If alpha, beta are the roots of the equationx^2-3x+1=0 then the equation whose roots are 1/(alpha-2) , 1/ (beta-2) will be?

Answer is : x^2-x-1=0

1 Answer
Jul 18, 2017

x^2-x-1=0.

Explanation:

Knowing that, for the roots alpha and beta of the Quadr. Eqn.

ax^2+bx+c=0; alpha+beta=-b/a, and, alpha*beta=c/a.

In our case, we have, :. alpha+beta=3, and, alpha*beta=1.........(1).

We seek for the quadr. eqn. with roots, gamma and delta, where,

gamma=1/(alpha-2), and, delta=1/(beta-2).

We find, gamma+delta=1/(alpha-2)+1/(beta-2),

={(beta-2)+(alpha-2)}/{(alpha-2)(beta-2)},

=(alpha+beta-4)/{alphabeta-2(alpha+beta)+4},

=(3-4)/{1-2(3)+4}=(-1)/(-1),

rArr gamma+delta=1............(2).

Also, gamma*delta=1/(alpha-2)*1/(beta-2),

=1/{alphabeta-2(alpha+beta)+4}=(1)/(-1)=-1......(3).

Now, we know that, the quadr. eqn. having roots gamma, delta is,

x^2-(gamma+delta)x+(delta.gamma)=0.

Hence, the Desired Eqn., is x^2-x-1=0.

Enjoy Maths.!