Solving non linear inequalities?

Question

can someone explain this in more detail i got confused all 3 examples http://www.flickr.com/photos/26479899@N03/2783900305/

Answer 1

You're given that α < β.

If (x-α)(x-β) > 0, then one of two things must be true: both factors are positive, or else both factors are negative.

Both factors positive:

x-α > 0 and x-β > 0

x > α and x > β

Since α < β, the numbers x that satisfy both inequalities satisfy x > β

(It might help to choose specific numbers for α and β to make this clear.)

Both factors negative:

x-α < 0 and x-β < 0

x < α and x < β

Again, since α < β, the numbers x that satisfy both inequalities satisfy x < α

If (x-α)(x-β) < 0, then one of the factors is positive and the other factor is negative. There are two possibilities:

x-α < 0 and x-β > 0, which implies

x < α and x > β

But recall that α < β. If x < α and α < β, then x < β, which contradicts the requirement above that x > β. (Again, this should be easier to see if you choose specific numbers for α and β.) Similarly, if x > β and β > α then x > α, contradicting the requirement that x < α. So there are no solutions for the pair x-α < 0 and x-β > 0.

The other possibility is

x-α > 0 and x-β < 0, which implies

x > α and x < β

This is easy to see on a number line:

-----|-------------|-------------> x

......α.................β

All numbers x to the right of α satisfy x > α. All numbers x to the left of β satisfy x < β. The numbers x that are simultaneously greater than α and less than β are the numbers x between α and β. So the solution is

α < x < β

Finally, if (x-α)² ≥ 0, the conclusion written, that x = α only, is incorrect. All real values of x satisfy (x-α)² ≥ 0. I think what was probably meant is that if (x-α)² ≤ 0, then x = α only.

The square of a real number is non-negative, and is zero only if the real number is zero. x-α = 0 ⇒ x = α.