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    (root(x) + 8root(x))^2

    ie root(x) + 8(rootx) and the entire thing taken to power of 2
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    (Original post by jon2016)
    (root(x) + 8root(x))^2
    \displaystyle (a +b)^2 = a^2 + 2ab + b^2

    So here (\sqrt{x} + 8\sqrt{x})^2 = (\sqrt{x})^2 + 2\sqrt{x} \times 8 \sqrt{x} + 8^2 (\sqrt{x})^2.

    Now (\sqrt{x})^2 = \sqrt{x} \times \sqrt{x} = x. Can you take it from here?
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    (Original post by Zacken)
    \displaystyle (a +b)^2 = a^2 + 2ab + b^2

    So here (\sqrt{x} + 8\sqrt{x})^2 = (\sqrt{x})^2 + 2\sqrt{x} \times 8 \sqrt{x} + 8^2 (\sqrt{x})^2.

    Now (\sqrt{x})^2 = \sqrt{x} \times \sqrt{x} = x. Can you take it from here?
    Thanks
    What is the (a +b)^2 = a^2 + 2ab + b^2 rule called ? looks similar to difference of two squares , when would know to use it ?

    I was trying to do

    (a+b)(a+b) and just got a huge mess
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    (Original post by jon2016)
    Thanks
    What is the (a +b)^2 = a^2 + 2ab + b^2 rule called ? looks similar to difference of two squares , when would know to use it ?

    I was trying to do

    (a+b)(a+b) and just got a huge mess
    It comes from (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2.

    You would use this whenever you had (something + something else)^2.
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    How is this 'Undergraduate'? LOL
 
 
 
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