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July 28th, 2016, 01:32 AM   #1
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roots of the equation

If $\displaystyle 2a+b+c=0\;,$ Then at least one roots of the equation $\displaystyle 24ax^2+4bx+c=0$ lie between.

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$\displaystyle (a)\; [0,1]$

$\displaystyle (b)\; [0,2]$

$\displaystyle (c)\;\; [-1,0]$

$\displaystyle (d)\;\; \left[-\frac{1}{2},\frac{1}{2}\right]$
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July 28th, 2016, 04:45 AM   #2
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What have you done so far?

There is a psychological barrier to solving this problem. We learn the quadratic formula using a, b, and c as parameters. This problem involves a quadratic, but is using a, b, and c in a different sense. If that is part of your difficulty in attacking the problem, I suggest you use different variables. (Changing the variables is totally unnecessary mathematically; it is just a psychological way to reduce confusion.)

$2u + v + w = 0\ and\ 24ux^2 + 4vx + w = 0 \implies$

$w = -\ 2u - v\ and\ so\ 24ux^2 + 4vx - 2u - v= 0.$

I cannot tell whether you have given the exact wording of the problem. It is implied in how you have expressed the problem that it is asking about REAL roots. If a quadratic has real roots, its discriminant is non-negative. So

$(4v)^2 - 4(24u)(-\ 2u - v) \ge 0.$ Can you take it from here?
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July 28th, 2016, 06:19 AM   #3
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It is asking about real roots because the given ibtervals are real (assuming that $a$, $b$ and $c$ in the question are real).

In addition to JeffM1's suggestion, you should consider that the to roots $r_1$ and $r_2$ of the quadratic $ax^2 +bx +c$ satisfy
$$r_1 r_2 = \tfrac{c}{a}$$
and
$$r_1 + r_2 = -\tfrac{b}{a}$$
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Last edited by v8archie; July 28th, 2016 at 06:31 AM.
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July 28th, 2016, 07:15 AM   #4
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Thanks JeffM1 and v8archie, I have seems that we can use Rolles Theorem here, But i did not understand how can i apply, Thanks
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July 28th, 2016, 09:24 AM   #5
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Quote:
Originally Posted by panky View Post
Thanks JeffM1 and v8archie, I have seems that we can use Rolles Theorem here, But i did not understand how can i apply, Thanks
So are you Ok now?
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