# Help rationalize this numerator so that...

• Sep 23rd 2010, 07:26 PM
yess
Help rationalize this numerator so that...
rationalize this numerator so that

[(√2(x+delta x)-3) - (√2x-3)]/delta x = 2/[(√2(x+delta x)-3) + √2x-3)]
• Sep 23rd 2010, 08:02 PM
Soroban
Hello, yess!

Is there a typo on the problem?

Quote:

$\displaystyle \text{Rationalize this numerator so that:}$

.$\displaystyle \displaystyle \frac{\sqrt{2}(x+ \Delta x)-3) - (\sqrt{2}x-3)}{\Delta x} \;=\; \frac{2}{[\sqrt{2}(x+\Delta x)-3] + [\sqrt{2}x-3]}$

This is not true . . .
• Sep 23rd 2010, 08:16 PM
yess
Quote:

Originally Posted by Soroban
Hello, yess!

Is there a typo on the problem?

This is not true . . .

ohh oops the square root should be over the whole expression so square root of 2(x+delta x) -3 minus the square root of 2x-3 all over delta x equals 2 over the square root 2(x+delta x)-3 + square root of 2x-3.
• Sep 24th 2010, 03:50 AM
HallsofIvy
if you mean
$\displaystyle \frac{\sqrt{2(x+\Delta x)- 3}- \sqrt{2x- 3}}{\Delta x}$

you "rationalize the numerator" by multiplying by
$\displaystyle \frac{\sqrt{2(x+\Delta x)- 3}+ \sqrt{2x- 3}}{\sqrt{2(x+\Delta x)- 3}+ \sqrt{2x- 3}}$