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Math Help - find k

  1. #1
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    find k

    for six distinct real solutions of equation mod(x^2-5mod(x)+6)=k we have value of k as
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    Quote Originally Posted by prasum View Post
    for six distinct real solutions of equation \bigl|x^2-5|x|+6\bigr|=k we have value of k as
    Factorise the quadratic as x^2-5|x|+6 = (|x|-2)(|x|-3). So the equation becomes (|x|-2)(|x|-3) = \pm k.

    The quadratic function f(t) = (t-2)(t-3) has a minimum value 1/4 (when t=2.5). So the equation f(t) = c has two solutions if c>1/4, one solution if c=1/4, and no solutions if c<1/4. If t=|x| (and t>0), then x=\pm t, so there will be two values of x for each value of t.

    Now put that information together to find one value of k for which the equation f(|x|) = \pm k has six solutions.
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  3. #3
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    Quote Originally Posted by prasum View Post
    for six distinct real solutions of equation mod(x^2-5mod(x)+6)=k we have value of k as
    I expect that completing the square is the easiest way to go here, noting that \displaystyle |x|^2 = x^2.

    \displaystyle \begin{align*} \left||x|^2 - 5|x| + 6\right| &= k\\ \left| |x|^2 - 5|x| + \left(-\frac{5}{2}\right)^2 - \left(-\frac{5}{2}\right)^2 + 6\right| &= k \\ \left|\left(|x| - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{24}{4}\right| &= k \\ \left|\left(|x| - \frac{5}{2}\right)^2 - \frac{1}{4}\right| &= k\end{align*}

    Can you go from here?
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