- Thread starter
- #1

- Jan 29, 2012

- 661

I have given a link to the topic there so the OP can see my response.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....

- Thread starter Fernando Revilla
- Start date

- Thread starter
- #1

- Jan 29, 2012

- 661

I have given a link to the topic there so the OP can see my response.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....

- Thread starter
- #2

- Jan 29, 2012

- 661

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.

- Admin
- #3

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....

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.

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