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nilankshi
Group Title
obtain all the zeroes of 3x^4+6x^22x^210x5,if two zeroes are root5/3 and root 5/3
 2 years ago
 2 years ago
nilankshi Group Title
obtain all the zeroes of 3x^4+6x^22x^210x5,if two zeroes are root5/3 and root 5/3
 2 years ago
 2 years ago

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Lumenaire Group TitleBest ResponseYou've already chosen the best response.0
I assume you mean: 3x^4+6x^32x^210x5
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
the roots aree \(\large \sqrt{\frac{5}{3}}\) and \(\large \sqrt{\frac{5}{3}}\) right? or is the root applied on 5 only?
 2 years ago

Lumenaire Group TitleBest ResponseYou've already chosen the best response.0
is there a square root sign at all? I understood that 5/3 and 5/3 are the roots
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
divide the whole system by its two factors that are given (or try synthetic division) After two whole divisions by each factor, you 'll be left with a quadratic as the quotient, factorising which you'll get you the other two roots.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
he/she said root 5/3 and root 5/3..she already said zeroes so i assume root>radical
 2 years ago

nilankshi Group TitleBest ResponseYou've already chosen the best response.0
yes that is right..
 2 years ago

nilankshi Group TitleBest ResponseYou've already chosen the best response.0
root is on both..
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
\[\mathbb{yes}\]
 2 years ago

Lumenaire Group TitleBest ResponseYou've already chosen the best response.0
and is it an x^3 in the equation?
 2 years ago

Lumenaire Group TitleBest ResponseYou've already chosen the best response.0
(6x^2) should be 6x^3
 2 years ago

nilankshi Group TitleBest ResponseYou've already chosen the best response.0
yes it is in equion
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
\(\large x = \sqrt{\frac{5}{3}}\) \(\large x^2 = \frac{5}{3}\) \(\large 3x^2 = 5\) \(\large 3x^2  5 = 0\) as for the other zero... \(\large x = \sqrt{\frac{5}{3}}\) \(\large x^2 = \frac{5}{3}\) \(\large 3x^2 = 5\) \(\large 3x^2  5 = 0\) so two of the factors would be (3x^2  5)^2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
something doesnt feel right...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
hmm idk how to do this sorry... @apoorvk may know though
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
Sorry, Trance. :) and Nilakshi. I 'll work on the solution now
 2 years ago

TranceNova Group TitleBest ResponseYou've already chosen the best response.0
Thanks :) @Rohangrr if you want to talk about it, please make a new question.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.4
I would multiply out \[ (x\sqrt{5/3})(x+\sqrt{5/3}) = x^2\frac{5}{3}\] we can write this as \(3x^25\) (why? see below) divide this into the original equation to get \(x^2+2x+1 \) now factor this to get the remaining roots. * we are looking for \((x^2\frac{5}{3}) (x^2+2x+1) =0\) if we multiply both sides by 3 we get \( 3(x^2\frac{5}{3})(x^2+2x+1) =0\) \( (3x^25) (x^2+2x+1)=0\)
 2 years ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
Hmm, I got so bored, Phi already posted it. :/ Good job, Phi
 2 years ago
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