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Solve the equation: \[(2+\frac{ 3 }{ 2 })^2-(2+\frac{3 }{ x })-12=0\]
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@dumbcow you mind helping me with this?
@dumbcow now any hints on how to solve this one? lol
try to isolate the "3/x" 2+ 3/2 = 7/2
\[(\frac{7}{2})^2 -12-2 = \frac{3}{x}\]
so then its \[\frac{ 49 }{ 4 }-12-2=\frac{ 3 }{ x }\]
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which equals \[\frac{ -7 }{ 4 }=\frac{ 3 }{ x }\]
correct ... cross-multiply and solve for x
x=-12/7?
yep
alright thanks man
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:)
and i just noticed im an idiot -_-
lol the beginning (2+3/2)^2 is (2+3/x)^2
\[(2+\frac{ 3 }{ x })^2-(2+\frac{ 3 }{ x })-12=0\]
yeah i was afraid of that... i should have said something ok its not hard just make a substitution let u = 2 + 3/x --> u^2 - u - 12 = 0 solve for u
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