If α, β are the roots of the equation x2-4 x+8=0, then for any n ∈ N, α2 n+β2 n equals

Question

If α, β are the roots of the equation x2-4 x+8=0, then for any n ∈ N, α2 n+β2 n equals (A) 22 n+1 cos (n π/2) (B) 23 n cos (n π/2) (C) 23 n+1 cos (n π/2) (D) 23 n cos (n π/4)

Tardigrade Admin · Accepted Answer

Since, α, β are the roots of the equation x2-4 x+8=0 ∴ α, β=(4 ± √16-32/2)=(4 ± 4 i/2)=2 ± 2 i ⇒ α, β=2 √2((1/√2) ± i (1/√2)) =2 √2( cos (π/4) ± i sin (π/4)) Now, α2 n=(2 √2)2 n( cos (π/4)+i sin (π/4))2 n, n ∈ N =(2 √2)2 n( cos (n π/2)+i sin (n π/2)) Similarly, β2 n=(2 √2)2 n( cos (n π/2)-i sin (n π/2)) ∴ α2 n+β2 n=(2 √2)2 n ⋅ 2 cos (n π/2) =(23 / 2)2 n ⋅ 2 cos (n π/2) ⇒ α2 n+β2 n=23 n+1 cos (n π/2)

Q. If $α, β$ are the roots of the equation $x^{2} - 4 x + 8 = 0$ , then for any $n \in N, α^{2 n} + β^{2 n}$ equals

$2^{2 n + 1} cos \frac{nπ}{2}$

$2^{3 n} cos \frac{nπ}{2}$

$2^{3 n + 1} cos \frac{nπ}{2}$

$2^{3 n} cos \frac{nπ}{4}$

Q. If α,β are the roots of the equation x2−4x+8=0, then for any n∈N,α2n+β2n equals

22n+1cos2nπ​

23ncos2nπ​

23n+1cos2nπ​

23ncos4nπ​

Q. If $α, β$ are the roots of the equation $x^{2} - 4 x + 8 = 0$ , then for any $n \in N, α^{2 n} + β^{2 n}$ equals

$2^{2 n + 1} cos \frac{nπ}{2}$

$2^{3 n} cos \frac{nπ}{2}$

$2^{3 n + 1} cos \frac{nπ}{2}$

$2^{3 n} cos \frac{nπ}{4}$