If the equation `2^(2x)+a*2^(x+1)+a+1=0` has roots of opposite sign, then the exhaustive set of real values of `a` is (a)(−∞,0) (b)(−1,−2/3) (c)(−∞,−2/3) (d)(−1,∞)

Options

A. `(-oo,0)`
B. `(-1,(-2)/(3))`
C. `(-oo,(-2)/(3))`
D. `(-1,oo)`

Explanation

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

### Step 1: Substitute \(2^x\) with \(T\)
Let \(T = 2^x\). Then, we can rewrite the equation as:
\[ T^2 + a \cdot 2T + (a + 1) = 0 \]
This simplifies to:
\[ T^2 + 2aT + (a + 1) = 0 \]

### Step 2: Analyze the roots
For the roots of the quadratic equation \(T^2 + 2aT + (a + 1) = 0\) to be of opposite signs, the product of the roots must be negative. The product of the roots \(r_1\) and \(r_2\) can be found using Vieta's formulas:
\[ r_1 \cdot r_2 = \frac{a + 1}{1} \]
Thus, we require:
\[ a + 1 < 0 \quad \Rightarrow \quad a < -1 \]

### Step 3: Calculate the discriminant
Next, we need to ensure that the roots are real. For the roots to be real, the discriminant \(D\) of the quadratic must be non-negative:
\[ D = (2a)^2 - 4 \cdot 1 \cdot (a + 1) \geq 0 \]
Calculating the discriminant:
\[ D = 4a^2 - 4(a + 1) = 4a^2 - 4a - 4 \]
Setting the discriminant greater than or equal to zero:
\[ 4a^2 - 4a - 4 \geq 0 \quad \Rightarrow \quad a^2 - a - 1 \geq 0 \]

### Step 4: Solve the quadratic inequality
To solve \(a^2 - a - 1 \geq 0\), we first find the roots of the equation \(a^2 - a - 1 = 0\) using the quadratic formula:
\[ a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2} \]
The roots are:
\[ a_1 = \frac{1 + \sqrt{5}}{2}, \quad a_2 = \frac{1 - \sqrt{5}}{2} \]
The approximate values are:
\[ a_1 \approx 1.618, \quad a_2 \approx -0.618 \]

### Step 5: Determine the intervals
The quadratic \(a^2 - a - 1\) opens upwards, so it is non-negative outside the roots:
\[ a \leq \frac{1 - \sqrt{5}}{2} \quad \text{or} \quad a \geq \frac{1 + \sqrt{5}}{2} \]
Thus, we have:
\[ a \in (-\infty, \frac{1 - \sqrt{5}}{2}] \cup [\frac{1 + \sqrt{5}}{2}, \infty) \]

### Step 6: Combine conditions
From the condition \(a < -1\) and \(a \leq \frac{1 - \sqrt{5}}{2}\), we find that:
\[ a < -1 \quad \text{is the more restrictive condition.} \]

### Conclusion
Thus, the exhaustive set of real values of \(a\) such that the equation has roots of opposite signs is:
\[ (-\infty, -1) \]