MHB Solve Simple Inequality: x < 0 or x > 2

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The inequality (x - 1/2)² > x + 1/4 simplifies to x(x - 2) > 0, leading to the solution set x < 0 or x > 2. The discussion emphasizes completing the square as an effective method for solving quadratic inequalities. Participants note that responses may not appear immediately, which can lead to confusion about whether a question has been previously answered. The final conclusion is that the values of x satisfying the inequality are in the intervals (-∞, 0) and (2, +∞). This method provides a clear approach to solving similar quadratic inequalities.
Fernando Revilla
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I quote a question from Yahoo! Answers

Find the set of values of x for ( x - 1/2 )^2 > x + 1/4 . Answer: { x: x < 0 or x > 2 }
Please help me to show the solution...

I have given a link to the topic there so the OP can see my response.
 
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We have $$\left (x - \frac{1}{2}\right )^2 > x + \frac{1}{4}\Leftrightarrow x^2-x+\frac{1}{4}>x + \frac{1}{4}\\\Leftrightarrow x^2-2x=0\Leftrightarrow x(x-2)>0$$
Then, $$x(x-2)>0\Leftrightarrow (x>0\wedge x-2>0)\vee(x<0\wedge x-2<0)\\\Leftrightarrow (x>0\wedge x>2)\vee(x<0\wedge x<2)\Leftrightarrow (x>2)\vee(x<0)$$
That is, $x$ is a solution of the inequality iff $x\in(-\infty,0)\cup (2,+\infty)$.

P.S. Sorry, I didn´t notice that this question had been already solved there. I might have to make an appointment to visit an Alzheimer's specialist.
 
Fernando Revilla said:
...
P.S. Sorry, I didn´t notice that this question had been already solved there. I might have to make an appointment to visit an Alzheimer's specialist.

I've noticed sometimes the replies of others do not show up until after you have posted a response. So, you may put off that appointment for now. (Happy)
 
I find that the most direct way to solve quadratic inequalities is to complete the square.

[math]\displaystyle \begin{align*} \left( x - \frac{1}{2} \right) ^2 &> x + \frac{1}{4} \\ x^2 - x + \frac{1}{4} &> x + \frac{1}{4} \\ x^2 - 2x &> 0 \\ x^2 - 2x + (-1)^2 &> (-1)^2 \\ (x - 1)^2 &> 1 \\ |x - 1| &> 1 \\ x - 1 < -1 \textrm{ or } x - 1 &> 1 \\ x < 0 \textrm{ or } x &> 2 \end{align*}[/math]
 
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