Solve the following inequailities :
`4abs(x^(2)-1)+abs(x^(2)-4)ge6`

Answer

To solve the inequality \( 4|x^2 - 1| + |x^2 - 4| \geq 6 \), we will break it down into different cases based on the expressions inside the absolute values.

### Step 1: Identify critical points
The expressions inside the absolute values are \( x^2 - 1 \) and \( x^2 - 4 \). We need to find the points where these expressions are zero:
- \( x^2 - 1 = 0 \) gives \( x = \pm 1 \)
- \( x^2 - 4 = 0 \) gives \( x = \pm 2 \)

Thus, the critical points are \( -2, -1, 1, 2 \).

### Step 2: Create intervals
The critical points divide the number line into the following intervals:
1. \( (-\infty, -2) \)
2. \( [-2, -1) \)
3. \( [-1, 1) \)
4. \( [1, 2) \)
5. \( [2, \infty) \)

### Step 3: Analyze each interval
We will analyze the expression \( 4|x^2 - 1| + |x^2 - 4| \) in each of these intervals.

#### Interval 1: \( (-\infty, -2) \)
In this interval:
- \( x^2 - 1 \geq 0 \) (since \( x^2 \geq 4 \))
- \( x^2 - 4 \geq 0 \)

Thus, \( |x^2 - 1| = x^2 - 1 \) and \( |x^2 - 4| = x^2 - 4 \).
The inequality becomes:
\[ 4(x^2 - 1) + (x^2 - 4) \geq 6 \]
\[ 4x^2 - 4 + x^2 - 4 \geq 6 \]
\[ 5x^2 - 8 \geq 6 \]
\[ 5x^2 \geq 14 \implies x^2 \geq \frac{14}{5} \]
Thus, \( x \leq -\sqrt{\frac{14}{5}} \).

#### Interval 2: \( [-2, -1) \)
In this interval:
- \( x^2 - 1 \geq 0 \)
- \( x^2 - 4 < 0 \)

Thus, \( |x^2 - 1| = x^2 - 1 \) and \( |x^2 - 4| = -(x^2 - 4) = 4 - x^2 \).
The inequality becomes:
\[ 4(x^2 - 1) + (4 - x^2) \geq 6 \]
\[ 4x^2 - 4 + 4 - x^2 \geq 6 \]
\[ 3x^2 \geq 6 \implies x^2 \geq 2 \]
Thus, \( x \in [-2, -\sqrt{2}] \).

#### Interval 3: \( [-1, 1) \)
In this interval:
- \( x^2 - 1 < 0 \)
- \( x^2 - 4 < 0 \)

Thus, \( |x^2 - 1| = -(x^2 - 1) = 1 - x^2 \) and \( |x^2 - 4| = 4 - x^2 \).
The inequality becomes:
\[ 4(1 - x^2) + (4 - x^2) \geq 6 \]
\[ 4 - 4x^2 + 4 - x^2 \geq 6 \]
\[ 8 - 5x^2 \geq 6 \implies -5x^2 \geq -2 \implies x^2 \leq \frac{2}{5} \]
Thus, \( x \in [-\sqrt{\frac{2}{5}}, \sqrt{\frac{2}{5}}] \).

#### Interval 4: \( [1, 2) \)
In this interval:
- \( x^2 - 1 \geq 0 \)
- \( x^2 - 4 < 0 \)

Thus, \( |x^2 - 1| = x^2 - 1 \) and \( |x^2 - 4| = 4 - x^2 \).
The inequality becomes:
\[ 4(x^2 - 1) + (4 - x^2) \geq 6 \]
\[ 4x^2 - 4 + 4 - x^2 \geq 6 \]
\[ 3x^2 \geq 6 \implies x^2 \geq 2 \]
Thus, \( x \in [\sqrt{2}, 2) \).

#### Interval 5: \( [2, \infty) \)
In this interval:
- \( x^2 - 1 \geq 0 \)
- \( x^2 - 4 \geq 0 \)

Thus, \( |x^2 - 1| = x^2 - 1 \) and \( |x^2 - 4| = x^2 - 4 \).
The inequality becomes:
\[ 4(x^2 - 1) + (x^2 - 4) \geq 6 \]
\[ 4x^2 - 4 + x^2 - 4 \geq 6 \]
\[ 5x^2 - 8 \geq 6 \implies 5x^2 \geq 14 \implies x^2 \geq \frac{14}{5} \]
Thus, \( x \geq \sqrt{\frac{14}{5}} \).

### Step 4: Combine the results
The solution to the inequality is the union of the results from all intervals:
- From interval 1: \( (-\infty, -\sqrt{\frac{14}{5}}] \)
- From interval 2: \( [-2, -\sqrt{2}] \)
- From interval 3: \( [-\sqrt{\frac{2}{5}}, \sqrt{\frac{2}{5}}] \)
- From interval 4: \( [\sqrt{2}, 2) \)
- From interval 5: \( [\sqrt{\frac{14}{5}}, \infty) \)

Thus, the final solution is:
\[ x \in (-\infty, -\sqrt{\frac{14}{5}}] \cup [-2, -\sqrt{2}] \cup [-\sqrt{\frac{2}{5}}, \sqrt{\frac{2}{5}}] \cup [\sqrt{2}, 2) \cup [\sqrt{\frac{14}{5}}, \infty) \]