If α and β are the roots of the equation x 2+ px +2=0 and (1/α) and (1/β) are the roots of the equation 2 x2+2 q x+1=0, then (α-(1/α))(β-(1/β))(α+(1/β))(β+(1/α)) is equal to:

Question

If α and β are the roots of the equation x 2+ px +2=0 and (1/α) and (1/β) are the roots of the equation 2 x2+2 q x+1=0, then (α-(1/α))(β-(1/β))(α+(1/β))(β+(1/α)) is equal to: (A) (9/4)(9+ p 2) (B) (9/4)(9-q2) (C) (9/4)(9- p 2) (D) (9/4)(9+ q 2)

Tardigrade Admin · Accepted Answer

α, β are roots of x2+p x+2=0 ⇒ α2+ p α+2=0 β2+ p β+2=0 ⇒ (1/α), (1/β) are roots of 2 x 2+ px +1=0 But (1/α), (1/β) are roots of 2 x 2+2 qx +1=0 ⇒ p=2 q Also α+β=-p α β=2 (α-(1/α))(β-(1/β))(α+(1/β))(β+(1/α)) =((α2-1/α))((β2-1/β))((α β+1/β))((α β+1/α)) =((- p α-3)(- p β-3)(α β+1)2/(α β)2) =(9/4)( p α β+3 p (α+β)+9) =(9/4)(9- p 2)=(9/4)(9-4 q 2)

Q. If $α$ and $β$ are the roots of the equation $x^{2} + p x + 2 = 0$ and $\frac{1}{α}$ and $\frac{1}{β}$ are the roots of the equation $2 x^{2} + 2 q x + 1 = 0,$ then
$(α - \frac{1}{α}) (β - \frac{1}{β}) (α + \frac{1}{β}) (β + \frac{1}{α})$ is equal to:

$\frac{9}{4} (9 + p^{2})$

$\frac{9}{4} (9 - q^{2})$

$\frac{9}{4} (9 - p^{2})$

$\frac{9}{4} (9 + q^{2})$

Q. If α and β are the roots of the equation x2+px+2=0 and α1​ and β1​ are the roots of the equation 2x2+2qx+1=0, then (α−α1​)(β−β1​)(α+β1​)(β+α1​) is equal to:

49​(9+p2)

49​(9−q2)

49​(9−p2)

49​(9+q2)

Q. If $α$ and $β$ are the roots of the equation $x^{2} + p x + 2 = 0$ and $\frac{1}{α}$ and $\frac{1}{β}$ are the roots of the equation $2 x^{2} + 2 q x + 1 = 0,$ then
$(α - \frac{1}{α}) (β - \frac{1}{β}) (α + \frac{1}{β}) (β + \frac{1}{α})$ is equal to:

$\frac{9}{4} (9 + p^{2})$

$\frac{9}{4} (9 - q^{2})$

$\frac{9}{4} (9 - p^{2})$

$\frac{9}{4} (9 + q^{2})$