My Math Forum more square roots

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 July 16th, 2011, 06:09 PM #1 Newbie   Joined: Jul 2011 Posts: 20 Thanks: 0 more square roots I could really use some help on setting these two problems up: I need to rewrite 4/?7 as an equivilent fraction that does not have a square root in the denominator. I also need to rewrite it as a sum of two nonzero fractions with different denominators. Thanks!
 July 16th, 2011, 06:23 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,945 Thanks: 1136 Math Focus: Elementary mathematics and beyond Re: more square roots $\frac{4}{\sqrt{7}}\,=\,\frac{4}{\sqrt{7}}\,\cdot\, \frac{\sqrt{7}}{\sqrt{7}}\,=\,\cdots$ $\frac{3}{\sqrt{7}}\,+\,\frac{?}{7}\,=\,\cdots$
 July 16th, 2011, 06:59 PM #3 Newbie   Joined: Jul 2011 Posts: 20 Thanks: 0 Re: more square roots I understand the first one, but I am lost on the second....
 July 16th, 2011, 07:09 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,945 Thanks: 1136 Math Focus: Elementary mathematics and beyond Re: more square roots $\frac{3}{\sqrt{7}}\,+\,\frac{\sqrt{7}}{7}\,=\,\fra c{3}{\sqrt{7}}\,\cdot\,\frac{\sqrt{7}}{\sqrt{7}}\, +\,\frac{\sqrt{7}}{7}\,=\,\frac{3\sqrt{7}}{7}\,+\, \frac{\sqrt{7}}{7}\,=\frac{4\sqrt{7}}{7}\,=\,\frac {4}{\sqrt{7}}$

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