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![[Post New]](/templates/default/images/icon_minipost_new.gif) 26 Feb 2007 21:35:55 IST
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Hi all, plz help me with this problem If a,b,c,d,e,x are real and (a2+b2+c2+d2)x2 -2(ab+bc+cd+de)x+(b2+c2+d2+e2)<=0, then a,b,c,d,e are in a. AP b. GP c. HP d. none <= means less than or equal to.
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-Ramya |
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26 Feb 2007 22:07:13 IST
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For a quadratic expression ax2 +bx +c to be less than or equal to zero for all x,
'a' should be less than 0 and discriminant <=0
but 'a' here equals (a2+b2+c2+d2) which can never be less than 0.
therefore I think the question must be >=0 not <=0
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27 Feb 2007 02:59:05 IST
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ya, i know that, but i'm told that question is right, so the option has to be (d) right?. Bcoz if u go even by the graphical method, it cant be solved.
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-Ramya |
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it can be rearranged like this
a2x2+b2x2+c2x2+d2x2 -2abx-bcx-cdx-dex+b2+c2+d2+e2 and then
(ax-b)2+(bx-c)2+(cx-d)2+(dx-e)2 =0
then x=b/a=c/b=d/c=e/d
now you can match your options
it gives they are in GP
you are giving me a salute i think
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