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Title: Quadratic equation Post by Christine on Feb 24th, 2013, 12:19pm a, b, c are nonzero real numbers. m is a nonzero real root of the ax^2 + bx + c = 0 n ............................................ -ax^2 + bx + c = 0 Does 1/2(ax^2) + bx + c always have root between m and n? Prove it. Otherwise give a counterexample. |
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Title: Re: Quadratic equation Post by pex on Feb 24th, 2013, 1:02pm on 02/24/13 at 12:19:33, Christine wrote:
Yes. Let f1(x) = ax2 + bx + c, f2(x) = -ax2 + bx + c, and f3(x) = (1/2)ax2 + bx + c. [hideb]As f1 and f2 obviously cannot share any roots (f1(x) = f2(x) if and only if x=0, which is not a root because c is nonzero), there are four cases to consider: 1. f2(m) < 0 and f1(n) < 0 2. f2(m) < 0 and f1(n) > 0 3. f2(m) > 0 and f1(n) < 0 4. f2(m) > 0 and f1(n) > 0 Now consider the function f1 - f2. Clearly (f1 - f2)(x) = 2ax2, which never changes sign. This observation rules out cases 1 and 4. For the other cases, observe that f3 is a convex combination of the other two functions, f3 = (3/4)f1 + (1/4)f2, so that f3(x) is always "between" f1(x) and f2(x). So in case 2, f3(m) < 0 and f3(n) > 0, and by continuity there exists r between m and n such that f3(r) = 0. Case 3 is similar, but with f3(m) > 0 and f3(n) < 0.[/hideb] |
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Title: Re: Quadratic equation Post by ravibhole_1 on Apr 6th, 2013, 7:29pm Not always if the equation is perfect square where m=n Roots of 1/2(ax^2) + bx + c does not belong to the given criteria |
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Title: Re: Quadratic equation Post by Rosiethomas on Jul 1st, 2013, 3:36am Not always.. You can check by putting a = 1, b = 4, c = 4 and for formulae you can refer below URL removed. The answer above is wrong anyway. |
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