The sum of all possible values of x satisfying the equations 2^{4x^{2}}-2^{2 $x^{2} + x+16}+2^{2 x+30}=0 , is
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Official Correct Answer: C. The equation is: \(2^{4x^2} - 2^{2x^2 + x + 16} + 2^{2x + 30} = 0\). We need to find the sum of all possible values of \(x\) that satisfy this equation. The options are given, and the correct answer is supposed to be C, which is \(\frac{1}{2}\).
Hmm, okay. So, this looks like an exponential equation with different powers of 2. Since all terms are powers of 2, maybe I can manipulate them by taking logarithms or expressing them with a common base. But before jumping into that, if I can rewrite the equation in a way that makes it easier to handle.
First, let me note that all the exponents are functions of \(x\). Let's write them out:
1. The first term is \(2^{4x^2}\).
2. The second term is \(2^{2x^2 + x + 16}\).
3. The third term is \(2^{2x + 30}\).
Note that the exponents can be expressed in terms of \(x^2\) and \(x\). Maybe I can make a substitution to simplify things. Let me let \(y = 2^{x^2}\). Wait, but looking at the exponents, the first term is \(4x^2\), which is \((2x^2)^2\), so maybe that's not the best substitution. Alternatively, perhaps I can let \(t = 2^{x^2}\), but .
Wait, another approach: Factoring out the smallest exponent term. The smallest exponent is \(4x^2\), but actually, looking at the exponents:
- \(4x^2\)
- \(2x^2 + x + 16\)
- \(2x + 30\)
Hmm, not sure if factoring out the smallest is the way to go. Alternatively, maybe I can express all terms in terms of \(2^{2x^2}\) or something similar.
Wait, The first term is \(2^{4x^2} = (2^{2x^2})^2\).
The second term is \(2^{2x^2 + x + 16} = 2^{2x^2} \cdot 2^{x + 16}\).
The third term is \(2^{2x + 30}\).
So, if I let \(a = 2^{2x^2}\), then the equation becomes:
\(a^2 - a \cdot 2^{x + 16} + 2^{2x + 30} = 0\).
Hmm, that might be helpful. Let me write that down:
\(a^2 - a \cdot 2^{x + 16} + 2^{2x + 30} = 0\).
Now, this looks like a quadratic in terms of \(a\). So, perhaps I can solve for \(a\) using the quadratic formula.
But before that, if I can express \(2^{x + 16}\) and \(2^{2x + 30}\) in terms of another variable. Let me let \(b = 2^x\). Then:
\(2^{x + 16} = 2^{16} \cdot 2^x = 65536 \cdot b\).
Similarly, \(2^{2x + 30} = 2^{30} \cdot (2^x)^2 = 1073741824 \cdot b^2\).
So, substituting back into the equation:
\(a^2 - a \cdot 65536b + 1073741824b^2 = 0\).
Hmm, that seems a bit messy with those large coefficients. Maybe there's a better substitution or a different approach.
Wait, let me think again. The original equation is:
\(2^{4x^2} - 2^{2x^2 + x + 16} + 2^{2x + 30} = 0\).
Factoring this equation. Maybe I can factor by grouping.
Looking at the first two terms:
\(2^{4x^2} - 2^{2x^2 + x + 16}\).
Factor out \(2^{2x^2}\):
\(2^{2x^2}(2^{2x^2} - 2^{x + 16})\).
So, the equation becomes:
\(2^{2x^2}(2^{2x^2} - 2^{x + 16}) + 2^{2x + 30} = 0\).
Hmm, not sure if that helps. Alternatively, maybe I can write all terms with exponents in terms of \(x^2\) and \(x\), but I don't see a clear path.
Wait, another idea: Let me set \(t = 2^{x^2}\). Then, \(2^{4x^2} = t^4\), and \(2^{2x^2} = t^2\). But the exponent in the second term is \(2x^2 + x + 16\), so that would be \(t^2 \cdot 2^{x + 16}\). Similarly, the third term is \(2^{2x + 30}\).
So, substituting \(t = 2^{x^2}\), the equation becomes:
\(t^4 - t^2 \cdot 2^{x + 16} + 2^{2x + 30} = 0\).
Hmm, this still has both \(t\) and \(x\) in it, which complicates things. Maybe I need another substitution.
Wait, let me think about the exponents again. The exponents are \(4x^2\), \(2x^2 + x + 16\), and \(2x + 30\). Maybe I can express all exponents in terms of \(2x^2\). - \(4x^2 = 2 \cdot 2x^2\)
- \(2x^2 + x + 16 = 2x^2 + x + 16\)
- \(2x + 30 = 2x + 30\)
Not sure if that helps. Alternatively, maybe I can let \(u = 2^{2x^2}\), then \(2^{4x^2} = u^2\), and \(2^{2x^2 + x + 16} = u \cdot 2^{x + 16}\