Question
Solve :$2^{2x}+ 2^{x+2} - 4 \times 2^3= 0$
✓
Answer
$2^{2x}+ 2^{x+2} - 4 \times 2^3= 0$
$\Rightarrow ( 2^x)^2 + 2^x. 2^2 - 4 \times 2 \times 2 \times 2 = 0$
$\Rightarrow ( 2^x)^2 + 2^x. 2^2 - 32 = 0$
Putting $y = 2^x$
$\Rightarrow y^2 + 4y - 32 = 0$
$\Rightarrow y2 + 8y - 4y - 32 = 0$
$\Rightarrow y( y + 8 ) - 4( y + 8 ) = 0$
$\Rightarrow ( y + 8 )( y - 4 ) = 0$
$\Rightarrow y + 8 = 0$ or $y - 4 = 0$
$\Rightarrow y = - 8$ or $y = 4$
$\Rightarrow 2^x = - 8$ or $2^x = 4$
$\Rightarrow 2^x = 2^2$
$[ \because 2^x = - 8$ is not possible. $]$
$\Rightarrow x = 2.$
$\Rightarrow ( 2^x)^2 + 2^x. 2^2 - 4 \times 2 \times 2 \times 2 = 0$
$\Rightarrow ( 2^x)^2 + 2^x. 2^2 - 32 = 0$
Putting $y = 2^x$
$\Rightarrow y^2 + 4y - 32 = 0$
$\Rightarrow y2 + 8y - 4y - 32 = 0$
$\Rightarrow y( y + 8 ) - 4( y + 8 ) = 0$
$\Rightarrow ( y + 8 )( y - 4 ) = 0$
$\Rightarrow y + 8 = 0$ or $y - 4 = 0$
$\Rightarrow y = - 8$ or $y = 4$
$\Rightarrow 2^x = - 8$ or $2^x = 4$
$\Rightarrow 2^x = 2^2$
$[ \because 2^x = - 8$ is not possible. $]$
$\Rightarrow x = 2.$
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