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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If α,β and γ are the roots of the equation 233x−2+211x+2=222x+1+1, then 11(α+β+γ) is equal to

233x−2+211x+2=222x+1+1 ⇒(211x)34+211x⋅4=(211x)2⋅2+1 Let 211x=t Then, t34+4t=2t2+1 ⇒t3−8t2+16t−4=0 Let the roots of the equation be t1,t2,t3 t1t2t3=−da=4 ⇒211α⋅211β⋅211γ=4 ⇒211(α+β+γ)=22 ⇒11(α+β+γ)=2

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