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DLS

  • 2 years ago

How would you solve this expression?

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  1. DLS
    • 2 years ago
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    \[\Huge \sqrt{7+\sqrt{48}}=\sqrt{m}+\sqrt{n}\] solve for m^2+n^2

  2. marsss
    • 2 years ago
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    \[\sqrt{7+\sqrt{48}}=\sqrt{7+2\sqrt{12}}\] m+n=7 and m*n=12 m=3 n=4 or m=4 n=3 3^2+4^2=9+16=25

  3. DLS
    • 2 years ago
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    how did you get all that? m+n=7 and m*n=12 m=3 n=4 or m=4 n=3 3^2+4^2=9+16=25

  4. marsss
    • 2 years ago
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    \[(\sqrt{m}+\sqrt{n})^{2}=m+n+2\sqrt{mn}\] \[m+n+2\sqrt{mn}=7+2\sqrt{12 }\] do you understand?

  5. DLS
    • 2 years ago
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    oh hmm okay thanks

  6. marsss
    • 2 years ago
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    you're welcome :)

  7. DLS
    • 2 years ago
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    how can we compare btw,3 terms on one side 2 on other? :O

  8. marsss
    • 2 years ago
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    m+n is just one term like 3+4. it's 7. did you get?

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