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Math Help - Solve for k: (Two different types of problems)

  1. #1
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    Solve for k: (Two different types of problems)

    Hi all, thanks for taking time out of your day to have a look at my problems. I'm studying and need some help solving these two problems.

    I need to solve for k:

    Here are the problems:

    1) One root of 6x^2 - 3x +k = 0 is (1 - i√23)/4

    2) kx^2 - 3x + 4 = 0 one root is twice the other.

    I'm not really sure how to do these problems and am really greatful for your help! Btw, the "√" sign in the first problem is a square root sign (just in case someone got confused).

    Thanks!
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  2. #2
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    Quote Originally Posted by OurIslamic View Post

    1) One root of 6x^2 - 3x +k = 0 is (1 - i√23)/4

    Thanks!
    The other root is the complex conjugate. Take both linear roots and expand them back out. This should give you k.


    Quote Originally Posted by OurIslamic View Post


    2) kx^2 - 3x + 4 = 0 one root is twice the other.
    Solve this using the quadratic formula, x = \frac{3\pm\sqrt{9-16k}}{2k}

    Now ask what values for k will make one root twice the other?
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  3. #3
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    Quote Originally Posted by OurIslamic View Post
    2) kx^2 - 3x + 4 = 0 one root is twice the other.
    Using the quadratic:
    x = [3 + SQRT(9 - 16k)] / (2k) or [3 - SQRT(9 - 16k)] / (2k)

    one root = twice the other:
    3 + SQRT(9 - 16k) = 2[3 - SQRT(9 - 16k)]
    3 + SQRT(9 - 16k) = 6 - 2SQRT(9 - 16k)
    3SQRT(9 - 16k)] = 3
    SQRT(9 - 16k)] = 1
    9 - 16k = 1
    16k = 8
    k = 1/2
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  4. #4
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    [QUOTE=OurIslamic;468155]Hi all, thanks for taking time out of your day to have a look at my problems. I'm studying and need some help solving these two problems.

    I need to solve for k:

    Here are the problems:

    1) One root of 6x^2 - 3x +k = 0 is (1 - i√23)/4
    6(x- (1- i\sqrt{23})/4)(x- (1+ i\sqrt{23)/4
    = 6((x-1/4)+ i\sqrt{23}/4)((x- 1/4)- i\sqrt{23}/4)
    is easier to calculate.

    2) kx^2 - 3x + 4 = 0 one root is twice the other.
    [/tex]kx^2- 3x+ 4= k(x- a)(x- 2a) [/tex]
    Multiply that out and compare "corresponding coefficients to find both a and k.

    I'm not really sure how to do these problems and am really greatful for your help! Btw, the "√" sign in the first problem is a square root sign (just in case someone got confused).

    Thanks!
    Follow Math Help Forum on Facebook and Google+

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