# Homework Help: Polynomial > 0 proof

1. Jul 27, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
Show that if $a > 0$,
$ax^2 + 2bx +c >= 0$ for all values of $x$ iff $b^2 -ac <=0$

2. Relevant equations

3. The attempt at a solution
Well, I don't think this really makes any sense but away we go.

All I did was take $ax^2 + 2bx +c >= 0$ and compleate the square.

I got $a^2(x+b/a)^2 >= b^2 -ac$

I tried to think if this is telling me anything. I just said the square of a real number is greater then or equal to 0 ∴$b^2 -ac$ must be less then or equal to 0 for this equality to hold.
More then likely this is horse crap. What do you think?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 27, 2013

### hilbert2

You're doing that the wrong way around. You should start by assuming that ax2+bx+c<0 and show that this assumption leads to a logical contradiction.

3. Jul 27, 2013

### Jbreezy

Is this a common way to show something is by contradiction? Comes up a lot?

4. Jul 27, 2013

### HallsofIvy

Well, I wouldn't call it "horse crap"- don't know horses that well!

What you have done is pretty good! Yes, completing the square gives you $ax^2+ 2bx+ c= a(x^2+ (2b/a)xc+ b^2/a^2- b^2/a^2)+ c$$= a(x^2+(2b/a)x+ b^2/a^2)- b^2/a+ c$$= a(x+ b/a)^2+ c- b^2/a$ or $= a(x+ b/a)^2- (b^2/a- c)[/tex]. That will be larger than 0, [itex]a(x+ b/a)^2- (b^2/a- c)> 0$, as long as $a(x+ b/a)^2> b^2/a- c$. Since a> 0, the left side of that cannot be negative so that will certainly be true as long as the right side is negative: $b^2/a- c< 0$ and, since again a> 0, we can multiply by it to get $b^2- ac< 0$.

5. Jul 27, 2013

### Darth Frodo

Hint: Draw a sketch of the polynomial and see where the sketch cuts the x-axis.

6. Jul 27, 2013

$= a(x+ b/a)^2- (b^2/a- c)[/tex]. How do you get a-c on the denomator there? With just one a next to the a(x+b/a)? How did you get the subtraction sign? Is my reasoning wrong? I know the one dude said to show a contradiction. By starting with ax^2 + 2bx +c < 0 $a^2(x+b/a)^2 >= b^2 -ac$ My left side will also not be negative or less then 0 so my equality will hold as long as $b^2 -ac$ <=0. Is that what you were saying? Like I said I had a hard time to follow your post ...reading wise I meant. Thanks 7. Jul 27, 2013 ### Theorem. Take your time to walk through it. It basically is what you said! but in more detail. You didn't seem to be too confident with what you did, but you do indeed do have the right idea. Go through each step of what HallsofIvy wrote, and try and justify it. In response to Hilbert2's post: Sure, that is one way of doing it. But the OP's method is definitely NOT the wrong way of doing it. It is just a direct argument that is logically equivalent. I would suggest the OP do it by contradiction as an exercise, it might help them understand better but it is in no way necessary. 8. Jul 27, 2013 ### Jbreezy Thanks amigo! 9. Jul 27, 2013 ### ehild You have to prove that $ax^2 + 2bx +c >= 0$, you can not start the prof by assuming it is true. Completing the square is a good start. Rewrite the original expression $ax^2 + 2bx +c$ by completing square and show that it is a sum with each term greater or equal to zero. You have completed the square correctly $ax^2 + 2bx +c =a(x+b/a)^2+[c-b^2/a]$ Multiplying the whole expression by the positive $a$ does not change the sign. So you have to prove that $a^2(x+b/a)^2+[ca-b^2]$is not negative. You are right, the first therm is a square, so it is ≥0. What about the term in the brackets $[ac-b^2]$? Given that $b^2 -ac <=0$ ? ehild 10. Jul 27, 2013 ### Theorem. I don't think what the OP was doing was a formal proof. He simply started by assuming it was true to see what the statement would required. From that he can work backwards, which is a fairly good technique! I guess I just should have clarified that the argument still needed to be formalized : ) 11. Jul 27, 2013 ### Jbreezy Yeah I see so $[ac-b^2]$>0 and we are happy. So I learned never start the proof with the thing your trying to prove. And be formal! 12. Jul 27, 2013 ### hilbert2 This is exactly what I meant with my initial comment. You can not prove a statement by first assuming it is true and then proceeding to derive another true statement from it. In fact, it can be shown that it is possible to "prove" ANY statement if one starts by assuming that a false statement is true. Proof by contradiction is a common method to prove mathematical theorems, often it is simpler than a "direct" proof. It even has a fancy Latin name, "reductio ad absurdum". 13. Jul 27, 2013 ### Jbreezy Honestly, I was told that before not to start with what your trying to prove I guess I had a brain fart. Thanks for the advice. 14. Jul 27, 2013 ### Theorem. Sometimes it is useful to informally starting with the result in order to extract certain things the result would imply. Suppose you want to prove the statement A is true. Suppose that , informally, we assume A is true, and somehow arrive at the statement B using A (A=>B). Now we know that if we show that B cannot hold then A cannot hold also. We can use this to our advantage in a formal argument. But clearly, in a formal proof we can't start by assuming our result, then what are we proving! 15. Jul 28, 2013 ### ehild Yes, that method has show that (ax+b)^2 and ac -b^2 come into somehow. Neither of them can be negative. You say "I try their sum". Lot of proofs start with some true statement. So you can start your proof: (ax+b)2 ≥ 0 as it is a square. Given that b2-ac ≤ 0, so ac -b^2 ≥ 0. The sum of two non-negative expression is also non-negative, (ax+b)2+ac-b2≥0 Expand the square and simplify: a2x2+2axb+[STRIKE]b2[/STRIKE]+ac-[STRIKE]b2[/STRIKE]>0 But a>0 was also given, you can divide with it, the inequality holds, and you get the result you had to prove. Such proofs looks as if coming into the mind of the mathematician in his dreams. Sometimes it really happens. But is is a proof. ehild 16. Jul 28, 2013 ### Jbreezy Follow up using the results from above. I figured it would be best to post it in this thread. Prove Schwarz's inequlity by considering the expession $(a_1x +b_1)^2 + (a_2x +b_2)^2 + ...+ (a_nx +b_n)^2$, collecting terms and aplying ex. 4(which is the proof earlier in the thread) This is what I'm trying but don't give it away. $(a_1x +b_1)^2 + (a_2x +b_2)^2 + ...+ (a_nx +b_n)^2 = ((a_1x)^2 + 2a_1b_1x + b_1^2) + ((a_2x)^2 + 2a_2b_2x + b_2^2)+...+ ((a_nx)^2 + 2a_nb_nx + b_n^2)$ Then I said this could be writen in summation notation but let me take the $n^{th}$ term and use that for my general case. So this looks very similar to $ax^2 + 2bx +c >= 0$ So now I tried to apply the result that was generated above o get to Cauhy-Schwarz inequality. Which my book states as.... $(a_1b_1+ a_2b_2 +...+ a_nb_n)^2 <= (a_1^2 + ...a_n^2)(b_1^2 + ...b_n^2)$ I really did not get anywhere by applying the result from $((a_nx)^2 + 2a_nb_nx + b_n^2)$ I just did as we did for earlier. Compleated the square and got, $(x+b_n/a_n)^2 >= -b_n^2/a_n^2 + b_n^2/a_n^2$ So the right side is 0 and I'm not really sure about this now. It is not the Cauchy Schwarz inequality. 17. Jul 28, 2013 ### pasmith That won't work. You are trying to prove $$\left(\sum_{i=1}^n a_i b_i \right)^2 \leq \left(\sum_{i=1}^n a_i^2\right) \left( \sum_{i=1}^n b_i^2\right).$$ Follow through on your idea and write the initial expression as a sum: $$\sum_{i=1}^n (a_i x + b_i)^2 \geq 0.$$ Do you see how to turn that into the form [itex]Ax^2 + Bx + C \geq 0$?

18. Jul 28, 2013

### Jbreezy

Yeah, after I posted I began went back to trying to write it as a sum. I have run into an issue though. I don't want the answer just something to think about.

Because
$\sum_{i=1}^n (a_i x + b_i)^2 \ = x^2\sum_{i=1}^n (a_i)^2 + 2x\sum_{i=1}^n (a_ib_i) +\sum_{i=1}^n (b_i)^2$

Like it was said..
$x^2\sum_{i=1}^n (a_i)^2 + 2x\sum_{i=1}^n (a_ib_i)+\sum_{i=1}^n (b_i)^2 >= 0$
Just to manipulate it

$x^2\sum_{i=1}^n (a_i)^2 +\sum_{i=1}^n (b_i)^2 >= -2x\sum_{i=1}^n (a_ib_i)$

But like it was said I'm trying to prove...
$\left(\sum_{i=1}^n a_i b_i \right)^2 \leq \left(\sum_{i=1}^n a_i^2\right) \left( \sum_{i=1}^n b_i^2\right).$
Don't give it away!