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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Apr 2008 19:19:58 IST
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The equation has a)at least 1 real solution b)at least 3 real solutions c)exactly one real solution d)All of these
ans-c i am getting x= 2 and as the solutions of this equation (taking log on both sides)
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Apr 2008 19:33:53 IST
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Re:triplets
How do I use LATEX here?
Obviously there cannot be 2 real roots(it may have 3) to this equation as it is cubic.
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Let us learn to dream, gentlemen, and then perhaps we shall learn the truth.
- August Kekule |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Apr 2008 20:09:30 IST
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i am getting x=2 as a double root....
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Apr 2008 21:55:40 IST
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there are 3 roots 2, 2^-1/3, 1/4
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 21 Apr 2008 22:07:46 IST
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| take log with base 2 on both sides. let log2 x = t on taking log u get (3/4t2 + t - 5/4)t = 1/2 => 3t3 +4 t2 -5t -2 =0. => 3t3 +6 t2 -2 t2 -4t -t -2 =0. =>3t2(t+2) -2t(t+2) -(t+2) =0 =>(t+2)(3t2 - 2t -1)=0 =>(t+2)(3t+1)(t -1)=0 => t = 1,-1/3,-2. so log2 x =1,-1/3,-2. => x= 2,2-1/3,2-2. so 3 roots. | |
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