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satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2gimmick is to put \[x=w^2\] and write \[x^214x2=0\] solve for x

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2then go back. solve for x via \[x^214x=2\] \[(x7)^2=2+49\] \[(x7)^2=51\] \[x7=\pm\sqrt{51}\] \[x=7\pm\sqrt{51}\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2now \[7\sqrt{51}<0\] so you cannot have \[w^2=7\sqrt{51}\] with real numbers. therefore \[w^2=7+\sqrt{51}\] and finally \[w=\pm\sqrt{7+\sqrt{51}}\]
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