3/3sqrt2-sqrt3 rationalize the denominator and simplify can i get an explanation please

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3/3sqrt2-sqrt3 rationalize the denominator and simplify can i get an explanation please

Mathematics
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Is your original expression: a) \(\frac{3}{3\sqrt{2} - \sqrt{3}}\) b) \(\frac{3}{3\sqrt{2}} - \sqrt{3}\)
a
Ok, so multiply top and bottom by the conjugate \(3\sqrt{2} + \sqrt{3}\)

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Other answers:

top and bottom??
yes, anything you do to the denominator you must do to the numerator to keep the ratio the same.
Anything you multiply the denominator by that is. \[\frac{4}{3} = \frac{4*2}{3*2} = \frac{8}{6}\]
im still confused i wrote this \[3/3\sqrt2-\sqrt3 *3\sqrt2+\sqrt3/3\sqrt2+\sqrt3\]
\[3/3\sqrt2-\sqrt3 * 3\sqrt2+\sqrt3/3\sqrt2+\sqrt3\]i mean
You need to use parentheses
\[\frac{3}{3\sqrt{2} - \sqrt{3}} = \frac{3(3\sqrt{2} + \sqrt{3})}{(3\sqrt{2} - \sqrt{3})(3\sqrt{2} + \sqrt{3})}\]
And on bottom we get some nice simplification.
i dont understand how that simplifies
Foil it out.. \[(3\sqrt{2} - \sqrt{3})(3\sqrt{2} + \sqrt{3})\] \[=3\sqrt{2}*3\sqrt{2} - 3\sqrt{2}\sqrt{3} +3\sqrt{2}\sqrt{3} -\sqrt{3}*\sqrt{3}\] \[=9(2) + 0 - 3 = 15\]
Which cancels with the 3 up top
to give you \[\frac{3\sqrt{2} + \sqrt{3}}{5}\]
i am not grasping this
i dont understand how you got 15
i dont understand how you got 15
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: \[a^2 - b^2 \]\[= (3\sqrt{2})^2 - (\sqrt{3})^2 \]\[= 3^2\sqrt{2}^2 - \sqrt{3}^2 \]\[=9(2) - 3 = 18-3 = 15\]
i understand the foil method better thankyou i understand now

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