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rationalize the denominator of 14 divided by 3 + the square root of 2
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\[\frac{ 14 }{ 3+\sqrt2 }\] you need to use a conjugate here to get rid of the square root. So..\[\frac{ 14 }{ 3+\sqrt2 }*\frac{ 3-\sqrt2 }{ 3-\sqrt2 }\]\[\frac{ 14*(3-\sqrt2) }{ (3+\sqrt2)(3-\sqrt2) }\]
\[\frac{ 14(3-\sqrt2) }{ 9-3\sqrt2+3\sqrt2-2 }\]\[\frac{ 14(3-\sqrt2) }{ 9-2 }\]\[\frac{ 42-14\sqrt2 }{ 7 }\]
thank you !!
\[\frac{ 42 }{ 7}-\frac{ 14\sqrt2 }{ 7 }=6-2\sqrt2=2(3-\sqrt2)\]
no problem :)
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lol I think you did a little bit more than what they asked for
haha, well... simplest form, right? :)
i guess, that's the simplest :P
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