## anonymous 5 years ago 3/3sqrt2-sqrt3 rationalize the denominator and simplify can i get an explanation please

1. anonymous

Is your original expression: a) $$\frac{3}{3\sqrt{2} - \sqrt{3}}$$ b) $$\frac{3}{3\sqrt{2}} - \sqrt{3}$$

2. anonymous

a

3. anonymous

Ok, so multiply top and bottom by the conjugate $$3\sqrt{2} + \sqrt{3}$$

4. anonymous

top and bottom??

5. anonymous

yes, anything you do to the denominator you must do to the numerator to keep the ratio the same.

6. anonymous

Anything you multiply the denominator by that is. $\frac{4}{3} = \frac{4*2}{3*2} = \frac{8}{6}$

7. anonymous

im still confused i wrote this $3/3\sqrt2-\sqrt3 *3\sqrt2+\sqrt3/3\sqrt2+\sqrt3$

8. anonymous

$3/3\sqrt2-\sqrt3 * 3\sqrt2+\sqrt3/3\sqrt2+\sqrt3$i mean

9. anonymous

You need to use parentheses

10. anonymous

$\frac{3}{3\sqrt{2} - \sqrt{3}} = \frac{3(3\sqrt{2} + \sqrt{3})}{(3\sqrt{2} - \sqrt{3})(3\sqrt{2} + \sqrt{3})}$

11. anonymous

And on bottom we get some nice simplification.

12. anonymous

i dont understand how that simplifies

13. anonymous

Foil it out.. $(3\sqrt{2} - \sqrt{3})(3\sqrt{2} + \sqrt{3})$ $=3\sqrt{2}*3\sqrt{2} - 3\sqrt{2}\sqrt{3} +3\sqrt{2}\sqrt{3} -\sqrt{3}*\sqrt{3}$ $=9(2) + 0 - 3 = 15$

14. anonymous

Which cancels with the 3 up top

15. anonymous

to give you $\frac{3\sqrt{2} + \sqrt{3}}{5}$

16. anonymous

i am not grasping this

17. anonymous

i dont understand how you got 15

18. anonymous

i dont understand how you got 15

19. anonymous

This is what you had in the denominator right? $(3\sqrt{2} - \sqrt{3})(3\sqrt{2} + \sqrt{3})$ This is just like $(a - b)(a+b) = a^2 -ba + ab - b^2$ The middle two terms cancel each other out and you are left with $a^2 - b^2$ Where in your case, $$a=3\sqrt{2}$$ and $$b=\sqrt{3}$$ So: $a^2 - b^2$$= (3\sqrt{2})^2 - (\sqrt{3})^2$$= 3^2\sqrt{2}^2 - \sqrt{3}^2$$=9(2) - 3 = 18-3 = 15$

20. anonymous

i understand the foil method better thankyou i understand now