Maths year 11

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    (Original post by z_o_e)
    I get stuck on these steps


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    You cannot further simplify this so just leave it as it is.
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    (Original post by RDKGames)
    Well you have 3 x 3 x root7 x root7 so do whatever you think is right here.

    Like you see I don't know how to add 3 root 3 + 2 root 3?
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    (Original post by z_o_e)
    I'm not sure. I think root 7 stays and I multiply 3*3

    so 9 root 7
    Well you got the 9 right, but the root 7 doesn't stay because there are 2 of them. Think about it again.

    (Original post by z_o_e)
    Like you see I don't know how to add 3 root 3 + 2 root 3?
    All you're doing with addition like that is factoring out the root 3 out of any common terms with it then add up the numbers in bracket.

    For example, it would be the same thing if I had 3a+9a. I can factor out the a's then add 3 and 9 together - a(3+9) which gives me 12a.
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    (Original post by RDKGames)
    Well you got the 9 right, but the root 7 doesn't stay because there are 2 of them. Think about it again.



    All you're doing with addition like that is factoring out the root 3 out of any common terms with it then add up the numbers in bracket.

    For example, it would be the same thing if I had 3a+9a. I can factor out the a's then add 3 and 9 together - a(3+9) which gives me 12a.
    so 6 root 5?
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    (Original post by RDKGames)
    Well you got the 9 right, but the root 7 doesn't stay because there are 2 of them. Think about it again.

    So this must be 9 root 49
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    They are over here.

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    (Original post by z_o_e)
    They are over here.

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    Those are correct though you can simplify the first one down further.
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    (Original post by RDKGames)
    Those are correct though you can simplify the first one down further.
    Like this


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    (Original post by z_o_e)
    Like this


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    Yes but it's much easier to remember that the square of any root number is that number itself. Here's why:

    \sqrt{a} \cdot \sqrt{a}= (\sqrt{a})^2 = (a^{\frac{1}{2}})^2 = a^{\frac{2}{2}} = a^1 = a and you can see we use laws of indices as part of the proof and the fact that the square root of a number is the same as that number raised to the power of a half.
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    (Original post by RDKGames)
    Yes but it's much easier to remember that the square of any root number is that number itself. Here's why:

    \sqrt{a} \cdot \sqrt{a}= (\sqrt{a})^2 = (a^{\frac{1}{2}})^2 = a^{\frac{2}{2}} = a^1 = a and you can see we use laws of indices as part of the proof and the fact that the square root of a number is the same as that number raised to the power of a half.
    And then I added the like terms up



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    (Original post by z_o_e)
    And then I added the like terms up



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    Last 3 lines don't make sense (well, I mean you ignored the 1 for some reason, not sure where that one went). Attempt it again from the first line.
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    (Original post by RDKGames)
    Last 3 lines don't make sense (well, I mean you ignored the 1 for some reason, not sure where that one went). Attempt it again from the first line.
    Oh no I'm just adding the like terms up right now I left the one for the final answer.

    I wanted you to check if I correctly added the like terms correctly.

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    (Original post by z_o_e)
    Oh no I'm just adding the like terms up right now I left the one for the final answer.

    I wanted you to check if I correctly added the like terms correctly.

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    Oh then yeah you added them right. Got a bit misled since you worked from the question.
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    (Original post by RDKGames)
    Oh then yeah you added them right. Got a bit misled since you worked from the question.


    This was the second one


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    (Original post by z_o_e)
    This was the second one


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    First one: No. You add the terms correctly, yes, but you are not being careful in WHAT you are adding and please check what 9\sqrt{49} simplifies to. It is not what you have there.

    Second one: Where did the \sqrt5 come from?
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    (Original post by RDKGames)
    First one: No. You add the terms correctly, yes, but you are not being careful in WHAT you are adding and please check what 9\sqrt{49} simplifies to. It is not what you have there.

    Second one: Where did the \sqrt5 come from?
    yeah I'm struggling to add, subtract and simplify these types of questions..

    Shall I focus on that area first and then come back to these questions?

    Do you have any examples or anything?
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    (Original post by z_o_e)
    yeah I'm struggling to add, subtract and simplify these types of questions..

    Shall I focus on that area first and then come back to these questions?

    Do you have any examples or anything?
    Simplify:

    2\sqrt2 \cdot \sqrt2

    6\sqrt5 \cdot 2\sqrt5

    3\sqrt9

    7\sqrt4 - 2\sqrt9

    3\sqrt8 - 2\sqrt{64}
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    (Original post by RDKGames)
    Simplify:

    2\sqrt2 \cdot \sqrt2

    6\sqrt5 \cdot 2\sqrt5

    3\sqrt9

    7\sqrt4 - 2\sqrt9

    3\sqrt8 - 2\sqrt64
    I don't know how to simplify these :/

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    (Original post by z_o_e)
    I don't know how to simplify these :/

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    Remember your surd rules for multiplication and have a go. \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} use this for the first one to try and get the answer of 4.
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    (Original post by z_o_e)
    I don't know how to simplify these :/

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    Remember that: 2(root2) x (root2) is the same as 2 x root2 x root2. You can calculate root2 x root2, and then multiply by 2.

    You can do the same with some of the others, multiplying the terms inside roots together and simplifying (all but the last question simplify to whole numbers) and then multiply by the remaining numbers to get your answer. However, you can only do this when all terms are being multiplied.

    Once you get addition or subtraction in the expression, you cannot just multiply the roots across. Instead, you factor out the root and add/subtract the whole numbers according to the question.

    Surds (unsimplifiable roots) act like letters in an equation.
    2a x 3a = 2 x a x 3 x a = 2 x 3 x a x a = 6a^2.
    If you replace "a" with some root, you can do the same.
    Likewise, 5a + 3a = a(5 + 3) = 8a. Again, the root acts in the same way.

    You must be careful to make sure that you only add/subtract the same roots. 3a + 2b is in it's most simple form. Likewise, 3(root2) + 2(root5) is also in its simplest form. This is true because when you have different surds you cannot factor them out as they are different values.

    Now try and have a go at the problems that RDKGames wrote for you.
 
 
 
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