Let α and β be two roots of the equation x2 + 2x + 2 = 0, then α15 + β15 is equal to :

Question

Let α and β be two roots of the equation x2 + 2x + 2 = 0, then α15 + β15 is equal to : (A) 512 (B) - 512 (C) -256 (D) 256

Tardigrade Admin · Accepted Answer

We have (x + 1)2 + 1 = 0 ⇒ (x + 1)2 - (i)2 = 0 ⇒ (x + 1 + i) (x + 1 - i) = 0 ∴ x = - (1+i) - ( 1 -i) α15 + β15 = (α2)7 α + (β2)7 β = -128 (-i + 1 + i + 1) = - 256

Q. Let $α$ and $β$ be two roots of the equation $x^{2} + 2 x + 2 = 0$ , then $α^{15} + β^{15}$ is equal to :

512

- 512

-256

256

We have
$(x + 1)^{2} + 1 = 0$
$\Rightarrow (x + 1)^{2} - (i)^{2} = 0$
$\Rightarrow (x + 1 + i) (x + 1 - i) = 0$
$∴ x = - (1 + i) - (1 - i)$
$α^{15} + β^{15} = (α^{2})^{7} α + (β^{2})^{7} β$
= $- 128 (- i + 1 + i + 1)$
= $- 256$

Q. Let α and β be two roots of the equation x2+2x+2=0, then α15+β15 is equal to :

512

- 512

-256

256

We have (x+1)2+1=0 ⇒(x+1)2−(i)2=0 ⇒(x+1+i)(x+1−i)=0 ∴x=−(1+i)−(1−i) α15+β15=(α2)7α+(β2)7β = −128(−i+1+i+1) = −256

Q. Let $α$ and $β$ be two roots of the equation $x^{2} + 2 x + 2 = 0$ , then $α^{15} + β^{15}$ is equal to :

We have
$(x + 1)^{2} + 1 = 0$
$\Rightarrow (x + 1)^{2} - (i)^{2} = 0$
$\Rightarrow (x + 1 + i) (x + 1 - i) = 0$
$∴ x = - (1 + i) - (1 - i)$
$α^{15} + β^{15} = (α^{2})^{7} α + (β^{2})^{7} β$
= $- 128 (- i + 1 + i + 1)$
= $- 256$