Question
Solve the following quadratic equation:
$2^{2x} - 3.2^{(x+2)} + 32= 0$
$2^{2x} - 3.2^{(x+2)} + 32= 0$
✓
Answer
$2^{2x} - 3.2^{(x+2)} + 32$
$= 0 2^{2x} - 3.2^x.2^2 + 32$
$ = 0 y^2 - 12y + 32 = 0$
where $2^x = y$
$ \Rightarrow y^2 - 8y - 4y + 32 = 0 $
$\Rightarrow y(y - 8) - 4(y - 8) = 0 $
$\Rightarrow (y - 8)(y - 4) = 0 $
$\Rightarrow y - 8 = 0$ or y$ - 4 = 0$
$\Rightarrow y = 8$ or $y = 4$
$\Rightarrow 2^x = 8$
$\Rightarrow 2^x = (2)^3$
$\Rightarrow x = 3$
$\Rightarrow 2^x = 4$
$\Rightarrow 2^x = (2)^2$
$\Rightarrow x = 2$
Hence, $3$ and $2$ are the roots of given equation.
$= 0 2^{2x} - 3.2^x.2^2 + 32$
$ = 0 y^2 - 12y + 32 = 0$
where $2^x = y$
$ \Rightarrow y^2 - 8y - 4y + 32 = 0 $
$\Rightarrow y(y - 8) - 4(y - 8) = 0 $
$\Rightarrow (y - 8)(y - 4) = 0 $
$\Rightarrow y - 8 = 0$ or y$ - 4 = 0$
$\Rightarrow y = 8$ or $y = 4$
$\Rightarrow 2^x = 8$
$\Rightarrow 2^x = (2)^3$
$\Rightarrow x = 3$
$\Rightarrow 2^x = 4$
$\Rightarrow 2^x = (2)^2$
$\Rightarrow x = 2$
Hence, $3$ and $2$ are the roots of given equation.
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