Sum of the roots of the equation is `4^(x)- 3 (2 ^(x+3)) + 128 = 0`

Options

A. 5
B. 6
C. 7
D. 8

Answer

To find the sum of the roots of the equation \( 4^x - 3(2^{x+3}) + 128 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting \( 4^x \) in terms of \( 2^x \):
\[ 4^x = (2^2)^x = (2^x)^2 \]
Let \( y = 2^x \). Then, the equation becomes:
\[ y^2 - 3(2^{x+3}) + 128 = 0 \]
We can express \( 2^{x+3} \) as:
\[ 2^{x+3} = 2^x \cdot 2^3 = 8y \]
Now, substituting this back into the equation gives:
\[ y^2 - 3(8y) + 128 = 0 \]
This simplifies to:
\[ y^2 - 24y + 128 = 0 \]

### Step 2: Apply the quadratic formula
Now we will use the quadratic formula to find the roots of the equation \( y^2 - 24y + 128 = 0 \). The quadratic formula is given by:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -24 \), and \( c = 128 \).

### Step 3: Calculate the discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = (-24)^2 = 576 \]
\[ 4ac = 4 \cdot 1 \cdot 128 = 512 \]
Thus, the discriminant is:
\[ b^2 - 4ac = 576 - 512 = 64 \]

### Step 4: Find the roots
Now substituting back into the quadratic formula:
\[ y = \frac{24 \pm \sqrt{64}}{2} \]
Calculating \( \sqrt{64} \):
\[ \sqrt{64} = 8 \]
So we have:
\[ y = \frac{24 \pm 8}{2} \]
Calculating the two possible values for \( y \):
1. \( y = \frac{24 + 8}{2} = \frac{32}{2} = 16 \)
2. \( y = \frac{24 - 8}{2} = \frac{16}{2} = 8 \)

### Step 5: Find the sum of the roots
The roots of the equation \( 2^x = y \) are:
1. \( 2^x = 16 \) implies \( x = 4 \)
2. \( 2^x = 8 \) implies \( x = 3 \)

Thus, the sum of the roots \( x_1 + x_2 = 4 + 3 = 7 \).

### Final Answer
The sum of the roots of the equation is:
\[ \boxed{7} \]