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The number of real roots of $\big|x^2+4|x|+3\big| +2x-11=0$ is? I have tried by expanding modulus, and if I'm right I'm getting two real roots but I'm a bit confused because of the innermost modulus, please help

dxiv
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Nitin Jha
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    I'm getting two real roots It would help your question if you posted how you got those two roots, and what the values you found are. – dxiv Jul 23 '18 at 03:39
  • Are we consider $x \in \mathbb{R}$ or $x \in \mathbb{C}$? I presume the former, but just checking – Brevan Ellefsen Jul 23 '18 at 03:51

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If $x\ge0$ then we simply get $x (x + 6) = 8$ so that we get the positive solution $x = \sqrt{17}-3$.

If $x<0$ then we get $|x^2 - 4x+3|+2x=11$.

We break this up into two cases (consider $x^2 - 4x+3 = 0$):

  • If $2-\sqrt{15} \le x < 0$ then we get $-x^2 + 4x-3+2x=11$, which has only non-real roots (which thus can't satisfy $x<0$)
  • If $x < 2-\sqrt{15}$ then we get $x^2 - 4x+3+2x=11$ which has solutions $x = -2$ and $x=4$. Since $x < 2-\sqrt{15}$, only $x = -2$ is a valid solution.

We thus have two real solutions: $x = \sqrt{17}-3$ and $x = -2 \qquad \square$

Brevan Ellefsen
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First answer this question: Under what circumstances can $x^2 + 4\lvert x\rvert + 3$ be negative?

This will simplify the problem quite a bit.

Remember that a formula like this can be expressed as a piecewise polynomial function if you can figure out the values of $x$ at which anything inside an absolute value sign goes from negative to positive or vice versa. Within any interval where you do not have such a sign change, the function is polynomial. In this problem there are fewer pieces than you might expect.

David K
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$$\big|x^2+4|x|+3\big| +2x-11=0$$

For $x>0$ the equation is equivalent to $$x^2+4x+3+2x-11=x^2+6x -8=0$$

With solutions $x=-3\pm \sqrt {17}$

We keep the positive one. $ x=0$ is not a solution.

For $x<0$, the equation is equivalent to $$\big|x^2-4x+3\big| +2x-11=0$$ or $$ x^2-4x+3+2x-11=0$$

$$x^2-2x-8=0$$

The solutions to this last equation are $x=4$ or $x=-2$

We keep $x=-2$

Thus over all we have two solutions $x=-2$ or $x=-3+\sqrt {17}$

Mohammad Riazi-Kermani
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  • Two of those "solutions" aren't even solutions:

    $|(2)^2+4|(2)|+3\big| +2(2)-11= 19-11=8 \neq0$

    $|(4)^2+4|(4)|+3\big| +2(4)-11= 32 \neq0$

    Moreover, you missed the solution $\sqrt{17}-3$, which works since $|(\sqrt{17}-3)^2+4|(\sqrt{17}-3)|+3\big| +2(\sqrt{17}-3)-11 = (17 - 2\sqrt{17})+2(\sqrt{17}-3)-11 = 0$

     – Brevan Ellefsen Jul 23 '18 at 04:26
  • @BrevanEllefsen Thanks for the comment. I should have used quadratic formula instead of my mental math skills. – Mohammad Riazi-Kermani Jul 23 '18 at 04:38