Register to reply 
Problem on quadratic polynomial. 
Share this thread: 
#19
Dec2111, 06:28 AM

P: 758

This implies a<0 , isn't it ? 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. 


#20
Dec2111, 07:41 AM

HW Helper
Thanks
P: 10,780

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)=x^{2}; f(x)=x^{2}; f(x)=(x2)^{2}; f(x)=(x+2)^{2}; f(x)=(x2)(x+2); f(x)=(2x)(x+2); f(x)=(x1)(x3); f(x)=(1x)(x3); f(x)=(x1)(x+3); f(x)=(x+1)((x3); f(x)=(3x)(x+1). 


#21
Dec2111, 04:22 PM

HW Helper
P: 6,189

You seem to be picking things up nicely! ;) 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 ax^{2}? 


#22
Dec2111, 04:47 PM

P: 127

x1= b/2a + sqrt[b^24ac]/2a x2= b/2a  sqrt[b^24ac]/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. 


#23
Dec2211, 05:47 AM

P: 758

x=0 a=1 , b=0 , c=0 Sum of roots = product of roots = 0 a=1 , b=0 , c=0 Sum of roots = product of roots = 0 a=1 , b=4 , c=4 Sum of roots = Product of roots = 4 a=1 , b=4 , c=4 Sum of roots = 4 Product of roots = 4 a=1 , b=0 , c=4 Sum of roots =0 Product of roots = 4 a=1 , b=0 , c=4 Sum of roots =0 Product of roots = 4 a=1 , b=4 , c=3 Sum of roots =4 Product of roots = 3 a=1 , b=4 , c=3 Sum of roots =4 Product of roots = 3 a=1 , b=2 , c=3 Sum of roots =2 Product of roots =3 a=1 , b=2 , c=3 Sum of roots =2 Product of roots = 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)(x3) f(x) = x^{2}2x3 Let f(x) = y , y=x^{2}2x3 Wow ! I see that in my graph concavity is in upward direction and so a>0 , b<0 and c<0 ! then x can be negative or positive and c>0. yc = Y is negative , so Y=ax^{2}+bx then bx is positive or negative. If positive then a<0 , if negative then... If b=0 then Y=ax^{2} a<0 If b>0 then Y=ax^{2}+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 ! [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 ? 


#24
Dec2211, 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. 


#25
Dec2211, 09:34 AM

P: 758

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} b^{2}/4a^{2} > 0 So b^{2}>0 => b>0 Hence in b/2a >0 a < 0 !! Your method is easy I see. 


#26
Dec2211, 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 


#27
Dec2211, 11:19 PM

HW Helper
Thanks
P: 10,780

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 x^{2} alone. Which parabola is most similar to the one in your original post? ehild 


#28
Dec2311, 06:41 AM

P: 758




#29
Dec2311, 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 b^{2}>0 => b>0" in your previous post. b^{2} >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: A^{2} > B^{2} 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: x^{2} > 2 the answer is that: x<√2 or x>+√2 x^{2}<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 a^{2}=b^{2} here without a problem (as there was with 1<0 above) On the other hand if you have that a^{2}=b^{2} then you write that a= ±b. 


#30
Dec2311, 02:33 PM

HW Helper
Thanks
P: 10,780

If the maximum is really at x=0, what is b then? ehild 


#31
Dec2511, 07:49 AM

P: 758

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. 


#32
Dec2511, 08:16 AM

HW Helper
Thanks
P: 10,780

ehild 


#33
Dec2511, 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. 


#34
Dec2511, 11:51 AM

P: 758




Register to reply 
Related Discussions  
Finding quadratic maclaurin polynomial  Calculus & Beyond Homework  1  
Linear Algebra  Quadratic polynomial to Matrix  Calculus & Beyond Homework  4  
Quadratic equation, A.P. and G.P. related problem problem  Precalculus Mathematics Homework  2  
Quadratic equations and inequalities / applications of quadratic functions question  Precalculus Mathematics Homework  3  
Solving a Polynomial Equation using quadratic techniques.  Precalculus Mathematics Homework  21 