2/\sqrt{10+2\sqrt{21} - 1/\sqrt{12+2\sqrt{35} -1/\sqrt{9+2\sqrt{15}=

TS ICET 23 May 2019 Shift 1 Paper

Question : 80 of 200

Marks: +1, -0

2/√{10+2√{21}} - 1/√{12+2√{35}} -1/√{9+2√{15}}=

Solution:

√{(a+b)+2 √{a b}}=√{a}+√{b}

√{10+2 √{21}}=√{(7+3)+2 √{7 × 3}}=√{7}+√{3}

√{12+2 √{35}}=√{(7+5)+2 √{7 × 5}}=√{7}+√{5}

√{8+2 √{15}}=√{(5+3)+2 √{5 × 3}}=√{5}+√{3}

Required to find

\;{2}/{√{10+2 √{21}}}-\;{1}/{√{12+2 √{35}}}-\;{1}/{√{8+2 √{15}}}

=\;{2}/{√{7}+√{3}}-\;{1}/{√{7}+√{5}}-\;{1}/{√{5}+√{3}}

=\;{2}/{√{7}+√{3}} × \;{√{7}-√{3}}/{√{7}-√{3}}-\;{1}/{√{7}+√{5}} × \;{√{7}-√{5}}/{√{7}-√{5}}-

\;{1}/{√{5}+√{3}} × \;{√{5}-√{3}}/{√{5}-√{3}}

=\;{√{7}-√{3}}/{2}-\;{√{7}-√{5}}/{2}-\;{√{5}-√{3}}/{2}

=0

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