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If $\alpha$ and $\beta$ are the roots of the equation $2^{6 x}-3\left(2^{3 x+2}\right)+32$ $=0$ with $\beta < 1$, then $2 \alpha+3 \beta=$
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4
Since, $2^{6 x}-3\left(2^{3 x+2}\right)+32=0$
$\begin{aligned} & \Rightarrow\left(2^{3 x}\right)^2-12\left(2^{3 x}\right)+32=0 \\ & \text { Let } y=2^{3 x} \\ & \therefore y^2-12 y+32=0 \\ & (y-4)(y-8)=0 \\ & \Rightarrow y=4 \quad \text { or, } y=8 \\ & \Rightarrow 2^{3 x}=4 \quad \text { or, } 2^{3 x}=8 \\ & \Rightarrow 2^{3 x}=22 \quad \text { or, } 2^{3 x}=23\end{aligned}$
$\Rightarrow x=\frac{2}{3} \quad$ or, $x=1$
Hence, $\alpha=1 \& \beta=\frac{2}{3}$
Then, $2 \alpha+3 \beta=2 \times 1+3 \times \frac{2}{3}=4$
$\begin{aligned} & \Rightarrow\left(2^{3 x}\right)^2-12\left(2^{3 x}\right)+32=0 \\ & \text { Let } y=2^{3 x} \\ & \therefore y^2-12 y+32=0 \\ & (y-4)(y-8)=0 \\ & \Rightarrow y=4 \quad \text { or, } y=8 \\ & \Rightarrow 2^{3 x}=4 \quad \text { or, } 2^{3 x}=8 \\ & \Rightarrow 2^{3 x}=22 \quad \text { or, } 2^{3 x}=23\end{aligned}$
$\Rightarrow x=\frac{2}{3} \quad$ or, $x=1$
Hence, $\alpha=1 \& \beta=\frac{2}{3}$
Then, $2 \alpha+3 \beta=2 \times 1+3 \times \frac{2}{3}=4$
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