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

inequality

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.

Best Answer

$\lambda$ 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.

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