Q50 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

If $α$ and $β$ are the roots of the equation $x^{2} - 2 x + 4 = 0,$ then $α^{12} + β^{12} =$

Solution

$x^{2} - 2 x + 4 = 0$

and $α$ and $β$ are roots of this equation so

$α^{2} = 2 α - 4, β^{2} = 2 β - 4$

Now $α^{6} = {(2 α - 4)}^{3} = 8 α^{3} - 48 α^{2} + 96 α - 64$

$= 8 α (2 α - 4) - 48 (2 α - 4) + 96 α - 64$

$= 16 (2 α - 4) - 32 α + 128 = 64 = 2^{6}$

Similarly $β^{6} = 2^{6}$

Hence $α^{12} + β^{12} = 2^{12} + 2^{12}$ $= 2^{13}$

Hence, the correct answer is option: 'C'.