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    Question:

    Workings:

     \sqrt{(3-3p^2)^2 + 36p^2}
     = \sqrt{9p^4 + 18p^2 + 9}
     = 3p^2 + 3\sqrt{2}p + 3

    That is what I have ended up with. Why is this incorrect?
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    (Original post by NinCheng)
    Question:

    Workings:

     \sqrt{(3-3p^2)^2 + 36p^2}
     = \sqrt{9p^4 + 18p^2 + 9}
     = 3p^2 + 3\sqrt{2}p + 3

    That is what I have ended up with. Why is this incorrect?

    You cannot do that with square roots, \sqrt{a + b} \neq \sqrt{a} + \sqrt {b} in general.

    Instead, try factorising what you have under the square root sign.
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    (Original post by NinCheng)
    Question:

    Workings:

     \sqrt{(3-3p^2)^2 + 36p^2}
     = \sqrt{9p^4 + 18p^2 + 9}
     = 3p^2 + 3\sqrt{2}p + 3

    That is what I have ended up with. Why is this incorrect?
    As above. Do you think that \sqrt{2} = \sqrt{1 + 1} = \sqrt{1 } + \sqrt{1} = 1 + 1 = 2? Surely not.

    Instead, \sqrt{9p^4 + 18p^2 + 9} = 3\sqrt{p^4 + 2p^2 + 1} = 3\sqrt{(p^2)^2 + 2(p^2) + 1}
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    (Original post by 16Characters....)
    You cannot do that with square roots, \sqrt{a + b} \neq \sqrt{a} + \sqrt {b} in general.

    Instead, try factorising what you have under the square root sign.
    Which leaves me with  \sqrt{9(p^4 + 2p^2 + 1)} but now I'm clueless
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    (Original post by NinCheng)
    Which leaves me with  \sqrt{9(p^4 + 2p^2 + 1)} but now I'm clueless
    Can you see how to factorise that bracket? You might like to let y = p^2
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    line 2, take a common factor of 9 out of expression which becomes 3 outside sqrt. The function left inside the sqrt is a perfect square.

    Your fundamental error is that \sqrt{a+b} does NOT equal \sqrt a + \sqrt b which is what your "solution' assumes.

    if
    \sqrt{a+b} did equal \sqrt a + \sqrt b then try it with a and b = 1 which would predict that the square root of 2 was 2
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    (Original post by 16Characters....)
    Can you see how to factorise that bracket? You might like to let y = p^2
    Oh, so

     \sqrt{9(p^4 + 2p^2 + 1)}
     \sqrt{9(p^2 + 1)(p^2 + 1)}
     \sqrt {9(p^2+1)^2}
     = 3(p^2 + 1)
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    (Original post by NinCheng)
    Oh, so

     \sqrt{9(p^4 + 2p^2 + 1)}
     \sqrt{9(p^2 + 1)(p^2 + 1)}
     \sqrt {9(p^2+1)^2}
     = 3(p^2 + 1)
    Yep.
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    (Original post by 16Characters....)
    Yep.
    Thank you
 
 
 
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