Solve each of the following quadratic equations:
$3^{(x + 2)} + 3^{- x} = 10$

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Solution

Given;

$3^{x + 2} + 3^{- x} = 10 \Rightarrow 3^{x} {.3}^{2} + \frac{1}{3^{x}} = 10 \Rightarrow {9.3}^{x} + \frac{1}{3^{x}} = 10 \Rightarrow \frac{{9.3}^{x} {.3}^{x} + 1}{3^{x}} = 10 \Rightarrow 9 (3^{x})^{2} + 1 = {10.3}^{x} \Rightarrow 9 (3^{x})^{2} - {10.3}^{x} + 1 = 0$

Put $z = 3^{x}$

$9 z^{2} - 10 z + 1 = 0 \Rightarrow 9 z^{2} - 9 z - z + 1 = 0 \Rightarrow 9 z (z - 1) - 1 (z - 1) = 0 \Rightarrow (9 z - 1) (z - 1) = 0 \Rightarrow 9 z - 1 = 0 o r z - 1 = 0 \Rightarrow z = \frac{1}{9} o r z = 1$

$∴ 3^{x} = \frac{1}{9} o r 3^{x} = 1 \Rightarrow 3^{x} = 3^{- 2} o r 3^{x} = 3^{0} \Rightarrow x = - 2 o r 0$

Thus, the values of x are 0 or – 2.

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