Problem on quadratic polynomial.

[sankalpmittal]

Not quite right. Squaring is going to mix up everything in your mind, because even if you square -1 you will get +1, and then will you go tell that -1>0 because its square is (+1)>0?

Because that's what you do when you type me:"So b²>0 => b>0" in your previous post.b² >0 (for b in ℝ) means that b<0 or b>0 but surely not 0.

So be careful when squaring. Another thing is that you have:
+4 > -6
squaring:
16 > 36 ?Another example:
-1 < 0
squaring:
1<0 !and one last for another reason:
let's say you have one number A and one number B, for which you know that their squares will also follow the below rule:
A² > B²

What you can say for A and B?
the above rule tells you that:
A>B
or
A<-B

for example on this is to try find which number x is:
x² > 2
the answer is that:
x<-√2 or x>+√2

x²<2
gives
-√2 < x < +√2
even when you have equalities it is being difficult.
If you have tha a=b then you can say that a²=b² here without a problem (as there was with -1<0 above)
On the other hand
if you have that a²=b² then you write that a= ±b.

Ok , I was fully messed up until I discovered the correct answer by a correct method using your logic only !

Product of two roots or c/a< 0
And mean of two roots or -b/2a>0
Now on dividing the two I get :
i.e. c/a/-b/2a <0
2c/-b < 0 because +/- gives - , its obvious !
now b has to positive ! So b>0. As c/a <0 here you can see that a<0.

Your parabola in post #1 is not a real parabola but it has maximum at x=0. So it must be symmetric to the y axis. It would be useful to see the original picture or a better copy of the original.
If the maximum is really at x=0, what is b then?

ehild

If the maximum is really at x=0, then b>0 is what I concluded by my discovery on this aspect.

 

[ehild]

If the maximum is really at x=0, then b>0 is what I concluded by my discovery on this aspect.

The maximum or minimum of a quadratic function is at x=-b/2a. (Why?) If it is at x=0, b=0

ehild

 

[Morgoth]

I shall repeat only one thing:
The mean value you can only check if your axis is CALIBRATED. in a general axis where you don't know if x1,x2 you cannot use it.
if x1=-4 and x2=2 then the mean value is at -1<0.
if x1=-1 and x2=5 then the meave value is at +3>0.
when you don't know or have no clue where is each of the two, you CANNOT use it.

For example in your graph i get that the mean value is positive, but i cannot be sure because it is not so obvious.

 

[sankalpmittal]

The maximum or minimum of a quadratic function is at x=-b/2a. (Why?) If it is at x=0, b=0

ehild

Are you taking in account mean value of x ? If yes , then mean value of x is 0 as x is itself 0. Mean of x=-b/2a => 0=-b/2a => b=0.

I shall repeat only one thing:
The mean value you can only check if your axis is CALIBRATED. in a general axis where you don't know if x1,x2 you cannot use it.
if x1=-4 and x2=2 then the mean value is at -1<0.
if x1=-1 and x2=5 then the meave value is at +3>0.
when you don't know or have no clue where is each of the two, you CANNOT use it.

For example in your graph i get that the mean value is positive, but i cannot be sure because it is not so obvious.

I saw no reason to post this thing ! Anyways in my original graph which was drawn in exam paper I could easily see that +x is many times greater than magnitude of x in -x. So mean value will be positive.