4
$\begingroup$

Find the set of values of $\lambda$ for which the equation $|x^2-4|x|-12|=\lambda$ has 6 distinct real roots

My Approach:

$|x^2-4|x|-12|=\lambda$

Case 1:

$x^2-4|x|-12=\lambda$

  • If $x\geq 0$
    $x^2-4x-12=\lambda\cdots(i)$

  • If $x<0$ $x^2+4x-12=\lambda\cdots(ii)$

Case 2:

$x^2-4|x|-12=-\lambda$

  • If $x\geq 0$
    $x^2-4x-12=-\lambda\cdots(iii)$

  • If $x<0$ $x^2+4x-12=-\lambda\cdots(iv)$

Now, we know the equation has 6 distinct real roots. So either only 3 equations have real roots which are all distinct too, or, some roots are common. I don't now how to solve further and I need a hint to proceed.

  • 2
    Just sketch the graph of the function $|x^2-4|x|-12|$. In particular identify all of the local mins and maxs and the behavior at $\pm \infty$. Since the function is continuous with finitely many critical points, this will provide enough information to identify which values in the range are taken exactly $6$ times. – 2017-01-11

3 Answers 3

3

$\lambda$ must obviously be non-negative. Then assume

  • $x\ge0$ and

    1. $x^2-4x-12-\lambda=0$. The roots are $x=2\pm{\sqrt{16+\lambda}}$, but only the $+$ sign is valid. For all $\lambda$, one root.

    2. $x^2-4x-12+\lambda=0$. The roots are $x=2\pm{\sqrt{16-\lambda}}$. They are both positive and distinct for $12<\lambda<16$.

  • $x\le0$ and

    1. $x^2+4x-12-\lambda=0$. The roots are $x=-2\pm{\sqrt{16+\lambda}}$, but only the $-$ sign is valid. For all $\lambda$, one root.

    2. $x^2+4x-12+\lambda=0$. The roots are $x=-2\pm{\sqrt{16-\lambda}}$. They are both negative and distinct for $12<\lambda<16$.

In conclusion, for $12<\lambda<16$, there are three positive and three negative roots.

  • 1
    clearly the best answer at the moment, plotting shows (at best) that there are at least 6 roots but not that there are exactly 16 roots for these values of $\lambda$. – 2017-01-11
  • 1
    @Surb: I agree that graphical solutions are never complete and rigorous proofs. In this case, anyway, the function is piecewise quadratic and this might be used to bound the number of roots. – 2017-01-11
3

Hint: enter image description here

$$|x^2-4|x|-12|=\lambda$$

$$y=|x^2-4|x|-12|$$ $$y=\lambda$$ Answer: $$12<\lambda<16$$

3

you can plot the $|x^2-4|x|-12|$

enter image description here

then you easy to find $$12<\lambda<16$$