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Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0@mukushla @experimentX @Ishaan94

alanli123
 2 years ago
Best ResponseYou've already chosen the best response.0then look at x intercepts?

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0i wabt to solve this manually not by graph

alanli123
 2 years ago
Best ResponseYou've already chosen the best response.0Ok use the rational roots theorem and plug in possible root values

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.3my guess is only one. and that one root should be negative.

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.3i used descartes rule of signs.

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0hw u guys are using that rule can u teach me that @Ishaan94

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0i think that method is very easy!

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.3i can link you, http://mathworld.wolfram.com/DescartesSignRule.html. yes it's quite easy.

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0it says file not found @Ishaan94

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.3haha really? try again, http://mathworld.wolfram.com/DescartesSignRule.html

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0@Ishaan94 that helps a lot.....i am happy that i learnt new concept ....thxx

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0since there is no change of sign it does not have any + ve roots
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