The sum of all possible values of x satisfying the equation $$2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$$, is
It is given thatĀ $$2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$$, which can be written as:
=>$$\left(2^{2x^2}\right)^2-2^{2x^2}\cdot2^{x+15}\cdot2^1+\left(2^{x+15}\right)^{^2}=0$$
=>Ā $$\left(2^{2x^2}-2^{x+15}\right)^{^2}=0$$
=>Ā $$2^{2x^2}-2^{x+15}=0$$ (SinceĀ $$\left(a-b\right)^2\ =\ 0\ =>\ a-b\ =0$$)
=>Ā $$2x^2\ =\ x+15$$
=>Ā $$2x^2-x-15=0$$Ā
=>Ā $$2x^2-6x+5x-15=0$$
=>Ā $$2x\left(x-3\right)+5\left(x-3\right)=0$$
=>Ā $$\left(2x+5\right)\left(x-3\right)\ =\ 0$$
Hence, the possible values of x areĀ $$-\frac{5}{2}$$, andĀ $$3$$, respectively.
Therefore, the sum of the possible values isĀ $$\left(3-\frac{5}{2}\right)=\frac{1}{2}$$
The correct option is D
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