1. ## inequality

How do I solve this?

$.01 > \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+1}}$

I just can't seem to figure it out, thanks

2. Originally Posted by 11rdc11
How do I solve this?

$.01 > \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+1}}$

I just can't seem to figure it out, thanks
you want $\frac 1{\sqrt {x}} - \frac 1{\sqrt{x + 1}} - 0.01 < 0$

$\Rightarrow \frac {\sqrt{x + 1} - 1 - 0.01 \sqrt{x} \sqrt{x + 1}}{\sqrt{x(x + 1)}} < 0$

now, for the fraction to be less than zero, we need the numerator and the denominator to be different signs. so, WLOG, just solve for

$\sqrt{x + 1} - 1 - 0.01 \sqrt{x} \sqrt{x + 1} < 0$

and

$\sqrt{x(x + 1)} > 0$

when you find the required x-values, plot them on a number line, and test values in all the regions to see if the satisfy the original inequality, this will give you your intervals of solutions

3. Thanks a million

${\sqrt{x(x + 1)}} = \sqrt{x}\sqrt{x+1}$

Guess I got to work on my algebra

Quick question though shouldn't the numerator be

${\sqrt{x + 1} - \sqrt{x} - 0.01 \sqrt{x} \sqrt{x + 1}}$

4. I'm still having problems solving the numerator. Can someone show me how to do it?

I use ln to simply it but still a bit confused.

5. Hello,
Originally Posted by 11rdc11
Quick question though shouldn't the numerator be

${\sqrt{x + 1} - \sqrt{x} - 0.01 \sqrt{x} \sqrt{x + 1}}$
Yes it should.

Ok, here are the steps I've used, for the brute force method

$\sqrt{x+1}<\sqrt{x}+0.01 \sqrt{x} \sqrt{x+1}$

Square the inequation (it is possible because both sides are positive). There is an x that will cancel out.
After that, group all the terms that don't contain any square root in one side, and the stuff with the square root the other side.
Let X=x(x+1) and then, square again.

You'll have a quadratic inequality. Remember that it's the opposite sign of a within the interval formed by the roots.

Then, you'll have to solve twice a quadratic.... well, you'll see ^^'