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Problem on quadratic polynomial.

by sankalpmittal
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sankalpmittal
#19
Dec21-11, 06:28 AM
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Quote Quote by I like Serena View Post
Suppose we pick the parabola y=2x2.
It goes up doesn't it?

What would it take for the parabola to go down?
Concavity of parabola is in the direction of -y axis. In y=2x2 , x2 is positive and so is 2 , so this equation is not satisfied if we compare it with the parabola going downwards as y is negative. You are asking me to use parabolic equation... y=ax2 where x2 is positive and y is negative , so "a" has to be negative.

This implies a<0 , isn't it ?

(I like your picture btw. )
Thanks !
Quote Quote by NascentOxygen View Post
You could start by returning to my question which so far has not been answered:
Answered , I think. See the above reply to ILS' post in my post.

Quote Quote by ehild View Post
You do overcomplicate the problem. Look at NascentOxygen's big formula:

y=ax2+bx+c

and compare with your drawing:
The curve intersects the y axis at a positive y value.The points on the y axis correspond to x=0. Therefore

c>0

A general parabola can be obtained from the standard y=X2 one by shifting it along x, y and multiplying by a number a as

[tex]y=a^2+bx+c=a\left((x+\frac{b}{2a})^2+(c/a-\frac{b^2}{4a^2})\right)[/tex]

If a is a negative number, the parabola is upside down with respect to the standard one. The standard one is open upward. Your parabola is open downward, so
a<0

b/2a shows the shift of the parable along the x direction. As I see, your parabola is not shifted.

ehild
Ahh !! One more easier method ! Parabola is yielding real and distinct roots. So one value of x is positive and another is of course negative. So product of the two roots is negative as well.

P or Product of two roots = c/a
P = c/a
Since "P" is negative and c>0 so of course a<0 ! I couldn't help laughing !

Can this problem be solved without using parabolic equation ?
BTW , how can I say that b<0 i.e. option D is wrong ?

Thanks in advance.
ehild
#20
Dec21-11, 07:41 AM
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Sankalpmittal,
This is a little exercise for you, to get the feeling about quadratic equations and parabolas. Look at the following quadratic functions and sketch them.
What is the product of the roots and the sum of the roots of the f(x)=0 quadratic equations? Find a, b, c for all of them. Find the position of the peaks. Discover how the position of the peak is related to the roots.

f(x)=x2;
f(x)=-x2;

f(x)=(x-2)2;
f(x)=(x+2)2;

f(x)=(x-2)(x+2);
f(x)=(2-x)(x+2);

f(x)=(x-1)(x-3);
f(x)=(1-x)(x-3);

f(x)=(x-1)(x+3);
f(x)=(x+1)((x-3);
f(x)=(3-x)(x+1).
I like Serena
#21
Dec21-11, 04:22 PM
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Quote Quote by sankalpmittal View Post
Concavity of parabola is in the direction of -y axis. In y=2x2 , x2 is positive and so is 2 , so this equation is not satisfied if we compare it with the parabola going downwards as y is negative. You are asking me to use parabolic equation... y=ax2 where x2 is positive and y is negative , so "a" has to be negative.

This implies a<0 , isn't it ?

(Snip)
Yep!

You seem to be picking things up nicely! ;)


Quote Quote by sankalpmittal View Post
Can this problem be solved without using parabolic equation ?
BTW , how can I say that b<0 i.e. option D is wrong ?

Erm... all the answers are related to the parabolic equation...
So how would you deduce an answer without it?
I think the trick is to understand what the parameters mean geometrically.
Ehild's exercises should help with that.

And suppose you had simply y=bx+c.
What does it mean if either b<0, b=0, or b>0?
What happens if you add the extra term ax2?
Morgoth
#22
Dec21-11, 04:47 PM
P: 127
"Can this problem be solved without using parabolic equation ?
BTW , how can I say that b<0 i.e. option D is wrong ?"
from your solutions x1,x2 you can see that b<0 would mean that:

x1= -b/2a + sqrt[b^2-4ac]/2a
x2= -b/2a - sqrt[b^2-4ac]/2a

[x1+x2]= -b/a => [x1+x2]/2 = -b/2a
so where would the mean value of your solutions (y(x)=0) lie? on which side of the x axis acording to b and a signs?

Hope this can help a little bit more.
sankalpmittal
#23
Dec22-11, 05:47 AM
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Quote Quote by ehild View Post
Note : f(x) = 0 for all equations given :
So here ax2+bx+c = 0

1. f(x)=x2;
Product of roots = c/a ; Sum of roots = -b/a
x=0
a=1 , b=0 , c=0
Sum of roots = product of roots = 0

2. f(x)=-x2;
x=0
a=-1 , b=0 , c=0
Sum of roots = product of roots = 0

3. f(x)=(x-2)2;
x=2
a=1 , b=-4 , c=4
Sum of roots = Product of roots = 4

4. f(x)=(x+2)2;
x=-2
a=1 , b=4 , c=4
Sum of roots = -4
Product of roots = 4

5. f(x)=(x-2)(x+2);
x= 2 , -2
a=1 , b=0 , c=-4
Sum of roots =0
Product of roots = -4

6. f(x)=(2-x)(x+2);
x= 2 , -2
a=-1 , b=0 , c=4
Sum of roots =0
Product of roots = -4

7. f(x)=(x-1)(x-3);
x= 1 , 3
a=1 , b=-4 , c=3
Sum of roots =4
Product of roots = 3

8. f(x)=(1-x)(x-3);
x= 1 , 3
a=-1 , b=4 , c=-3
Sum of roots =4
Product of roots = 3

9. f(x)=(x-1)(x+3);
x= 1 , -3
a=1 , b=2 , c=-3
Sum of roots =-2
Product of roots =-3

10. f(x)=(x+1)(x-3);
x= -1 , 3
a=1 , b=-2 , c=-3
Sum of roots =2
Product of roots = -3

11. f(x)=(3-x)(x+1).
x= -1 , 3
a=-1 , b=2 , c=3
Sum of roots =2
Product of roots = -3

Now I am taking your 10th quadratic equation to plot the graph which is attached with the post , a bit too untidy:
f(x)=(x+1)(x-3)
f(x) = x2-2x-3
Let f(x) = y ,
y=x2-2x-3

Wow ! I see that in my graph concavity is in upward direction and so a>0 , b<0 and c<0 !

Quote Quote by I like Serena View Post
Yep!

You seem to be picking things up nicely! ;)
Really ?!


Erm... all the answers are related to the parabolic equation...
So how would you deduce an answer without it?
I think the trick is to understand what the parameters mean geometrically.
Ehild's exercises should help with that.

And suppose you had simply y=bx+c.
What does it mean if either b<0, b=0, or b>0?
What happens if you add the extra term ax2?
if b<0 then
then x can be negative or positive and c>0.
y-c = Y is negative , so
Y=ax2+bx
then bx is positive or negative. If positive then a<0 , if negative then...
If b=0 then
Y=ax2
a<0
If b>0
then
Y=ax2+bx
then again same problem arises as in case b<0 !

How about this method :
P or Product of two roots = c/a
P = c/a
Since "P" is negative and c>0 so of course a<0 !

Quote Quote by Morgoth View Post
from your solutions x1,x2 you can see that b<0 would mean that:

x1= -b/2a + sqrt[b^2-4ac]/2a
x2= -b/2a - sqrt[b^2-4ac]/2a

[x1+x2]= -b/a => [x1+x2]/2 = -b/2a
so where would the mean value of your solutions (y(x)=0) lie? on which side of the x axis acording to b and a signs?

Hope this can help a little bit more.
Mean value lies in +x axis according to parabola in post 1.
[x1+x2]/2>0
So -b/2a >0
b>0
if we solve like
1/2a >0
then it would be like a>0/0 which is indeterminate.
So b>0
Now ,
-b/2a >0
Now this implies a<0 !!

Am I correct ?
Attached Thumbnails
graphingii_html_588251c8.png  
Morgoth
#24
Dec22-11, 06:31 AM
P: 127
no you are not correct.
when you said b>0 and said 1/2a>0 why did you then type a>0/0? And don't forget the inequality properties. What happens to the inequality sign when you multiply both sides with a (-).

(-1)*b/2a = -b/2a>0 means that b/2a<0 (two negative numbers [-1 and b/2a<0] multiplied will give you a positive one).

so if a>0, b<0.
if a<0, b>0.
Otherwise if they had the same sign you would have [itex]\frac{+}{+}[/itex]= + >0 or [itex]\frac{-}{-}[/itex]= + >0

And you cannot use what I typed above in a diagram that doesn't have the x axis calibrated correctly. I just said that for one more calculus way to find the sign of b, knowing the one of a, and not to answer your question but try to give you one additional help of understanding.
sankalpmittal
#25
Dec22-11, 09:34 AM
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Quote Quote by Morgoth View Post
no you are not correct.
when you said b>0 and said 1/2a>0 why did you then type a>0/0? And don't forget the inequality properties. What happens to the inequality sign when you multiply both sides with a (-).

(-1)*b/2a = -b/2a>0 means that b/2a<0 (two negative numbers [-1 and b/2a<0] multiplied will give you a positive one).

so if a>0, b<0.
if a<0, b>0.
Otherwise if they had the same sign you would have [itex]\frac{+}{+}[/itex]= + >0 or [itex]\frac{-}{-}[/itex]= + >0

And you cannot use what I typed above in a diagram that doesn't have the x axis calibrated correctly. I just said that for one more calculus way to find the sign of b, knowing the one of a, and not to answer your question but try to give you one additional help of understanding.
Oops !
My method is wrong but not the result.
I know [x1+x2]/2>0
So -b/2a >0
This means that -b/2a as a whole is a positive value , isn't it. So if I square on both sides then signs will obviously not change !
(-b/2a)2 >(0)2
b2/4a2 > 0
So b2>0 => b>0
Hence in -b/2a >0
a < 0 !!
Your method is easy I see.
Morgoth
#26
Dec22-11, 10:21 AM
P: 127
better than squaring- when you have something like:

-(something)> (else)
do:
(-1)*(-something)<(-1)*(else)
so (something)<-(else)

if (else)=0 and something here will be (something)=b/2a
you have:
-b/2a>0
means
b/2a<0

you have understood how to find a's sign so you know that a<0 or a>0 so respectively b>0 or b<0
ehild
#27
Dec22-11, 11:19 PM
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Quote Quote by sankalpmittal View Post


Now I am taking your 10th quadratic equation to plot the graph which is attached with the post , a bit too untidy:
f(x)=(x+1)(x-3)
f(x) = x2-2x-3
Let f(x) = y ,
y=x2-2x-3

Wow ! I see that in my graph concavity is in upward direction and so a>0 , b<0 and c<0 !

Really ?!
Why are you surprised?

I send you a bunch of parabolas to get familiar with them. Notice that they are symmetric, the axis of symmetry is parallel to the y axis and goes through the peak, and the peak is halfway between the roots. Notice that the concavity depends on the sign of the multiplier of x2 alone.
Which parabola is most similar to the one in your original post?

ehild
Attached Thumbnails
parabola1.jpg   parabola2.jpg  
sankalpmittal
#28
Dec23-11, 06:41 AM
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Quote Quote by Morgoth View Post
better than squaring- when you have something like:

-(something)> (else)
do:
(-1)*(-something)<(-1)*(else)
so (something)<-(else)

if (else)=0 and something here will be (something)=b/2a
you have:
-b/2a>0
means
b/2a<0

you have understood how to find a's sign so you know that a<0 or a>0 so respectively b>0 or b<0
Morgoth , I know rules of inequalities. By squaring on both sides I can easily get that b>0 and a<0. If I mess up with - sign here then condition become not so obvious : either a>0 and b<0 , OR a<0 and b>0.

Quote Quote by ehild View Post
Why are you surprised?

I send you a bunch of parabolas to get familiar with them. Notice that they are symmetric, the axis of symmetry is parallel to the y axis and goes through the peak, and the peak is halfway between the roots. Notice that the concavity depends on the sign of the multiplier of x2 alone.
Which parabola is most similar to the one in your original post?

ehild
I guess that the parabola in somewhat light blue colour is the approximate match of my parabola in post 1 : http://www.physicsforums.com/attachm...5&d=1324617111
Morgoth
#29
Dec23-11, 10:26 AM
P: 127
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.
ehild
#30
Dec23-11, 02:33 PM
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Quote Quote by sankalpmittal View Post

I guess that the parabola in somewhat light blue colour is the approximate match of my parabola in post 1 : http://www.physicsforums.com/attachm...5&d=1324617111
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
sankalpmittal
#31
Dec25-11, 07:49 AM
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Quote Quote by Morgoth View Post
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.

Quote Quote by ehild View Post
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
#32
Dec25-11, 08:16 AM
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Quote Quote by sankalpmittal View Post

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
#33
Dec25-11, 11:43 AM
P: 127
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
#34
Dec25-11, 11:51 AM
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Quote Quote by ehild View Post
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.

Quote Quote by Morgoth View Post
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.


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