The solution to the equation 4^{(x^2 + 2)} - | vedclass

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(A) Given equation: $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$Let $y = 2^{(x^2 + 2)}$. Then the equation becomes $y^2 - 9y + 8 = 0$.Factoring the

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$x = \pm 1$

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(A) Given equation: $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$Let $y = 2^{(x^2 + 2)}$. Then the equation becomes $y^2 - 9y + 8 = 0$.Factoring the quadratic: $(y - 8)(y - 1) = 0$,so $y = 8$ or $y = 1$.Case $1$: $y = 8$ $\Rightarrow 2^{(x^2 + 2)} = 2^3$ $\Rightarrow x^2 + 2 = 3$ $\Rightarrow x^2 = 1$ $\Rightarrow x = \pm 1$.Case $2$: $y = 1$ $\Rightarrow 2^{(x^2 + 2)} = 2^0$ $\Rightarrow x^2 + 2 = 0$ $\Rightarrow x^2 = -2$,which has no real solution.Thus,the solutions are $x = 1$ and $x = -1$.

The solution to the equation $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$ is:

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