-4(2)(7)}\]

Answer 17

b=10?

Answer 18

Nope, we are still working on the values for the moment :)

a = 2
b = ?????
c = 7

And your 'b' value...just think you have an equation that LOOKS like \(\large ax^2 + bx + 0\)

*Notice how everythin is added!*

Well in YOUR equation you have \(\large 2x^2 - 10x + 7 = 0\]

Look how here we end up subtracting something!

Meaning your b is actually -10! not 10!

Answer 19

I made a typo up there...should be \(\large ax^2 + bx + c = 0\)

Answer 20

lol so you got this i guess

Answer 21

Ughh tbh this is distracitng \(\large ax^2 + bx + c = 0\)
Lol why it doesn't shows in Latex way >.>
anyways, i dont' know what b is, i give up

Answer 22

So we have

a=2
= -10
c = 7

Now plug those into \(\large x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Answer 23

wait it worked john, on my post not ur D:

Answer 24

Ooooh it was -10 >.>
anyways

Answer 25

mmhmm :)

So now onto the quadratic formula with your known values :)

Answer 26

wait why isn't it $$\\{a^2+b^2=c^2}$$

Answer 27

I got \[\sqrt{-10+4(2)(7)}\]

Answer 28

@First_world_PROBLEMS that is the Pythagorean theorm, we are working with quadratics here :)

@AloneS Not quite...also, dont focus on 1 part of the formula..you have the whole thing to work with

\[\large x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[\large x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(2)(7)}}{2(2)}\]

Answer 29

@First_world_PROBLEMS That's the pythagorean theorem. Quite different to the quadratic formula.

Answer 30

lol ooooooooooooooooooooohhhhhhhhhhhhhhhhhhhhhhhh....ok i knew that >.>

Answer 31

I'm dead.Sucha forest...
i got \[\frac{ 10\sqrt{44} }{ 4}\]

Answer 32

ohh i can d that

Answer 33

No i mean\[\frac{ 10\pm \sqrt{44} }{ 4 }\]

Answer 34

Remember you get 2 answers @AloneS type em both out for me :)

Answer 35

that square root can be split and simplified further

Answer 36

OOOOOH AGAIN WEELL-DONE
Fine fine fine i'll mke it
\[\frac{ 10+\sqrt{44} }{ 4 },\frac{ 10-\sqrt{44} }{ 4 }\]

Answer 37

Perfect!

Now! as @UsukiDoll has pointed out...that \(\large \sqrt{44}\) CAN be simplified further

Answer 38

Oh nooo

Answer 39

yeeeeesssssssssss that's math life also after the split you can reduce the fraction too and split that up

Answer 40

Usikii it is tortuing, literally i sit on each probelm 15 min each Oooh

Oh lol nvm it can be broken to 11 and 4

Answer 41

Haha it's not bad at all

Hint \(\large \sqrt{44} = \sqrt{4\times 11} = \sqrt{4}\sqrt{11}\)

What do you notice? *think about what is under the square root

Answer 42

oO.MG. 30 MINUTES I'M DOING THIS QUESTION D:
Well-donneee i see they become negative

Answer 43

I just dealt with real analysis today. I'll teach you torture. Let f:D->R be continuous. Prove If D is bounded then f(D) is bounded

Answer 44

5

Answer 45

Lol beat me to it! so perfect! now you would have
\[\large \frac{10 + 2\sqrt{11}}{4}, \frac{10-2\sqrt{11}}{4}\]

And you ca break that up and simplify further! :D haha

Answer 46

@UsukiDoll that looks like fun :D I just had to deal with my fluids final today >.< who wants to derive Navier Stokes for flow through a rough pipe :DD

Answer 47

Oooh JOHN WHEN IS IT ENDING?
I ca?
john to what do i simplify it to?

Answer 48

surely there's a shorter way

Answer 49

split the fractions and reduce

Answer 50

both are divisible by 2

Answer 51

Uh-oh, 11 aND 2? usiki?

Answer 52

Well you have

\[\large \frac{10 + 2\sqrt{11}}{4}, \frac{10 - 2\sqrt{11}}{4}\]

Break that up into

\[\large \frac{10}{4} + \frac{2\sqrt{11}}{4}, \frac{10}{4} - \frac{2\sqrt{11}}{4} \]

Which gives you

\[\large \frac{5}{2}+\frac{\sqrt{11}}{2},\frac{5}{2}-\frac{\sqrt{11}}{2}\]

or

\[\large \frac{5 + \sqrt{11}}{2}, \frac{5 - \sqrt{11}}{2}\]

Answer 53

^ #winning

Answer 54

That's all O_O

Answer 55

Also @First_world_PROBLEMS it really is a short process, just a long post haha

In general: in a nutshell

1) Find your coefficients, here a = 2, b = -10, c = 7
2) Plug them into the quadratic formula \(\large \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
3) \(\large \frac{-(-10) \pm \sqrt{(-10)^2 - 4(2)(7)}}{2(2)}\)
4) \(\large \frac{10 \pm \sqrt{44}}{4}\)

Then it's just simplifying

Answer 56

^ this user has a lot of patience... I don't lol

Answer 57

And @AloneS mmhmm that's all lol

Answer 58

Haha I'm studying for a final so ya know...procrastination at it's finest :D @UsukiDoll

Answer 59

my math professors curve so even if I eff up (which I try not to) I still pass so :P

corruption at its finest.

Answer 60

surely there's a shorter way

Answer 61

Ugh, I need the curve to be ever in my favor tomorrow >.< *When they force an ME major to take a PURELY EE based class* -sigh-

Answer 62

have you heard of Abstract Algebra?

Answer 63

https://www.google.com/imgres?imgurl=https%3A%2F%2Fs-media-cache-ak0.pinimg.com%2F236x%2F64%2F19%2F2d%2F64192d7292e5a657076d5b7d0830741b.jpg&imgrefurl=https%3A%2F%2Fwww.pinterest.com%2Fjoanthuffman%2Fevil-math%2F&docid=pd-hgkfl8SqzkM&tbnid=hgxPBbqs9ySFgM%3A&w=236&h=236&bih=445&biw=911&ved=0ahUKEwijgob6oNbMAhXHwj4KHcyUDekQMwgnKAEwAQ&iact=mrc&uact=8

Answer 64

Took it 2 semesters ago :D I blew through almost every math class offered here so far XD

Answer 65

WaY TO GOOOOOO THANK YOU Ughhhh sucha torture

Answer 66

blew through?

Answer 67

@AloneS the question is...did you understand it? because if you ask another Quadratic problem I'm going to make you tell me every step >:D

Answer 68

i got lost at the start.. idk why im still here

Answer 69

oh I got an even better one... if there's another quadratic problem, let's prove that f+g has a limit at x_0

Answer 70

0_0

Answer 71

^lol :D

Answer 72

You're here to spam @First_world_PROBLEMS ?
O.MG. @johnweldon1993 I lost my breath when i read it :3
okay i'll return back to this post to look at the steps. I have the last question ;)

Answer 73

NOPE BYE lol

Answer 74

Alright ladys and gents! I really SHOULD focus on studying XD that 8am final isn't coming any slower >.<

Answer 75

I had a 7:30 am exam...twice for foreign language

Answer 76

...that is disgusting on so many levels >.< lol

Answer 77

Finally i got thrie the steps.
Thanks everyone ;) (who helped._.)

Answer 78

i wanted to but then i got lost and these redirects are sooo anoying