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

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

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Answer 1 · 2017-07-18T18:55:11

#x^2-x-1=0.#

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.!