Question
Solve the following quadratic equation:
$3^{(x+2)} + 3^{-x} = 10$
$3^{(x+2)} + 3^{-x} = 10$
✓
Answer
$3^{(x+2)} + 3^{-x} = 10$
$3^x.3^2 + 3^{-x} = 10$
$\Rightarrow\text{9y}+\frac{1}{\text{y}}=10$ where $3^x = y$
$\Rightarrow 9y^2 - 10y + 1 = 0$
$\Rightarrow 9y^2 - 9y - y + 1 = 0$
$\Rightarrow 9y(y - 1) - 1(y - 1) = 0$
$\Rightarrow (9y - 1)(y - 1) = 0$
$\Rightarrow 9y - 1 = 0 or y - 1 = 0$
$\Rightarrow\text{y}=\frac{1}{9}$ or y = 1
If $3^\text{x}=\frac{1}{9}$
$\Rightarrow 3x = (3)^{-2}$
$\Rightarrow x = -2$
$If 3x = 1 = 30$
$\Rightarrow x = 0$
Hence, -2, 0 are the roots of given equation.
$3^x.3^2 + 3^{-x} = 10$
$\Rightarrow\text{9y}+\frac{1}{\text{y}}=10$ where $3^x = y$
$\Rightarrow 9y^2 - 10y + 1 = 0$
$\Rightarrow 9y^2 - 9y - y + 1 = 0$
$\Rightarrow 9y(y - 1) - 1(y - 1) = 0$
$\Rightarrow (9y - 1)(y - 1) = 0$
$\Rightarrow 9y - 1 = 0 or y - 1 = 0$
$\Rightarrow\text{y}=\frac{1}{9}$ or y = 1
If $3^\text{x}=\frac{1}{9}$
$\Rightarrow 3x = (3)^{-2}$
$\Rightarrow x = -2$
$If 3x = 1 = 30$
$\Rightarrow x = 0$
Hence, -2, 0 are the roots of given equation.
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