# Inequality problem.

• Jun 3rd 2009, 06:57 AM
a69356
Inequality problem.
Solve the below inequality :-

$\displaystyle {(x^2 - 2x - 3)}^{1/2} < 1$

Any help would be appreciated.

Thanks
• Jun 3rd 2009, 07:17 AM
HallsofIvy
Quote:

Originally Posted by a69356
Solve the below inequality :-

(x^2 - 2x - 3)^(1/2) < 1

Any help would be appreciated.

Thanks

The square root of any number larger than 1 is itself larger than 1.
That means that to have square root less than 1, the number itself must be less than 1.

So we have to solve $\displaystyle x^2- 2x- 3< 1$ which is the same as $\displaystyle x^2- 2x< 4$.

Adding 1 to both sides, to "complete the square", $\displaystyle x^2- 2x+ 1< 5$ which is the same $\displaystyle (x-1)^2< 5$. Now, $\displaystyle (x-1)^2= 5$ when $\displaystyle x- 1= \pm\sqrt{5}$ or $\displaystyle x= 1\pm\sqrt{5}$. Since x= 0, which is between those values, satisfies the inequality, all numbers between them and only numbers between them satisfy the inequality. (For continuous functions, inequalities can only reverse at points where they are [b]equalties.)
• Jun 3rd 2009, 02:06 PM
dhiab
Quote:

Originally Posted by a69356
Solve the below inequality :-

$\displaystyle {(x^2 - 2x - 3)}^{1/2} < 1$

Any help would be appreciated.

Thanks

HELLO :
they have :
http://www.gnux.be/latex/data/b3f659...958d7fa308.png
an important condition is :
http://www.gnux.be/latex/data/d340fd...8b9705e866.png