Question

The number of distinct real roots of the equation \[ |x| \, |x + 2| - 5|x + 1| - 1 = 0 \] is _________.

Answer 1

Correct Answer: 3

Consider the different cases based on the value of \(x\).

Case 1: \(x \geq 0\)

\[ x^2 + 2x - 5x - 1 = 0 \implies x^2 - 3x - 6 = 0 \]

The roots are given by:

\[ x = \frac{3 \pm \sqrt{9 + 24}}{2} = \frac{3 \pm \sqrt{33}}{2} \]

Since \(x \geq 0\), one positive root exists.

Case 2: \(-1 \leq x < 0\)

\[ -x^2 - 2x - 5x - 1 = 0 \implies x^2 + 7x + 6 = 0 \]

The roots are:

\[ x = -1, \, x = -6 \]

Only \(x = -1\) is within the range.

Case 3: \(-2 \leq x < -1\)

\[ x^2 - 2x + 5x - 1 = 0 \implies x^2 - 3x - 4 = 0 \]

The roots are:

\[ x = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2} \]

No root lies in the range.

Case 4: \(x < -2\)

\[ x^2 + 7x + 4 = 0 \]

The roots are:

\[ x = \frac{-7 \pm \sqrt{49 - 16}}{2} = \frac{-7 \pm \sqrt{33}}{2} \]

One root lies in the range.

Total number of distinct real roots: 3

Answer 2

Step 1: Identify breakpoints and split into intervals
The expression involves \(|x|\), \(|x+1|\), \(|x+2|\). Breakpoints are \(x=-2,-1,0\). Analyze on the intervals: \((-\infty,-2)\), \([-2,-1)\), \([-1,0)\), \([0,\infty)\).

Step 2: Region x ≥ 0
Here \(|x|=x,\ |x+2|=x+2,\ |x+1|=x+1\). The equation becomes \[ x(x+2)-5(x+1)-1=0 \;\Longrightarrow\; x^2-3x-6=0. \] Roots: \(x=\dfrac{3\pm\sqrt{33}}{2}\). Only \(x=\dfrac{3+\sqrt{33}}{2}\) satisfies \(x\ge 0\). Hence one root here.

Step 3: Region -1 \le x < 0
Here \(|x|=-x,\ |x+2|=x+2,\ |x+1|=x+1\). The equation becomes \[ -(x^2+2x)-5(x+1)-1=0 \;\Longrightarrow\; x^2+7x+6=0. \] Roots: \(x=-1,-6\). Only \(x=-1\) lies in \([-1,0)\). So \(x=-1\) is a root.

Step 4: Region -2 \le x < -1
Here \(|x|=-x,\ |x+2|=x+2,\ |x+1|=-(x+1)\). The equation becomes \[ -(x^2+2x)+5(x+1)-1=0 \;\Longrightarrow\; x^2-3x-4=0. \] Roots: \(x=4,-1\). Only \(x=-1\) belongs to this region (as the endpoint), which we already have. No new root added.

Step 5: Region x < -2
Here \(|x|=-x,\ |x+2|=-(x+2),\ |x+1|=-(x+1)\). The equation becomes \[ x(x+2)+5(x+1)-1=0 \;\Longrightarrow\; x^2+7x+4=0. \] Roots: \(x=\dfrac{-7\pm\sqrt{33}}{2}\). Only \(x=\dfrac{-7-\sqrt{33}}{2}\) satisfies \(x<-2\). Hence one root here.

Step 6: Count distinct real roots
We have three distinct solutions: \[ x=-1,\quad x=\frac{-7-\sqrt{33}}{2},\quad x=\frac{3+\sqrt{33}}{2}. \] Therefore, the number of distinct real roots is \(3\).

Final answer

3