by sankalpmittal
P: 700
 Quote by I like Serena 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 by NascentOxygen 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 by ehild 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 $$y=a^2+bx+c=a\left((x+\frac{b}{2a})^2+(c/a-\frac{b^2}{4a^2})\right)$$ 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 ?

 HW Helper Thanks P: 9,819 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).
HW Helper
P: 6,189
 Quote by sankalpmittal 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 by sankalpmittal 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?
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.
P: 700
 Quote by ehild 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 by I like Serena 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 !

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

 Quote by Morgoth 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

 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 $\frac{+}{+}$= + >0 or $\frac{-}{-}$= + >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.
P: 700
 Quote by Morgoth 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 $\frac{+}{+}$= + >0 or $\frac{-}{-}$= + >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.
 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
HW Helper
Thanks
P: 9,819
 Quote by sankalpmittal 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

P: 700
 Quote by Morgoth 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 by ehild 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
 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.
HW Helper
Thanks
P: 9,819
 Quote by sankalpmittal 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
P: 700
 Quote by Morgoth 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 by ehild 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.
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Thanks
P: 9,819
 Quote by sankalpmittal 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
 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.
P: 700
 Quote by ehild 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 by 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.
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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