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    Hi im stuck on a question could someone please help me out:

    Find the value of k such that the equation 2x^2−kx+2=0
    (The x is a letter not multiplication) :rofl3::excited:
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    Have you tried completing the square?
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    Can you do that with the k?
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    Anything else that goes with the question?
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    (Original post by akay8)
    Anything else that goes with the question?
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    I got x=(k/4)+/-sqrt((k^2/16)-1)
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    2x^2 - kx + 2 = 0

    Equal roots hence b^2-4ac=0

    k^2 - 4(2)(2) = 0 --> k^2 - 16 = 0 --> what is k (finish it off)
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    What level is this, GCSE Higher?
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    (Original post by MWills99)
    I got x=(k/4)+/-sqrt((k^2/16)-1)
    How did you get that ?
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    (Original post by peeked)
    What level is this, GCSE Higher?
    It’s national 5, I think it’s the same as GCSE though
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    Giving x the value of 1 and k the value of 6 seems to work out pretty fine but i guess that would be too easy.
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    (Original post by thekidwhogames)
    2x^2 - kx + 2 = 0

    Equal roots hence b^2-4ac=0

    k^2 - 4(2)(2) = 0 --> k^2 - 16 = 0 --> what is k (finish it off)
    Ohh is it this formula
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    (Original post by Bananasplitxxx)
    Ohh is it this formula
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    that formula is the quadratic formula, you use it to find 'x' if you already know a,b and c. In this case you don't know what b equals. I don't think you need the formula for this question.
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    (Original post by Bananasplitxxx)
    Ohh is it this formula
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    Are you in GCSE?

    That is the quadratic formula. To have equal root that means the thing under the root = 0 to cause one repeated solution e.g. b^2 - 4ac = 0.
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    (Original post by Bananasplitxxx)
    How did you get that ?
    The discriminant is (b^2 - 4ac)

    If this expression is equal to 0, then the solution only has one real root.

    a = 2
    b = -k
    c = 2

    So:

    (-k)^2 -4 x 2 x 2 = 0
    k^2 -16 = 0
    k^2 = 16 [now square root both sides]
    k = +-4

    EDIT: k = +-4 due to the square root. Just remembered!
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    (Original post by Bananasplitxxx)
    How did you get that ?
    So you start with 2x^2-kx+2=0
    Then you divide by 2: x^2-(k/2)x+1=0
    The complete the square: (x-(k/4))^2-((K^2)/16)+1=0
    Then solve for x: (x-(k/4))^2=((k^2)/16)-1
    (x-(k/4))=+/-sqrt(((k^2)/16)-1)
    x=(k/4)+/-sqrt(((k^2)/16)-1)
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    (Original post by Mehru1214)
    The discriminant is (b^2 - 4ac)

    If this expression is equal to 0, then the solution only has one real root.

    a = 2
    b = -k
    c = 2

    So:

    (-k)^2 -4 x 2 x 2 = 0
    k^2 -16 = 0
    k^2 = 16 [now square root both sides]
    k = 4
    Ohh thanks I finally get it now
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    (Original post by thekidwhogames)
    2x^2 - kx + 2 = 0

    Equal roots hence b^2-4ac=0

    k^2 - 4(2)(2) = 0 --> k^2 - 16 = 0 --> what is k (finish it off)
    Wait so with a quadratic with equal roots will the equation b^2-4ac=0 always be correct?
    Because in this situation does it make x -4 or 4?
    I never done 'equal roots' problems before but does 'equal roots' just mean that the two x values are the same number (so like -4 and 4)
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    Is it k=+-4 then?
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    (Original post by peeked)
    Wait so with a quadratic with equal roots will the equation b^2-4ac=0 always be correct?
    Because in this situation does it make x -4 or 4?
    I never done 'equal roots' problems before but does 'equal roots' just mean that the two x values are the same number (so like -4 and 4)

    If a quadratic has equal roots thn b^2-4ac=0 is always true e.g. (x-2)^2
    If a quadratic has 2 real distinct roots then b^4-4ac>0 and no roots b^2-4ac<0

    Think about it. If the thing under the root, the discriminant, is 0 then x=-b+-0/2a --> x=-b/2a (one equal root).

    If discriminant > 0 then you have 2 values for the root and hence 2 solutions. If <0, no real root hence no real solutoin.
 
 
 
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