The book I'm working though has just showed how to find the intersection points of functions containing moduli by graphing them, visually spotting the intersection points, and then solving algebraically. I've done the exercises and am happy with that. Now it's showing another method to solve without graph sketching. I've taken a picture of the text:

Solving Equations Involving Modulus Signs

The first exercise is asking to solve the equation $|x^2 - 1| - 1 = 3x - 2$, so I start by adding 1 to both sides, squaring, and simplifying to get $x^4 - 11x^2 + 6x = 0$. I now have no idea what to do! Can someone give he any hints on how to approach this please?

  • $\begingroup$ Find two obvious roots, and you will be back to a second degree equation you can solve. One obvious root is really obvious. For the other one, try integer values of $x$ which divide $6$ the lowest term coefficient. $\endgroup$ – Did Dec 18 '12 at 12:27
  • $\begingroup$ You have $|x^2-1|=3x-1$ after adding $1$. On the left there is a modulus, so the right side has to be positive: $3x-1\geq 0$. You do not have to square both sides. Make a distinction between $x^2-1\geq 0$ and $x^2-1\leq 0$ and that makes it solvable. $\endgroup$ – Bart Michels Dec 18 '12 at 12:31
  • $\begingroup$ @did Actually the form $|x^2-1|=3x-1$ should make a second root "almost" obvious by examination. $\endgroup$ – Mark Bennet Dec 18 '12 at 12:34
  • $\begingroup$ @MarkBennet Yup, solving $\color{red}{x}x-1=\color{red}{3}x-1$... $\endgroup$ – Did Dec 18 '12 at 13:02

An easier, though perhaps lengthier method. Divide by cases:

$$(1)\;\;x^2-1\geq 0\Longleftrightarrow x\leq -1\,\,\vee\,\,x\geq 1:\Longrightarrow |x^2-1|=x^2-1\Longrightarrow$$

$$x^2-1-1=3x-2\Longrightarrow x^2-3x=0\Longleftrightarrow x=0\,\,\vee\,\,\color{red}{x=3}$$

$$(2)\;\;x^2-1<0\Longleftrightarrow -1<x<1:\Longrightarrow |x^2-1|=1-x^2\Longrightarrow$$

$$1-x^2-1=3x-2\Longrightarrow x^2+3x-2=0\Longrightarrow x_{1,2}=\frac{-3\pm\sqrt{17}}{2}\Longrightarrow \color{red}{x=\frac{-3+\sqrt{17}}{2}}$$

The solutions are the red ones (why did we ban the other apparent solutions?)

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    $\begingroup$ Thank you for your small step by step answer. It was just what I needed to see the different branches of the problem. I understand the $x = 0$ solution was banned because we were in the branch where $x^2 - 1\geq 0$, and likewise, the solution $x=\frac{-3 - \sqrt{17}}{2}$ was banned because it didn't satisfy the inequality $-1 < x < 1$. $\endgroup$ – PeteUK Dec 18 '12 at 17:38


Case 1: $|x|<1$



$x=\frac{-3\pm \sqrt{17}}{2}$

Case 2: $|x|\geq 1$



$x=0$ or $x=3$

Therefore, $x=\frac{\sqrt{17}-3}{2}$or $x=3$.


Answer based on did's comment. In this case you can use your quartic equation $$\begin{equation*} x^{4}-11x^{2}+6x=0 \end{equation*}$$ and factor it $$\begin{eqnarray*} 0=x^{4}-11x^{2}+6x &=&x(x^{3}-11x+6)=x\left( x-3\right) \left( x^{2}+3x-2\right) \\ &=&x\left( x-3\right) (x-\frac{-3+\sqrt{17}}{2})(x-\frac{-3-\sqrt{17}}{2}). \end{eqnarray*}$$

The first factor, $x$, is easy $$\begin{equation*} x^{4}-11x^{2}+6x=x(x^{3}-11x+6). \end{equation*}$$

As for the second, $x-3$, you need to find a root of $x^{3}-11x+6=0$. By the rational root theorem you should try factors of the constant term $6$, i.e $ x=\pm 1,\pm 2,\pm 3,\pm 6$. From these $x=3$ is a root. By polynomial long division you get the 2nd degree polynomial $$\begin{equation*} \frac{x^{3}-11x+6}{x-3}=x^{2}+3x-2, \end{equation*}$$ whose roots are $\frac{-3\pm \sqrt{17}}{2}$. From the four possible roots $$\begin{equation*} x_{1}=0,\quad x_{2}=3,\quad x_{3}=\frac{-3+\sqrt{17}}{2},\quad x_{4}=\frac{-3-\sqrt{17}}{2}, \end{equation*}$$ you can check that only $x_{2}$ and $x_{3}$ are roots of the original equation $$\begin{equation*} \left\vert x^{2}-1\right\vert -1=3x-2. \end{equation*}$$

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    $\begingroup$ Thanks for the solution and expanding on did's comment which confused me at first (though I now see it made perfect sense!). I managed to follow your solution. I now see clearly the obvious first factor, and how to find the second factor to arrive at the quadratic. QUESTION: What's the difference between the rational root theorem and the factor theorem? $\endgroup$ – PeteUK Dec 18 '12 at 18:04
  • $\begingroup$ @PeteUK You are welcome. By the factor theorem a polynomial $P(x)$ has a factor $(x-x_1)$ iff $x_1$ is a root of $P(x)$ i.e. $P(x_1)=0$. The rational root theorem, which is applied to a polynomial with integer coefficients only, states the conditions its rational roots must satisfy. It is a test to find rational roots if there is any. $\endgroup$ – Américo Tavares Dec 18 '12 at 18:41

Let's go back for a moment. For an arbitrary example, notice that $|x| = 2 \Longleftrightarrow x = \pm 2$. Similarly, we have two cases:

THE FIRST CASE: $$\begin{aligned}x^2 - 1 &= 3x - 1 \\ x(x - 3)& = 0 \\ \Longleftrightarrow \boxed {x= 0 \ {\rm or} \ x = 3}\end{aligned}$$

*THE SECOND CASE:*$$\begin{aligned}x^2 - 1 &= -3x + 1 \\ \Longleftrightarrow \boxed{x = {-3 \pm \sqrt{17} \over 2}} \end{aligned}$$We have four solutions.

Notice that two solutions will not be taken in the real answer. Guess!


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