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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Choose the incorrect relation(s) from the following}\hspace{.33em}\\~\\ & i.)\ \sqrt{6}+\sqrt{2}=\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & ii.)\ \sqrt{6}+\sqrt{2}<\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & iii.)\ \sqrt{6}+\sqrt{2}>\sqrt{5}+\sqrt{3}\hspace{.33em}\\~\\ & a.)\ i.) \ \text{and}\ iii.)\hspace{.33em}\\~\\ & b.)\ ii.) \ \text{and}\ iii.)\hspace{.33em}\\~\\ & c.)\ i.) \hspace{.33em}\\~\\ & d.)\ ii.) \hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    i have to solve this with in a minute

  3. mathivh
    • one year ago
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    If a calculator is allowed you could just calculate the decimal values of both sides and check if they are equal.

  4. mathmath333
    • one year ago
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    lol no calculator is allowed

  5. ganeshie8
    • one year ago
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    In view of arriving at a contradiction, lets suppose that \(\sqrt{6}+\sqrt{2}=\sqrt{5}+\sqrt{3}\). Square both sides and get \(6+2+2\sqrt{12}=5+3+2\sqrt{15}\). \(\implies \sqrt{12}=\sqrt{15}\) which is wrong. So the initial assumption is wrong.

  6. mathmath333
    • one year ago
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    ok

  7. ganeshie8
    • one year ago
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    Alternatively you may work it like this : \[\large{\begin{align} \sqrt{6}+\sqrt{2}~~&\stackrel{?}{}~~\sqrt{5}+\sqrt{3}\\~\\ (\sqrt{6}+\sqrt{2})^2~~&\stackrel{?}{}~~(\sqrt{5}+\sqrt{3})^2\\~\\ 6+2+2\sqrt{12}~~&\stackrel{?}{}~~ 5+3+2\sqrt{15}\\~\\ \sqrt{12}~~&\stackrel{?}{}~~ \sqrt{15}\\~\\ \sqrt{12}~~&\lt~~ \sqrt{15} \end{align}}\] since the numbers are positive, the implications will work in the reverse direction also. read it from bottom to top.

  8. mathmath333
    • one year ago
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    option 1 is incorrect

  9. ganeshie8
    • one year ago
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    i and iii are wrong

  10. mathmath333
    • one year ago
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    thnks

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