Solve the following quadratic equation:
$2^{2 x} - {3.2}^{(x + 2)} + 32 = 0$

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Solution

$2^{2 x} - {3.2}^{x + 2} + 32 = 0 (2^{x})^{2} - 3 \times 2^{x} \times 2^{2} + 32 = 0 (2^{x})^{2} - 12 \times 2^{x} + 32 = 0 L e t a = 2^{x} a^{2} - 12 a + 32 = 0 a^{2} - 8 a - 4 a + 32 = 0 a (a - 8) - 4 (a - 8) = 0 (a - 8) (a - 4) = 0 a = 8 o r a = 4$

Case (i):

$a = 8 \Rightarrow 2^{x} = 8 \Rightarrow 2^{x} = 2^{3} \Rightarrow x = 3$

Case (ii):

$a = 4 \Rightarrow 2^{x} = 4 \Rightarrow 2^{x} = 2^{2} \Rightarrow x = 2$

So x = 3 or 2

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