| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Apr 2008 21:48:58 IST
|
|
|
|
|
|
|
5 Apr 2008 23:02:58 IST
|
|
|
![x^3+ax+b=0 \\ \\ \mbox{Critical points:} \ 3x^2+a=0 \ \Rightarrow x=\pm \sqrt{\frac{-a}{3}} \\ \\ \mbox{Now let the two critical pts be represented as} \ x_1,x_2 \\ \\ \mbox{Now for only one real root} \ f(x_1)*f(x_2)\leq0 \\ \\ \Rightarrow (\frac{-a}{3}*\sqrt{\frac{-a}{3}}+a*\sqrt{\frac{-a}{3}}+b)(b-((\frac{-a}{3})* \sqrt{\frac{-a}{3}}+a*\sqrt{\frac{-a}{3}}) \leq0 \\ \\ \Rightarrow b^2 - [\frac{-4a^3}{27}] \leq0 \\ \\ \Rightarrow 4a^3+27b^2\leq0](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/6/b/c/6bc6eed6da8dceb3283d58ca3958209689b93180.gif)
|
this reply: 20 points
(with 4 
in 4 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
5 Apr 2008 23:10:42 IST
|
|
|
@sboosy, For one real root , shouldn't it be  ?
Also , if a>0, the equation will always have one real root.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
6 Apr 2008 19:10:43 IST
|
|
|
i think it should be '<' sign only...because graph should be above the x-axis before x1 and below the axis after it...or vice versa...so they have opposite signs...
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
6 Apr 2008 19:13:36 IST
|
|
|
sboosy is right
nadeem think again !!!
|
all the best ... |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
6 Apr 2008 20:16:35 IST
|
|
|
I still don't understand.
Consider x3 - 2x +2
Here 4a3 + 27 b2 = - 32 + 108 > 0 . But the equation has only 1 real root.
You can also verify that if it is less than 0 , the equation will have 3 roots.
|
this reply: 7 points
(with 1
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
7 Apr 2008 07:53:20 IST
|
|
|
Firstly, in  , if  then because the derivative  and the fact that a cubic equation has atleast one real root, this equation will have only one real root. So, the discussion is for the case when  Now, as sboosy pointed out, the critical points are  and  Now, if you draw an approximate graph, you can easily see that if  the equation has three real roots and if  , we have only one real root.
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif){kind=icon}
1
![[Thumb - GraphPickup.png]](/upload/2008/4/6/ee8da5233a9894cbed526f8f3a4f4ced_26686.png_thumb)
