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Thread: Find Values of k

  1. #1
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    Find Values of k

    For which values of k will the roots of the equation

    x^2 = 2x(3k + 1) - 7(2k + 3) be equal?

    x^2 = 6xk + 2 - 14k - 21

    x^2 = 6xk - 14k -19

    Is this a new start?
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  2. #2
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    Re: Find Values of k

    Quote Originally Posted by mathdad1965 View Post
    For which values of k will the roots of the equation

    x^2 = 2x(3k + 1) - 7(2k + 3) be equal?

    x^2 = 6xk + 2 - 14k - 21

    x^2 = 6xk - 14k -19

    Is this a new start?
    no, check your algebra

    you should get

    $x^2 = (2+6k)x -(14k+21)$
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  3. #3
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    Re: Find Values of k

    Hint: A quadratic of the form $\displaystyle \begin{align*} a\,x^2 + b\,x + c = 0 \end{align*}$ has exactly one root (or two equal roots) when the discriminant $\displaystyle \begin{align*} b^2 - 4\,a\,c = 0 \end{align*}$.
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    Re: Find Values of k

    Quote Originally Posted by romsek View Post
    no, check your algebra

    you should get

    $x^2 = (2+6k)x -(14k+21)$
    How did you get +21 when -7 3 = -21?
    How did you get 14k when -7 2k = -14k?
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  5. #5
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    Re: Find Values of k

    Quote Originally Posted by Prove It View Post
    Hint: A quadratic of the form $\displaystyle \begin{align*} a\,x^2 + b\,x + c = 0 \end{align*}$ has exactly one root (or two equal roots) when the discriminant $\displaystyle \begin{align*} b^2 - 4\,a\,c = 0 \end{align*}$.
    Are you saying to rewrite x^2=(2+6k)x−(14k+21) in the form ax^2 + bx + c and then use the discriminant by setting it to zero?
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  6. #6
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    Re: Find Values of k

    Yes
    Thanks from mathdad1965
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  7. #7
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    Re: Find Values of k

    Quote Originally Posted by Prove It View Post
    Yes
    Ok. I will solve it tomorrow.
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  8. #8
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    Re: Find Values of k

    $x^2=2x(3k+1)-7(2k+3)$

    $x^2=6kx+2x-14k-21$

    $x^2=(6k+2)x-(14k+21)$

    $x^2-(6k+2)x+(14k+21)=0$

    For roots to be equal

    $b^2-4ac=0$

    $[-(6k+2)]^2-41(14k+21)=0$

    $36k^2+4+24k-56k-84=0$

    $36k^2-32k-80=0$

    $4(9k^2-8k-20)=0$

    $9k^2-8k-20=0$

    $(k-2)(9k+10)=0$

    $k=2,\;-\dfrac{10}{9}$
    Thanks from mathdad1965
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  9. #9
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    Re: Find Values of k

    Quote Originally Posted by deesuwalka View Post
    $x^2=2x(3k+1)-7(2k+3)$

    $x^2=6kx+2x-14k-21$

    $x^2=(6k+2)x-(14k+21)$

    $x^2-(6k+2)x+(14k+21)=0$

    For roots to be equal

    $b^2-4ac=0$

    $[-(6k+2)]^2-41(14k+21)=0$

    $36k^2+4+24k-56k-84=0$

    $36k^2-32k-80=0$

    $4(9k^2-8k-20)=0$

    $9k^2-8k-20=0$

    $(k-2)(9k+10)=0$

    $k=2,\;-\dfrac{10}{9}$
    Very impressive math work.
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  10. #10
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    Re: Find Values of k

    Quote Originally Posted by mathdad1965 View Post
    x^2 = 2x(3k + 1) - 7(2k + 3)
    x^2 = 6xk + 2 - 14k - 21
    Should be:
    x^2 = 6xk + 2x - 14k - 21
    That's why you got mixed up and didn't understand Romsek's results.
    Thanks from mathdad1965
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  11. #11
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    Re: Find Values of k

    Quote Originally Posted by DenisB View Post
    Should be:
    x^2 = 6xk + 2x - 14k - 21
    That's why you got mixed up and didn't understand Romsek's results.
    Ok.
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