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