For real roots, the solution of the equation `2^(2x+2)-6^(x)-2xx3^(2x+2)=0` is

Options

A. `-1`
B. `-2`
C. 3
D. none

Answer

To solve the equation \(2^{2x+2} - 6^x - 2 \cdot 3^{2x+2} = 0\), we will follow these steps:

### Step 1: Rewrite the equation
The original equation can be rewritten as:
\[ 2^{2x+2} - 6^x - 2 \cdot 3^{2x+2} = 0 \]
We can express \(2^{2x+2}\) as \(4 \cdot 2^{2x}\) and \(6^x\) as \(2^x \cdot 3^x\). Thus, we have:
\[ 4 \cdot 2^{2x} - 2^x \cdot 3^x - 2 \cdot 3^{2x+2} = 0 \]

### Step 2: Substitute variables
Let \(u = 2^x\) and \(v = 3^x\). Then, we can rewrite the equation as:
\[ 4u^2 - 2uv - 2v^2 \cdot 9 = 0 \]
This simplifies to:
\[ 4u^2 - 2uv - 18v^2 = 0 \]

### Step 3: Rearrange to standard quadratic form
Rearranging gives us:
\[ 4u^2 - 2uv - 18v^2 = 0 \]

### Step 4: Solve the quadratic equation
This is a quadratic equation in \(u\). We can use the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 4\), \(b = -2v\), and \(c = -18v^2\).

Calculating the discriminant:
\[ D = (-2v)^2 - 4 \cdot 4 \cdot (-18v^2) = 4v^2 + 288v^2 = 292v^2 \]

Now substituting back into the quadratic formula:
\[ u = \frac{2v \pm \sqrt{292v^2}}{8} \]
\[ u = \frac{2v \pm \sqrt{292}v}{8} \]
\[ u = \frac{v(2 \pm \sqrt{292})}{8} \]

### Step 5: Substitute back for \(u\) and \(v\)
Recall that \(u = 2^x\) and \(v = 3^x\):
\[ 2^x = \frac{3^x(2 \pm \sqrt{292})}{8} \]

### Step 6: Solve for \(x\)
Now we can express this in terms of \(x\):
\[ \frac{2^x}{3^x} = \frac{2 \pm \sqrt{292}}{8} \]
\[ \left(\frac{2}{3}\right)^x = \frac{2 \pm \sqrt{292}}{8} \]

Taking logarithms:
\[ x = \frac{\log\left(\frac{2 \pm \sqrt{292}}{8}\right)}{\log\left(\frac{2}{3}\right)} \]

### Step 7: Determine real roots
Since \(2^x\) and \(3^x\) are always positive, we need to check the values of \(2 \pm \sqrt{292}\). The negative root will not yield a valid solution since it would result in a negative ratio.

### Conclusion
Thus, the real root of the equation is:
\[ x = -2 \]