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can u solve it?
solve for x?
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?
2(x-1)^3=x+1 noo sorry not really can u tell me how to do it just by factoring etc?
lol that's what im thinking too
@myininaya pls help here
Hey @lgbasallote what was wrong with your way?
It seems to be awesome! :)
i don't know what to do next lol
Try doing a little of testing possible rational zeros for what you have
Synthetic division will come in handy
you heard her @louis413 ;)
lol im only in alg/trig in high schooll.. loll wut is that?
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
algebra/trig? which one?
well im takling the algebra2/trig course right now at summer school
synthetic division and looking at possible rational zeros are in algebra
ok so what have you learned then?
factoring... quadratic formula completeing the squares etc
And is your equation written correctly?
yes can it be solved like 2(x-1)^2(x-1)-1(X-1)=0
I can only see to find a real approximation by looking at the graph and seeing where it crosses the x-axis
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 ?
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)
hope i was able to explain well..