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Math Help - prove !!!

  1. #1
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    prove !!!

    3<(13)^1/2< 4, deduce that 0<(13)^1/2 -3<1

    b) [(13^1/2)+3]^4 + [(13^1/2)-3)]^4 =1904

    c)1903<[(13^1/2)+3]^4 <1904




    here is another question

    expresss [(p^1/2 +q(r^1/2)]^2 in the form =a+b(c^1/2). without evaluating the square root or using table or calculator,show that 10^1/2 +2(2^1/2) <6
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by qweiop90 View Post
    3<(13)^1/2< 4, deduce that 0<(13)^1/2 -3<1

    b) [(13^1/2)+3]^4 + [(13^1/2)-3)]^4 =1904

    c)1903<[(13^1/2)+3]^4 <1904




    here is another question

    expresss [(p^1/2 +q(r^1/2)]^2 in the form =a+b(c^1/2). without evaluating the square root or using table or calculator,show that 10^1/2 +2(2^1/2) <6
    Square each term of the first problem

    3\leq\sqrt{13}\leq{4}

    \Rightarrow{9\leq{13}\leq{16}}
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  3. #3
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    stil couldnt get

    9<13<16

    frm the question : (13^1/2+3)^4+(13^1/2-3)^4=1904

    how do you get? i had squared, but i stil couldnt get
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by qweiop90 View Post
    stil couldnt get

    9<13<16

    frm the question : (13^1/2+3)^4+(13^1/2-3)^4=1904

    how do you get? i had squared, but i stil couldnt get
    Ok sorry, I misread

    3\leq\sqrt{13}\leq{4}

    \Rightarrow{3-3\leq\sqrt{13}-3\leq{1}}

    \Rightarrow{0\leq\sqrt{13}-3\leq{1}}

    That is what you wanted right?
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  5. #5
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    the first question i already solved. yet, by using the same method, i stil couldnt prove the 2nd and 3rd question
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