The equation `2^(2x) + (a - 1)2^(x+1) + a = 0` has roots of opposite
sing, then exhaustive set of values of a is

Options

A. ` a in (- 1, 0 )`
B. `a lt 0`
C. ` a in (-infty, 1//3)`
D. ` a in (0, 1//3)`

Explanation

To solve the equation \(2^{2x} + (a - 1)2^{x+1} + a = 0\) and find the values of \(a\) such that the roots are of opposite signs, we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2^{2x} + (a - 1)2^{x+1} + a = 0 \]
We can express \(2^{2x}\) as \((2^x)^2\) and \(2^{x+1}\) as \(2 \cdot 2^x\). Let \(t = 2^x\). Then the equation becomes:
\[ t^2 + (a - 1) \cdot 2t + a = 0 \]

### Step 2: Identify the coefficients
The equation can be rewritten as:
\[ t^2 + 2(a - 1)t + a = 0 \]
Here, the coefficients are:
- \(A = 1\) (coefficient of \(t^2\))
- \(B = 2(a - 1)\) (coefficient of \(t\))
- \(C = a\) (constant term)

### Step 3: Condition for roots of opposite signs
For the roots of the quadratic equation to be of opposite signs, the product of the roots must be negative. The product of the roots \(r_1\) and \(r_2\) is given by:
\[ r_1 r_2 = \frac{C}{A} = \frac{a}{1} = a \]
Thus, for the roots to be of opposite signs, we require:
\[ a < 0 \]

### Step 4: Condition for real roots
Additionally, for the roots to be real, the discriminant must be non-negative:
\[ D = B^2 - 4AC = [2(a - 1)]^2 - 4 \cdot 1 \cdot a \geq 0 \]
Calculating the discriminant:
\[ D = 4(a - 1)^2 - 4a = 4[(a - 1)^2 - a] \]
Expanding this:
\[ D = 4[a^2 - 2a + 1 - a] = 4[a^2 - 3a + 1] \]
Setting the discriminant greater than or equal to zero:
\[ a^2 - 3a + 1 \geq 0 \]

### Step 5: Solve the quadratic inequality
To solve \(a^2 - 3a + 1 \geq 0\), we first find the roots using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]
Let \(r_1 = \frac{3 - \sqrt{5}}{2}\) and \(r_2 = \frac{3 + \sqrt{5}}{2}\). The quadratic \(a^2 - 3a + 1\) is positive outside the interval \((r_1, r_2)\).

### Step 6: Combine conditions
We have two conditions:
1. \(a < 0\)
2. \(a \leq \frac{3 - \sqrt{5}}{2}\) or \(a \geq \frac{3 + \sqrt{5}}{2}\)

Since we need \(a < 0\), we focus on the first condition. The exhaustive set of values of \(a\) is:
\[ a < 0 \]

### Final Answer
Thus, the exhaustive set of values of \(a\) such that the equation has roots of opposite signs is:
\[ \boxed{(-\infty, 0)} \]