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simplify 6/√14
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\[{6 \over \sqrt14}= {6 \over \sqrt 14} \times { \sqrt 14 \over \sqrt 14} = {{6 \times \sqrt14} \over 14}={{3\times \sqrt 14} \over 7}\]
rationalize the denominator
exactly what akhsay showed you
i am always exact :P :P jkjk
Thanks guys(:
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no one is a loser :D
\[6\over \sqrt{14}\]multiply both numerator and denominator by\[\sqrt{14}\]getting\[6\sqrt{14}\over \sqrt{14}\sqrt{14}\]getting \[3\sqrt{14}\over7\]
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