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Topic: Quadratic equation (Read 4760 times) |
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Christine
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Posts: 159
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Quadratic equation
« on: Feb 24th, 2013, 12:19pm » |
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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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pex
Uberpuzzler
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Posts: 880
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Re: Quadratic equation
« Reply #1 on: Feb 24th, 2013, 1:02pm » |
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on Feb 24th, 2013, 12:19pm, Christine wrote: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? |
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Yes. Let f1(x) = ax2 + bx + c, f2(x) = -ax2 + bx + c, and f3(x) = (1/2)ax2 + bx + c.
| hidden: | 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. |
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ravibhole_1
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Posts: 6
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Re: Quadratic equation
« Reply #2 on: Apr 6th, 2013, 7:29pm » |
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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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Rosiethomas
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Posts: 4
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Re: Quadratic equation
« Reply #3 on: Jul 1st, 2013, 3:36am » |
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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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| « Last Edit: Jul 2nd, 2013, 8:55am by Grimbal » |
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