If $α, β$ and $γ$ are the roots of the equation $2^{33 x - 2} + 2^{11 x + 2} = 2^{22 x + 1} + 1,$ then $11 (α + β + γ)$ is equal to

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Solution

$2^{33 x - 2} + 2^{11 x + 2} = 2^{22 x + 1} + 1$
$\Rightarrow \frac{(2^{11 x})^{3}}{4} + 2^{11 x} \cdot 4 = (2^{11 x})^{2} \cdot 2 + 1$
Let $2^{11 x} = t$
Then, $\frac{t^{3}}{4} + 4 t = 2 t^{2} + 1$
$\Rightarrow t^{3} - 8 t^{2} + 16 t - 4 = 0$
Let the roots of the equation be $t_{1}, t_{2}, t_{3}$
$t_{1} t_{2} t_{3} = \frac{- d}{a} = 4$
$\Rightarrow 2^{11 α} \cdot 2^{11 β} \cdot 2^{11 γ} = 4$
$\Rightarrow 2^{11 (α + β + γ)} = 2^{2}$
$\Rightarrow 11 (α + β + γ) = 2$

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