Solve Simple Inequality: x < 0 or x > 2

[Fernando Revilla]

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.

 

[Fernando Revilla]

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.

 

[MarkFL]

...
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)

 

[Prove It]

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]