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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Aug 2007 19:03:11 IST
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if c is positive and 2ax2+3bx+5c=0 doest not have any real roots then prove that 2a-3b+5c>0
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8 Aug 2007 19:39:11 IST
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D<0
9b^2-40ac<0
9b^2<40ac
here c , b^2 are +ve therefore a should also be +ve
rt(40ac)>3b
now by a.m. - g.m. inequality 2a+5c/2>rt(10ac)
i.e. 2a+5b>rt(40ac)>3b
therefore 2a-3b+5c>0
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this reply: 5 points
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