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(x+3)^2  (3x1)^2 = x^3 + 4?
i've tried this three times and I still get the wrong answer!
 2 years ago
 2 years ago
(x+3)^2  (3x1)^2 = x^3 + 4? i've tried this three times and I still get the wrong answer!
 2 years ago
 2 years ago

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a123Best ResponseYou've already chosen the best response.0
\[(x+3)^2  (3x  1)^2 = x^3 + 4\]
 2 years ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
unless you know the cubic formula, your best bet is to use a graphing calculator to find the approximate solutions.
 2 years ago

a123Best ResponseYou've already chosen the best response.0
we're not allowed to use a calculator. the answer is supposed to come out to be  2/3
 2 years ago

a123Best ResponseYou've already chosen the best response.0
but I don't know how to get there
 2 years ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
ok so there's a typo somewhere in the problem then, my guess is that x^3 is one of the typos
 2 years ago

a123Best ResponseYou've already chosen the best response.0
wait, I made a mistake typing out the problem. it's actually \[(x+3)^3\] as the first one
 2 years ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
oh ok thx, one sec
 2 years ago

a123Best ResponseYou've already chosen the best response.0
so \[(x+3)^3  (3x  1)^2 = x^3 + 4\]
 2 years ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
\[\large (x+3)^3 (3x1)^2 = x^3 + 4\] \[\large x^3+9x^2+27x+27 (9x^26x+1) = x^3 + 4\] \[\large x^3+9x^2+27x+27 9x^2+6x1 = x^3 + 4\] \[\large x^3+9x^2+27x+27 9x^2+6x1  x^3  4=0\] \[\large 33x+22=0\] \[\large 33x=22\] \[\large x=\frac{22}{33}\] \[\large x=\frac{2}{3}\]
 2 years ago

a123Best ResponseYou've already chosen the best response.0
thank you! i see exactly where I went wrong.
 2 years ago
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