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mathtard

  • 4 years ago

w^4-14w^2-2=0 solve

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  1. anonymous
    • 4 years ago
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    gimmick is to put \[x=w^2\] and write \[x^2-14x-2=0\] solve for x

  2. anonymous
    • 4 years ago
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    then go back. solve for x via \[x^2-14x=2\] \[(x-7)^2=2+49\] \[(x-7)^2=51\] \[x-7=\pm\sqrt{51}\] \[x=7\pm\sqrt{51}\]

  3. anonymous
    • 4 years ago
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    now \[7-\sqrt{51}<0\] so you cannot have \[w^2=7-\sqrt{51}\] with real numbers. therefore \[w^2=7+\sqrt{51}\] and finally \[w=\pm\sqrt{7+\sqrt{51}}\]

  4. mathtard
    • 4 years ago
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    thank you!

  5. anonymous
    • 4 years ago
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    yw

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