prove-that-any-real-root-of-the-equation-x-6n-4x-2n-4-n-N-0-verify-gt-2-1-2n-

Question Number 150883 by mathdanisur last updated on 16/Aug/21

prove that any real root α of the

equation : x^{6 n} = {4 x}^{2 n} + 4; n \in N - {0}

verify : ∣ α ∣ > \sqrt[2 n]{2}

Answered by mindispower last updated on 16/Aug/21

X^{6 n} - 4 X^{4 n} - 4 = 0

X^{2 n} = y > 0

\Rightarrow y^{3} - 4 y^{2} - 4 = 0

f (y) = y^{3} - 4 y^{2} - 4

f^{'} (y) = y (3 y^{2} - 8), f^{'} (y) > 0, y > 2 \sqrt{\frac{2}{3}} = β

f (0) = - 4, f (β) < f (0), f (x) < 0, \forall x \in [0, β [

f (2) = - 12

\Rightarrow f (y) = 0, y > 2

\Rightarrow∣ y ∣= y > 2

∣ X^{2 n} ∣> 2 \Rightarrow∣ X ∣> \sqrt[2 n]{2}

Commented by mathdanisur last updated on 16/Aug/21

Thank you Ser, but {4 x}^{4 n} no {4 x}^{2 n}

\Rightarrow y^{3} - {4 y}^{2} . ? or 4 y