- #1
odolwa99
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My final answer matches that of the textbook, but do I need to change the < to > at any point as I solve this? I ask because, if I assume x is 1 (until I solve for x), then the question statement and parts of the solution are made untrue.
Solve the following for x [itex]\in\mathbb{R}:[/itex] [itex]\frac{x}{2x-1}<-2[/itex]
[itex]\frac{x(2x-1)^2}{2x-1}<-2(2x-1)^2[/itex]
[itex]x(2x-1)<-2(4x^2-4x+1)[/itex]
[itex]2x^2-x<-8x^2+8x-2[/itex]
[itex]10x^2-9x+2<0[/itex]
[itex](5x-2)(2x-1)=0[/itex]
[itex]x=\frac{2}{5}[/itex] or [itex]\frac{1}{2}[/itex]
[itex]\frac{2}{5}<x<\frac{1}{2}[/itex]
Homework Statement
Solve the following for x [itex]\in\mathbb{R}:[/itex] [itex]\frac{x}{2x-1}<-2[/itex]
Homework Equations
The Attempt at a Solution
[itex]\frac{x(2x-1)^2}{2x-1}<-2(2x-1)^2[/itex]
[itex]x(2x-1)<-2(4x^2-4x+1)[/itex]
[itex]2x^2-x<-8x^2+8x-2[/itex]
[itex]10x^2-9x+2<0[/itex]
[itex](5x-2)(2x-1)=0[/itex]
[itex]x=\frac{2}{5}[/itex] or [itex]\frac{1}{2}[/itex]
[itex]\frac{2}{5}<x<\frac{1}{2}[/itex]