If α and β are roots of the equation, x2 - 4√2kx + 2e4 ln k -1 = 0 for some k, and α2+β2 = 66, then α3 + β3 is equal to :

Question

If α and β are roots of the equation, x2 - 4√2kx + 2e4 ln k -1 = 0 for some k, and α2+β2 = 66, then α3 + β3 is equal to : (A) 248√2 (B) 280√2 (C) -32√2 (D) -280√2

Tardigrade Admin · Accepted Answer

x2 -4√2kx+2k4 - 1 = 0 α+β = 2√2k αβ = 2k4 -1 ⇒ α 2+β 2 = 66 (α +β )2 -2 α β = 66 32 k2-2 (2k4-1) = 66 2 (2k4) - 32 k2 + 64 = 0 4 (k2-4)2 = 0 ⇒ k2 = 4 ⇒ k = 2 α 3+β 3 = (α +β )3-3 αβ(α +β ) = (α +β )(α2 +β2-αβ ) = (8√2) (66-31) = 280 √2

Q. If $α$ and $β$ are roots of the equation, $x^{2} - 42 k x + 2 e^{4 l n k} - 1 = 0$ for some k, and $α^{2} + β^{2} = 66$ , then $α^{3} + β^{3}$ is equal to :

$2482$

$2802$

$- 322$

$- 2802$

Q. If α and β are roots of the equation, x2−42​kx+2e4lnk−1=0 for some k, and α2+β2=66, then α3+β3 is equal to :

2482​

2802​

−322​

−2802​

x2−42​kx+2k4−1=0 α+β=22​k αβ=2k4−1 ⇒α2+β2=66 (α+β)2−2αβ=66 32k2−2(2k4−1)=66 2(2k4)−32k2+64=0 4(k2−4)2=0⇒k2=4⇒k=2 α3+β3=(α+β)3−3αβ(α+β) =(α+β)(α2+β2−αβ) =(82​)(66−31)=2802​

Q. If $α$ and $β$ are roots of the equation, $x^{2} - 42 k x + 2 e^{4 l n k} - 1 = 0$ for some k, and $α^{2} + β^{2} = 66$ , then $α^{3} + β^{3}$ is equal to :

$2482$

$2802$

$- 322$

$- 2802$