July 28th, 2016, 01:32 AM  #1 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16  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. Options:: $\displaystyle (a)\; [0,1]$ $\displaystyle (b)\; [0,2]$ $\displaystyle (c)\;\; [1,0]$ $\displaystyle (d)\;\; \left[\frac{1}{2},\frac{1}{2}\right]$ 
July 28th, 2016, 04:45 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
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 nonnegative. So $(4v)^2  4(24u)(\ 2u  v) \ge 0.$ Can you take it from here? 
July 28th, 2016, 06:19 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra 
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 $$r_1 + r_2 = \tfrac{b}{a}$$ Last edited by v8archie; July 28th, 2016 at 06:31 AM. 
July 28th, 2016, 07:15 AM  #4 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16 
Thanks JeffM1 and v8archie, I have seems that we can use Rolles Theorem here, But i did not understand how can i apply, Thanks

July 28th, 2016, 09:24 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  

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