Find number of real solutions of the equation `x(x^2+abs(x)+5abs(x-1)-6abs(x-2)=0`

Options

A. 2
B. 3
C. 4
D. 6

Answer

To find the number of real solutions of the equation

\[ x(x^2 + |x| + 5|x - 1| - 6|x - 2|) = 0, \]

we will analyze the equation by breaking it down into cases based on the values of \(x\) since the equation contains absolute values.

### Step 1: Factor out \(x\)

The equation can be factored as follows:

\[ x = 0 \quad \text{or} \quad x^2 + |x| + 5|x - 1| - 6|x - 2| = 0. \]

### Step 2: Case 1: \(x \geq 2\)

In this case, all absolute values are positive:

\[ |x| = x, \quad |x - 1| = x - 1, \quad |x - 2| = x - 2. \]

Substituting these into the equation gives:

\[ x^2 + x + 5(x - 1) - 6(x - 2) = 0. \]

Simplifying this:

\[ x^2 + x + 5x - 5 - 6x + 12 = 0 \implies x^2 + 0x + 7 = 0. \]

This simplifies to:

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

This has no real solutions since \(x^2 = -7\) is not possible for real \(x\).

### Step 3: Case 2: \(1 \leq x < 2\)

Here, we have:

\[ |x| = x, \quad |x - 1| = x - 1, \quad |x - 2| = 2 - x. \]

Substituting these into the equation gives:

\[ x^2 + x + 5(x - 1) - 6(2 - x) = 0. \]

Simplifying this:

\[ x^2 + x + 5x - 5 - 12 + 6x = 0 \implies x^2 + 12x - 17 = 0. \]

Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-12 \pm \sqrt{144 + 68}}{2} = \frac{-12 \pm \sqrt{212}}{2} = \frac{-12 \pm 2\sqrt{53}}{2} = -6 \pm \sqrt{53}. \]

Calculating the roots:

\[ x_1 = -6 + \sqrt{53}, \quad x_2 = -6 - \sqrt{53}. \]

Since \(-6 - \sqrt{53}\) is negative, we only consider \(x_1\). We need to check if \(-6 + \sqrt{53}\) is in the interval \([1, 2)\).

Calculating \(\sqrt{53} \approx 7.28\):

\[ x_1 \approx -6 + 7.28 \approx 1.28. \]

Since \(1.28\) is in the interval \([1, 2)\), we have one solution from this case.

### Step 4: Case 3: \(0 \leq x < 1\)

In this case, we have:

\[ |x| = x, \quad |x - 1| = 1 - x, \quad |x - 2| = 2 - x. \]

Substituting these into the equation gives:

\[ x^2 + x + 5(1 - x) - 6(2 - x) = 0. \]

Simplifying this:

\[ x^2 + x + 5 - 5x - 12 + 6x = 0 \implies x^2 + 2x - 7 = 0. \]

Using the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{4 + 28}}{2} = \frac{-2 \pm \sqrt{32}}{2} = \frac{-2 \pm 4\sqrt{2}}{2} = -1 \pm 2\sqrt{2}. \]

Calculating the roots:

\[ x_1 = -1 + 2\sqrt{2}, \quad x_2 = -1 - 2\sqrt{2}. \]

Since \(-1 - 2\sqrt{2}\) is negative, we only consider \(x_1\). We need to check if \(-1 + 2\sqrt{2}\) is in the interval \([0, 1)\).

Calculating \(2\sqrt{2} \approx 2.83\):

\[ x_1 \approx -1 + 2.83 \approx 1.83. \]

Since \(1.83\) is not in the interval \([0, 1)\), there are no solutions from this case.

### Step 5: Case 4: \(x < 0\)

In this case, we have:

\[ |x| = -x, \quad |x - 1| = 1 - x, \quad |x - 2| = 2 - x. \]

Substituting these into the equation gives:

\[ x^2 - x + 5(1 - x) - 6(2 - x) = 0. \]

Simplifying this:

\[ x^2 - x + 5 - 5x - 12 + 6x = 0 \implies x^2 + 0x - 7 = 0. \]

This simplifies to:

\[ x^2 - 7 = 0 \implies x^2 = 7 \implies x = \pm \sqrt{7}. \]

Since we are in the case \(x < 0\), we take \(x = -\sqrt{7}\).

### Conclusion

Now, we have identified the following solutions:

1. \(x = 0\) from Step 1.
2. \(x = -6 + \sqrt{53}\) from Step 3.
3. \(x = -\sqrt{7}\) from Step 4.

Thus, the total number of real solutions is **3**.