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lgbasalloteBest ResponseYou've already chosen the best response.0
here's my thought let x  1 = a therefore a +1 = x 2a^3 = a + 1 + 1 2a^3 = a + 2 2a^3  a  2 = 0 does that give you any ideas?
 one year ago

louis413Best ResponseYou've already chosen the best response.0
2(x1)^3=x+1 noo sorry not really can u tell me how to do it just by factoring etc?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
lol that's what im thinking too
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
@myininaya pls help here
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
Hey @lgbasallote what was wrong with your way?
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
It seems to be awesome! :)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i don't know what to do next lol
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
Try doing a little of testing possible rational zeros for what you have
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
Synthetic division will come in handy
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
you heard her @louis413 ;)
 one year ago

louis413Best ResponseYou've already chosen the best response.0
lol im only in alg/trig in high schooll.. loll wut is that?
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
oh wait i think i made a mistake anyways... let me try again but yeah that is what i would do test to see if any possible rational zeros work
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
algebra/trig? which one?
 one year ago

louis413Best ResponseYou've already chosen the best response.0
well im takling the algebra2/trig course right now at summer school
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
synthetic division and looking at possible rational zeros are in algebra
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
ok so what have you learned then?
 one year ago

louis413Best ResponseYou've already chosen the best response.0
factoring... quadratic formula completeing the squares etc
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
And is your equation written correctly?
 one year ago

louis413Best ResponseYou've already chosen the best response.0
yes can it be solved like 2(x1)^2(x1)1(X1)=0
 one year ago

myininayaBest ResponseYou've already chosen the best response.2
I can only see to find a real approximation by looking at the graph and seeing where it crosses the xaxis
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
would you like to go for a trignometric solution? might seem complicated but is easy.. involves inverse function just a bit..and sin 3x etc formullas ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
well.. in 2a^3  a  2 = 0 let a = m sin y => 2m^3 sin^3 y  msin y 2 =0 multiplying both sides by (2/m^3) => 4sin^3 y + 2sin y/(m^2) + 4/(m^3) =0 our aim is to convert this in the form of 4sin^3 y + 3 siny which will convert it into sin 3y and simplify things we have to chose a particular value of m for that purpose in our case 2/m^2 = 3 or m = sqrt(3/2) and sqrt(3/2) thus we have these 2 eqns : 4sin^3 y + 3sin y + 8sqrt(2)/(3sqrt(3)) = 0 and 4sin^3 y + 3sin y  8sqrt(2)/(3sqrt(3)) = 0 => sin 3y =  8sqrt(2)/(3sqrt(3)) and =>sin 3y = 8sqrt(2)/(3sqrt(3)) now using any scientific calculator, we may find value of y,then a(=msin y) and then finally x(=a+1)
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.3
hope i was able to explain well..
 one year ago
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