# Thread: find k

1. ## find k

for six distinct real solutions of equation mod(x^2-5mod(x)+6)=k we have value of k as

2. Originally Posted by prasum
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.

3. Originally Posted by prasum
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?