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    I’m stuck on these questions on surds, I think they’re like grade 8/9
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    I’ve worked out the first one but the method was more trial and error, I’ve also partially worked out the second one the same way e=5 I think?
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    Sorry, this is as far as I could get! I would suggest asking your maths teacher as they should explain how to do it-going to give it to my teacher so if I manage to do it I’ll let you know how
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    (Original post by Tamzincovell01)
    Sorry, this is as far as I could get! I would suggest asking your maths teacher as they should explain how to do it-going to give it to my teacher so if I manage to do it I’ll let you know how
    Yeah this is literally the same stage as me, and then I get confused, I think I’ve kind of done d1,2 and nearly 3 but it’s quite hard
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    (Original post by peeked)
    Yeah this is literally the same stage as me, and then I get confused, I think I’ve kind of done d1,2 and nearly 3 but it’s quite hard
    For D2, expand and simplify the LHS by collecting like terms, and then equate with the RHS. You need to make sure that you consider the integer and surd parts separately.

    Spoiler:
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    {1+sqrt(e)}{3+sqrt(e)} = 3 + 3sqrt(e) + sqrt(e) + e = (3+e) + 4sqrt(e) = f + 4sqrt(5)
    Thus, e = 5 and f = 3+e = 3+5 = 8


    The same concept applies for D1.

    Spoiler:
    Show

    {a+sqrt(b)}^2 = a^2 + (2a)sqrt(b) + b = (a^2 + b) + (2a)sqrt(b) = 49 + 12sqrt(b)
    So, 2a = 12 ---> a = 6
    Then, a^2 + b = (6)^2 + b = 36 + b = 49 ---> b = 13
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    (Original post by mupsman2312)
    For D2, expand and simplify the LHS by collecting like terms, and then equate with the RHS. You need to make sure that you consider the integer and surd parts separately.

    Spoiler:
    Show

    {1+sqrt(e)}{3+sqrt(e)} = 3 + 3sqrt(e) + sqrt(e) + e = (3+e) + 4sqrt(e) = f + 4sqrt(5)
    Thus, e = 5 and f = 3+e = 3+5 = 8


    The same concept applies for D1.

    Spoiler:
    Show

    {a+sqrt(b)}^2 = a^2 + (2a)sqrt(b) + b = (a^2 + b) + (2a)sqrt(b) = 49 + 12sqrt(b)
    So, 2a = 12 ---> a = 6
    Then, a^2 + b = (6)^2 + b = 36 + b = 49 ---> b = 13
    Wow thanks this was really helpful do you have any idea how to do d3 or d4 though as they really hard
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    (Original post by peeked)
    Wow thanks this was really helpful do you have any idea how to do d3 or d4 though as they really hard
    Glad to have helped.

    D3 and D4 are quite similar, too - you're really just setting-up two simultaneous equations in each question. Try using the same method as in D1 and D2, and see how far you can get (only then should you check these two spoilers):

    Spoiler:
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    {5+3sqrt(2)}^2 = 25 + 30sqrt(2) +18 = 43 + 30sqrt(2) = p + q/sqrt(8)
    So, p = 43 and q/sqrt(8) = 30sqrt(2) ---> q = 30sqrt(2) * sqrt(8) = 30sqrt(2*8) = 30sqrt(16) = 30*4 = 120



    For D4, try to simplify the LHS as much as possible before equating to the RHS.

    Spoiler:
    Show


    {sqrt(x)+sqrt(8x)}^2 = x + 2sqrt(8x^2) + 8x = 9x + 2sqrt(4x^2)*sqrt(2)
    = 9x + 2*(2x)*sqrt(2) = 9x + (4x)sqrt(2) = 54 + (y)sqrt(2)
    So, 9x = 54 ---> x = 6
    Then, (4x)sqrt(2) = (y)sqrt(2) ---> y = 4x = 4*6 = 24

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    if you expand D3 you get a number, p, and a surd with root 2. as root 8 is 2 times root 2, divide the number in front of the surd you find by 2 and you’ll find q
    for D4 expand it first and then i think you’re gonna have a part without surds and a part with surds in root 2. make the part without surds equal to 54, and the part with surds equal to y and you should be able to solve it simultaneously!
 
 
 
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