The equation 4^{(x^2 + 2)} - 9 cdot 2^{(x^2 + | vedclass

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(D) Given equation: $4^{(x^2 + 2)} - 9 \cdot 2^{(x^2 + 2)} + 8 = 0$Let $y = 2^{(x^2 + 2)}$. Since $4^{(x^2 + 2)} = (2^2)^{(x^2 + 2)} = (2^{(x^2 + 2)})

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$x = 1$ and $x = -1$

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

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

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