Find the set of all real 'a' such that `5a^(2)-3a-2,a^(2)+a-2` and `2a^(2)+a-1` are the lenghts of the sides of a triangle?

Answer

To find the set of all real 'a' such that \(5a^2 - 3a - 2\), \(a^2 + a - 2\), and \(2a^2 + a - 1\) are the lengths of the sides of a triangle, we need to apply the triangle inequality conditions. The triangle inequality states that for any triangle with sides \(x\), \(y\), and \(z\):

1. \(x + y > z\)
2. \(y + z > x\)
3. \(z + x > y\)

Let's denote:
- \(x = 5a^2 - 3a - 2\)
- \(y = a^2 + a - 2\)
- \(z = 2a^2 + a - 1\)

We will check each triangle inequality condition step by step.

### Step 1: Check \(x + y > z\)

\[ (5a^2 - 3a - 2) + (a^2 + a - 2) > (2a^2 + a - 1) \]

Combine like terms:

\[ 5a^2 + a^2 - 2a - 4 > 2a^2 + a - 1 \]

This simplifies to:

\[ 6a^2 - 2a - 4 > 2a^2 + a - 1 \]

Subtract \(2a^2 + a - 1\) from both sides:

\[ 6a^2 - 2a - 4 - 2a^2 - a + 1 > 0 \]

This simplifies to:

\[ 4a^2 - 3a - 3 > 0 \]

### Step 2: Solve the quadratic inequality \(4a^2 - 3a - 3 > 0\)

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

\[ a = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} \]

Calculating the discriminant:

\[ = \frac{3 \pm \sqrt{9 + 48}}{8} = \frac{3 \pm \sqrt{57}}{8} \]

The roots are:

\[ a_1 = \frac{3 + \sqrt{57}}{8}, \quad a_2 = \frac{3 - \sqrt{57}}{8} \]

The inequality \(4a^2 - 3a - 3 > 0\) holds for:

\[ a < \frac{3 - \sqrt{57}}{8} \quad \text{or} \quad a > \frac{3 + \sqrt{57}}{8} \]

### Step 3: Check \(y + z > x\)

\[ (a^2 + a - 2) + (2a^2 + a - 1) > (5a^2 - 3a - 2) \]

Combine like terms:

\[ 3a^2 + 2a - 3 > 5a^2 - 3a - 2 \]

This simplifies to:

\[ 3a^2 + 2a - 3 - 5a^2 + 3a + 2 > 0 \]

This simplifies to:

\[ -2a^2 + 5a - 1 > 0 \]

### Step 4: Solve the quadratic inequality \(-2a^2 + 5a - 1 > 0\)

Using the quadratic formula:

\[ a = \frac{-5 \pm \sqrt{5^2 - 4 \cdot (-2) \cdot (-1)}}{2 \cdot (-2)} \]

Calculating the discriminant:

\[ = \frac{-5 \pm \sqrt{25 - 8}}{-4} = \frac{-5 \pm \sqrt{17}}{-4} \]

The roots are:

\[ a_1 = \frac{5 + \sqrt{17}}{4}, \quad a_2 = \frac{5 - \sqrt{17}}{4} \]

The inequality \(-2a^2 + 5a - 1 > 0\) holds for:

\[ \frac{5 - \sqrt{17}}{4} < a < \frac{5 + \sqrt{17}}{4} \]

### Step 5: Check \(z + x > y\)

\[ (2a^2 + a - 1) + (5a^2 - 3a - 2) > (a^2 + a - 2) \]

Combine like terms:

\[ 7a^2 - 2a - 3 > a^2 + a - 2 \]

This simplifies to:

\[ 7a^2 - 2a - 3 - a^2 - a + 2 > 0 \]

This simplifies to:

\[ 6a^2 - 3a - 1 > 0 \]

### Step 6: Solve the quadratic inequality \(6a^2 - 3a - 1 > 0\)

Using the quadratic formula:

\[ a = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \]

Calculating the discriminant:

\[ = \frac{3 \pm \sqrt{9 + 24}}{12} = \frac{3 \pm \sqrt{33}}{12} \]

The roots are:

\[ a_1 = \frac{3 + \sqrt{33}}{12}, \quad a_2 = \frac{3 - \sqrt{33}}{12} \]

The inequality \(6a^2 - 3a - 1 > 0\) holds for:

\[ a < \frac{3 - \sqrt{33}}{12} \quad \text{or} \quad a > \frac{3 + \sqrt{33}}{12} \]

### Step 7: Combine the results

Now we need to find the intersection of the intervals obtained from all three inequalities:

1. From \(4a^2 - 3a - 3 > 0\): \(a < \frac{3 - \sqrt{57}}{8}\) or \(a > \frac{3 + \sqrt{57}}{8}\)
2. From \(-2a^2 + 5a - 1 > 0\): \(\frac{5 - \sqrt{17}}{4} < a < \frac{5 + \sqrt{17}}{4}\)
3. From \(6a^2 - 3a - 1 > 0\): \(a < \frac{3 - \sqrt{33}}{12}\) or \(a > \frac{3 + \sqrt{33}}{12}\)

### Final Result

The final set of values for \(a\) will be the intersection of these intervals.