Question

calculus 5, help please
Check directly near which points, we can solve the equation F(x,y) = y^2+y+3x+1 =0

Answer 1

@xapproachesinfinity

Answer 2

I got \(y = -\dfrac{1}{2}\pm\dfrac{\sqrt3}{2}\sqrt{-1-4x}\)

Answer 3

hmm how?

Answer 4

just quadratic equation for y

Answer 5

oh i see.
so you are saying any two points satisfying that solution is good!?

Answer 6

I think the solution is the whole parabola. However, if \(x= \dfrac{-1}{4}\), y = -1/2, then \(F(x,y)\neq 0\)

Answer 7

My question: 1) How to argue that the whole parabola is the solution but that point?
2) Are there any other points? How to check?

Answer 8

hmm what if your method was wrong lol

Answer 9

how about if you differentiate

Answer 10

Read the question, kid. It says :"check directly"

Answer 11

If you want to take differentiate, hold on, next problem, we will have some to play with. hehehehe

Answer 12

hmm well depends what is the meaning of directly here

Answer 13

may be they want us to do trial and error hahaha

Answer 14

ok, let check
\(F(x,f(x) )= ( -\dfrac{1}{2}\pm\dfrac{\sqrt3}{2}\sqrt{-1-4x})^2+(-\dfrac{1}{2}\pm\dfrac{\sqrt3}{2}\sqrt{-1-4x})+3x+1\)

Answer 15

is it =0 for all x?

Answer 16

hmm let's see

Answer 17

well apparently not for any x
take x=0

Answer 18

it is =0, kid

Answer 19

did you check it

Answer 20

yup

Answer 21

but you are operating in complex set

Answer 22

Let check + one,
\( ( -\dfrac{1}{2}+\dfrac{\sqrt3}{2}\sqrt{-1-4x})^2+(-\dfrac{1}{2}+\dfrac{\sqrt3}{2}\sqrt{-1-4x})+3x+1\)

Answer 23

I got 0

Answer 24

here is something i did
let x=0
\[f(0,y)=y^2+y=0 \Longrightarrow y=-1 \] disregareded y=0
\[f(x,0)=3x+1=0 \Longrightarrow x=\frac{-1}{3}\]
the point (-1/3, -1) works fine
but don't how to continue and argue this

Answer 25

hey, if x =0, f(0,y) = y^2+y+1

Answer 26

eh blindness my friend haha

it is still the same though y=-1 lol

Answer 27

oh no lol

Answer 28

darn it thought it worked lol

Answer 29

hahaha... kid, some girl took of your soul?? you are here but your heart and your mind follow her. hahaha..

Answer 30

hahaha, kind of lol

Answer 31

eh darn, i give up.
why don't use the derivatives, though i don't know how will that be useful

Answer 32

hehehe, thanks any way, kid, want to have something else?

Answer 33

how did you get 0 for x=0 in your equation

Answer 34

i didn't get zero

Answer 35

your equatoin fails for x=0, checked didn't get 0

Answer 36

hey, kid, redo, it is =0

Answer 37

\(( -\dfrac{1}{2}+\dfrac{\sqrt3}{2}\sqrt{-1})^2+(-\dfrac{1}{2}+\dfrac{\sqrt3}{2}\sqrt{-1})+3x+1\)

Answer 38

i'am sure it is not 0 lol

Answer 39

ok, watch what the old man do

Answer 40

\((-\dfrac{1}{2})^2+2(\dfrac{\sqrt3}{2}\sqrt{-1})(-\dfrac{1}{2})+(\dfrac{\sqrt3}{2}\sqrt{-1})^2 -\dfrac{1}{2}+\dfrac{\sqrt3}{2}\sqrt{-1}+1\)

Answer 41

i did that!

Answer 42

old you are just wasting time, it is not zero haha

Answer 43

you missed something there i guess

Answer 44

old man*

Answer 45

look at this hehe
https://books.google.com/books?id=LiRLJf2m_dwC&pg=PA253&lpg=PA253&dq=check+directly+which+points+solve+the+equation+f%28x,y%29%3D0&source=bl&ots=dT7irE9Og4&sig=rNg9UXwp1hAIEqftucb4iWSfEi0&hl=en&sa=X&ei=IDsLVbu-HsWkgwSXn4T4BQ&ved=0CB4Q6AEwAA#v=onepage&q=check%20directly%20which%20points%20solve%20the%20equation%20f(x%2Cy)%3D0&f=false

Answer 46

I'M CHEATING WITH GOOGLE

Answer 47

i don't even want to look at that page hahaha
i will get confused lol
things i didn't do yet

Answer 48

http://www.wolframalpha.com/input/?i=%28+- \dfrac{1}{2}%2B\dfrac{\sqrt3}{2}\sqrt{-1}%29^2%2B%28-\dfrac{1}{2}%2B\dfrac{\sqrt3}{2}\sqrt{-1}%29%2B1

Answer 49

hehe nothing appeared

Answer 50

you can copy and paste on Wolfram, it says 0

Answer 51

http://www.wolframalpha.com/input/?i=%28+- \dfrac{1}{2}%2B\dfrac{\sqrt3}{2}\sqrt{-1}%29^2%2B%28-\dfrac{1}{2}%2B\dfrac{\sqrt3}{2}\sqrt{-1}%29%2B1

Answer 52

ha, you can do it by yourself

Answer 53

yes looks like i did square something lol

Answer 54

it is zero

Answer 55

ha!! this stubborn kid!!! hehehe

Answer 56

didn't*

Answer 57

hey old man my mind is not working lol

Answer 58

gotta go study for programming lol

Answer 59

ok, break, I need rest also. thanks for being here

Answer 60

np! good luck with this :)