The value of 'a' for which the equation `3x^(2)+2(a^(2)+1)x+(a^(2)-3a+2)=0` has roots of opposite sign, lies in

Options

A. (1, 2)
B. (`- infty`, 1)
C. (2, `infty`)
D. `(-infty, infty)`

Answer

To determine the value of 'a' for which the quadratic equation \(3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0\) has roots of opposite signs, we can follow these steps:

### Step 1: Understand the condition for opposite sign roots
For a quadratic equation \(ax^2 + bx + c = 0\), the roots will have opposite signs if the product of the roots is negative. The product of the roots can be expressed as \(\frac{c}{a}\).

### Step 2: Identify coefficients
In our equation:
- \(a = 3\)
- \(b = 2(a^2 + 1)\)
- \(c = a^2 - 3a + 2\)

### Step 3: Set up the inequality for opposite sign roots
We need the product of the roots to be less than zero:
\[ \frac{c}{a} < 0 \]
This translates to:
\[ \frac{a^2 - 3a + 2}{3} < 0 \]

### Step 4: Eliminate the denominator
Multiplying both sides of the inequality by 3 (which is positive, so the direction of the inequality remains unchanged):
\[ a^2 - 3a + 2 < 0 \]

### Step 5: Factor the quadratic expression
Now, we can factor the quadratic:
\[ a^2 - 3a + 2 = (a - 1)(a - 2) \]
Thus, we have:
\[ (a - 1)(a - 2) < 0 \]

### Step 6: Determine the critical points
The critical points from the factors are:
- \(a = 1\)
- \(a = 2\)

### Step 7: Analyze the intervals
We will analyze the sign of the product \((a - 1)(a - 2)\) in the intervals determined by the critical points:
1. **Interval 1:** \( (-\infty, 1) \)
- Choose \(a = 0\): \((0 - 1)(0 - 2) = (−1)(−2) = 2\) (positive)

2. **Interval 2:** \( (1, 2) \)
- Choose \(a = 1.5\): \((1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25\) (negative)

3. **Interval 3:** \( (2, \infty) \)
- Choose \(a = 3\): \((3 - 1)(3 - 2) = (2)(1) = 2\) (positive)

### Step 8: Conclusion
The inequality \((a - 1)(a - 2) < 0\) holds true in the interval:
\[ (1, 2) \]
Thus, the values of 'a' for which the quadratic equation has roots of opposite signs lie in the interval \( (1, 2) \).