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    Expand and simplify (7 + 3ROOT 2)(5 − 2 ROOT2).


    I got the answer to be -11 but i'm not tooo sure !



    Also, does anyone know how to explain (iv) ,the last bit of this question, from the JAN2012 C1 OCREMEI past paper? Thaks everybody!

    (iii) Show that, where the line y = 2x + k intersects the circle,
    5x^2 + (4k − 4) x + k^2 − 16 = 0. [3]
    (iv) Hence find the values of k for which the line y = 2x + k is a tangent to the circle.
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    (Original post by alex7892)
    Expand and simplify (7 + 3ROOT 2)(5 − 2 ROOT2).


    I got the answer to be -11 but i'm not tooo sure !



    Also, does anyone know how to explain (iv) ,the last bit of this question, from the JAN2012 C1 OCREMEI past paper? Thaks everybody!

    (iii) Show that, where the line y = 2x + k intersects the circle,
    5x^2 + (4k − 4) x + k^2 − 16 = 0. [3]
    (iv) Hence find the values of k for which the line y = 2x + k is a tangent to the circle.
     (7+3\sqrt2)(5-2\sqrt2) \not= -11

    Show your working.
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    Sorry looks like I made a mistake. Would the answer be:

    35-14root5+15root2-6root10 ?
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    (Original post by alex7892)
    Also, does anyone know how to explain (iv) ,the last bit of this question, from the JAN2012 C1 OCREMEI past paper? Thaks everybody!

    (iii) Show that, where the line y = 2x + k intersects the circle,
    5x^2 + (4k − 4) x + k^2 − 16 = 0. [3]
    (iv) Hence find the values of k for which the line y = 2x + k is a tangent to the circle.

    We know a tangent intersects the circle at only one point, hence the quadratic,  5x^2 + (4k - 4) x + k^2 -16 = 0.  , will have repeated roots.
    So solve,  b^2 - 4ac =0
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    ..Ok, thanks Raheem, but what would be A, B and C?
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    (Original post by alex7892)
    Sorry looks like I made a mistake. Would the answer be:

    35-14root5+15root2-6root10 ?
    No.

     (7+3\sqrt2)(5-2\sqrt2) = 7\times 5 - 7 \times 2\sqrt2 +5 \times 3 \sqrt2 - 3\sqrt2 \times 2\sqrt2

    Remember,  \sqrt2 \times \sqrt2 = 2
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    (Original post by alex7892)
    ..Ok, thanks Raheem, but what would be A, B and C?
    Do you know about the discriminant?

    For repeated roots it is,  b^2 -4ac =0

     5x^2 + (4k - 4) x + k^2 -16 = 0
     a=5 \ \ \ b=4k-4 \ \ \ c=k^2-16
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    (Original post by raheem94)
    No.

     (7+3\sqrt2)(5-2\sqrt2) = 7\times 5 - 7 \times 2\sqrt2 +5 \times 3 \sqrt2 - 3\sqrt2 \times 2\sqrt2

    Remember,  \sqrt2 \times \sqrt2 = 2
    Oh so sorry, I posted the wrong question, and my answer was for the other question haha I mean (7+3root2) (5-2root5)
    Sorry!! And what would A, and C be then that is confusing
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    Okay thanks for the discriminant thing, just quite weired using k2-16 for c etc... never seen that before...
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    (Original post by alex7892)
    ..but what would be A, B and C?
    Quadratics have the form ax^2 + bx + c = 0

    a,b and c are the coefficients of each term in the general quadratic equation. So, 'a' is whatever comes before an x^2, 'b' is whatever comes before an 'x', and 'c' is anything that doesn't have an 'x' in it.

    (extra: technically, c is the coefficient of x^0, but as x^0 = 1 theres no need to write it.)
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    Thaaanks everyone
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    (Original post by alex7892)
    Oh so sorry, I posted the wrong question, and my answer was for the other question haha I mean (7+3root2) (5-2root5)
    Sorry!! And what would A, and C be then that is confusing
     (7+3 \sqrt2) (5-2\sqrt5) \not= -11

    By the way, are you sure you have typed out the right question now?
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    Okay was it the second answer I gave? and yesss...
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    (Original post by alex7892)
    Sorry looks like I made a mistake. Would the answer be:

    35-14root5+15root2-6root10 ?
    Yes, this is correct
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    Thanks a lot Raheem going to personal rep you in a minute, always sorting out my maths problems aren't you! lol
  16. Offline

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    (Original post by alex7892)
    Thanks a lot Raheem going to personal rep you in a minute, always sorting out my maths problems aren't you! lol
    I answer a lot of questions on the maths forum, so it is difficult to keep track of the people, except very active members.

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