If $\alpha$ and $\beta$ are the roots of the equation $2^{6 x}-3\left(2^{3 x+2}\right)+32$ $=0$ with $\beta

Question

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 $2 \alpha+3 \beta=$

Aicharya · Accepted Answer

Since, $2^{6 x}-3\left(2^{3 x+2}ight)+32=0$ $\begin{aligned} & \Rightarrow\left(2^{3 x}ight)^2-12\left(2^{3 x}ight)+32=0 \ & 	ext { Let } y=2^{3 x} \ & 	herefore y^2-12 y+32=0 \ & (y-4)(y-8)=0 \ & \Rightarrow y=4 \quad 	ext { or, } y=8 \ & \Rightarrow 2^{3 x}=4 \quad 	ext { or, } 2^{3 x}=8 \ & \Rightarrow 2^{3 x}=22 \quad 	ext { or, } 2^{3 x}=23\end{aligned}$ $\Rightarrow x=\frac{2}{3} \quad$ or, $x=1$ Hence, $\alpha=1 \& \beta=\frac{2}{3}$ Then, $2 \alpha+3 \beta=2 	imes 1+3 	imes \frac{2}{3}=4$

If $\alpha$ and $\beta$ are the roots of the equation $2^{6 x}-3\left(2^{3 x+2}\right)+32$ $=0$ with $\beta…

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