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A Problem on quadratic functions

  1. Jul 26, 2013 #1
    1. The problem statement, all variables and given/known data

    a) Let f(x) = ax^2+ bx + c; ∀x ∈ ℝ; where a(≠ 0), b and c ∈ ℝ are constants,
    (i) Prove that f(k) =(64+(b^2−4ac)^2)/(64a) ; where k = −(b/2a)+(1/a)+(b^2−4ac)/(8a).
    (ii) Hence, without using graphs and without using your knowledge on quadratic
    functions and equations, prove that (∀x ∈ ℝ f(x) > 0) [itex]\Rightarrow[/itex] (a > 0 and b^2 − 4ac< 0).

    2. Relevant equations

    3. The attempt at a solution
    i have already proven the first part but i m stuck in the 2nd part that says prove from "hence" . otherwise i can show that there exist M in R such that for all x in R ,M<=f(x) and M=inf(f(x)), and there exist x' in R such that f(x')=M, so i can take an argument that x' to be in real and if this happens (a>0 and b^2-4ac<0) ,M>0, so f(x)<0, but on that approach i m not using any above proven things to get my second answer(hence). if any one can help would be great. sorry for my english.
  2. jcsd
  3. Jul 26, 2013 #2
    You want to prove that the statement ∀x ∈ ℝ f(x) > 0) [itex]\Rightarrow[/itex] (a > 0 and b^2 − 4ac< 0) is a true statement. In this paticular case, if I were you I would start by thinking about when/under what conditions this statement isn't true. Do you know when an implication (p implies q) isn't true?

    Then I would try to find a contradiction. Are you familiar with proof by contradiction? If not, I would be happy to explain.

    This probably isn't the only way to do this, but it's the first solution I thought of and it seems well suited to this problem (I checked that the method works here).

  4. Jul 26, 2013 #3
    thanks for the reply , p[itex]\Rightarrow[/itex]q is false only when the p is true and q is false. the problem I'm having is i cant find a way to say it from using f(k). it just a constant right ?. i cant use it to talk show that for all x f(x) is greater than 0. you mean that I should show the p (for all x f(x)>0 ) implies the negation of (a>0 & b^2-4ac<0) proposition is always false right ?
  5. Jul 26, 2013 #4
    I'm sorry, I didn't get the emphasis on "hence"...I just suggested a less convoluted way of proving it (if you take (∀x ∈ ℝ f(x) > 0) to be true and (a > 0 and b^2 − 4ac< 0) to be false you can easily arrive at a contradiction). I'll look at it again, although off the top of my head I don't see a connection to the part you've already proved.

    Can you clarify the context here? What sort of level is this book/course? Have you recently covered a method that could be relevant?

    I'll try to help as much as I can, however there are plenty of people waaay smarter than me on these forums who might be better able to help, or see something I'm missing.
  6. Jul 26, 2013 #5
    its on Mathematics of Bsc Engineering , level 1. its came up with the real analysis topic of functions.
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