Question #255912

Prove that using proof by contradiction.

√2 + √6 < √15


1
Expert's answer
2021-10-25T16:27:14-0400

Assume the result is False; i.e. assume that 6+215.\sqrt{6}+\sqrt{2}\geq\sqrt{15}.

Then


6+215>0=>(6+2)2(15)2\sqrt{6}+\sqrt{2}\geq\sqrt{15}>0=>(\sqrt{6}+\sqrt{2})^2\geq(\sqrt{15})^2

=>6+212+215=>437=>6+2\sqrt{12}+2\geq15=>4\sqrt{3}\geq7

=>(43)2(7)2=>4849,=>(4\sqrt{3})^2\geq(7)^2=>48\geq49,

which is contradiction.

Hence we have proved that 6+2<15.\sqrt{6}+\sqrt{2}<\sqrt{15}.


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