If alpha and beta are the roots of the equati | vedclass

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(D) Given the equation: $2^{6x} - 3(2^{3x+2}) + 32 = 0$Since $2^{3x+2} = 2^{3x} 	imes 2^2 = 4 	imes 2^{3x}$,the equation becomes:$(2^{3x})^2 - 3(4 \

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$4$

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(D) Given the equation: $2^{6x} - 3(2^{3x+2}) + 32 = 0$Since $2^{3x+2} = 2^{3x} 	imes 2^2 = 4 	imes 2^{3x}$,the equation becomes:$(2^{3x})^2 - 3(4 	imes 2^{3x}) + 32 = 0$$(2^{3x})^2 - 12(2^{3x}) + 32 = 0$Let $y = 2^{3x}$. Then the equation is $y^2 - 12y + 32 = 0$.Factoring the quadratic: $(y - 4)(y - 8) = 0$.So,$y = 4$ or $y = 8$.Case $1$: $2^{3x} = 4 = 2^2 \implies 3x = 2 \implies x = \frac{2}{3}$.Case $2$: $2^{3x} = 8 = 2^3 \implies 3x = 3 \implies x = 1$.Given $\beta Therefore,$2\alpha + 3\beta = 2(1) + 3(\frac{2}{3}) = 2 + 2 = 4$.

If $\alpha$ and $\beta$ are the roots of the equation $2^{6x} - 3(2^{3x+2}) + 32 = 0$ with $\beta < 1$,then $2\alpha + 3\beta =$

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