# Let alpha and beta be the roots of the equation x^+px+1=0 and let gamma and delta be the roots of the equation x^2+(1/p)x+1=0. Given that `(alpha-gamma) (beta-gamma) (alpha-delta) (beta-delta) =...

Let alpha and beta be the roots of the equation x^+px+1=0 and let gamma and delta be the roots of the equation x^2+(1/p)x+1=0.

Given that

`(alpha-gamma) (beta-gamma) (alpha-delta) (beta-delta) = (gamma^2+p gamma+1) (delta^2+p delta+1)`

Deduce that;

`(alpha-gamma) (beta-gamma) (alpha-delta) (beta-delta) = (p-1/p)^2`

Posted on

`(alpha-gamma) (beta-gamma) (alpha-delta) (beta-delta) = (gamma^2+p gamma+1) (delta^2+p delta+1)`

RHS

`=(gamma xx delta)^2+(gamma^2) xx p xx delta +gamma^2+p xx gamma (delta)^2 + (p^2) xx gamma xx delta + p xx gamma +delta^2+p xx delta+1`

`=(gamma xx delta)^2+p xx gamma xx delta (gamma+delta) +p^2 xx gamma xx delta +p(gamma+delta)+gamma^2+delta^2+1`

`=(gamma xx delta)^2+p xx gamma xx delta (gamma+delta) +p^2 xx gamma xx delta +p(gamma+delta)+((gamma+delta)^2-2 xx gamma xx delta)+1`

We know that using `x^2+1/px+1 = 0` ;

`gamma+delta = -1/p`

RHS

`= 1+ p(-1/p)+p^2 (1)+p(-1/p)+(1/p^2-2)+1`

`=1-1+p^2-1+(1/p^2)-2+1`

`= p^2-2+1/p^2`

`=(p-1/p)^2`

So the answer is obtained as required.