How do I solve this?
$\displaystyle .01 > \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x+1}}$
I just can't seem to figure it out, thanks
you want $\displaystyle \frac 1{\sqrt {x}} - \frac 1{\sqrt{x + 1}} - 0.01 < 0$
$\displaystyle \Rightarrow \frac {\sqrt{x + 1} - 1 - 0.01 \sqrt{x} \sqrt{x + 1}}{\sqrt{x(x + 1)}} < 0$
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
$\displaystyle \sqrt{x + 1} - 1 - 0.01 \sqrt{x} \sqrt{x + 1} < 0$
and
$\displaystyle \sqrt{x(x + 1)} > 0$
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
Thanks a million
$\displaystyle {\sqrt{x(x + 1)}} = \sqrt{x}\sqrt{x+1}$
Guess I got to work on my algebra
Quick question though shouldn't the numerator be
$\displaystyle {\sqrt{x + 1} - \sqrt{x} - 0.01 \sqrt{x} \sqrt{x + 1}}$
Hello,
Yes it should.
Ok, here are the steps I've used, for the brute force method
$\displaystyle \sqrt{x+1}<\sqrt{x}+0.01 \sqrt{x} \sqrt{x+1}$
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 ^^'