Question

The number of real roots of the equation

[ x |x| - 5 |x + 2| + 6 = 0 ] ,  is:

Answer 1

We need to find the number of real roots of the equation \( x|x| - 5|x+2| + 6 = 0 \).

Solution

Step 1: Identify the cases based on the absolute values.

The expression involves absolute values, so we need to consider different cases for \( x \):

1. Case 1: \( x < -2 \)
2. Case 2: \( -2 \le x < 0 \)
3. Case 3: \( 0 \le x \)

Step 2: Solve Case 1: \( x < -2 \)

In this case:

• \( |x| = -x \)
• \( |x+2| = -(x+2) = -x - 2 \)

Substituting these into the equation:

\( x(-x) - 5(-x - 2) + 6 = 0 \)

This simplifies to:

\( -x^2 + 5x + 10 + 6 = 0 \Rightarrow -x^2 + 5x + 16 = 0 \)

Multiplying through by -1 gives:

\( x^2 - 5x - 16 = 0 \)

Step 3: Find the roots of the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\( x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 64}}{2} = \frac{5 \pm \sqrt{89}}{2} \)

The roots are:

\( x = \frac{5 + \sqrt{89}}{2} \) and \( x = \frac{5 - \sqrt{89}}{2} \)

We need to check if these roots are less than -2.

Calculating \( \frac{5 - \sqrt{89}}{2} \):

Since \( \sqrt{89} \approx 9.43 \):

\( \frac{5 - 9.43}{2} \approx \frac{-4.43}{2} \approx -2.215 \) (valid root)

The other root \( \frac{5 + \sqrt{89}}{2} \) is positive and not valid for this case.

Step 4: Solve Case 2: \( -2 \le x < 0 \)

In this case:

• \( |x| = -x \)
• \( |x+2| = x+2 \)

Substituting these into the equation:

\( x(-x) - 5(x+2) + 6 = 0 \)

This simplifies to:

\( -x^2 - 5x - 10 + 6 = 0 \Rightarrow -x^2 - 5x - 4 = 0 \)

Multiplying through by -1 gives:

\( x^2 + 5x + 4 = 0 \)

Step 5: Find the roots of the quadratic equation

Using the quadratic formula:

\( x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2} \)

The roots are:

\( x = \frac{-2}{2} = -1 \) and \( x = \frac{-8}{2} = -4 \)

Only \( x = -1 \) is valid in this case.

Step 6: Solve Case 3: \( 0 \le x \)

In this case:

• \( |x| = x \)
• \( |x+2| = x+2 \)

Substituting these into the equation:

\( x^2 - 5(x+2) + 6 = 0 \)

This simplifies to:

\( x^2 - 5x - 10 + 6 = 0 \Rightarrow x^2 - 5x - 4 = 0 \)

Step 7: Find the roots of the quadratic equation

Using the quadratic formula:

\( x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 16}}{2} = \frac{5 \pm \sqrt{41}}{2} \)

Both roots are valid since they are positive.

Conclusion

Summarizing the valid roots:

1. From Case 1: \( \frac{5 - \sqrt{89}}{2} \)
2. From Case 2: \( -1 \)
3. From Case 3: \( \frac{5 + \sqrt{41}}{2} \) and \( \frac{5 - \sqrt{41}}{2} \)

Thus, we have a total of 4 real roots.