If $α, β$ $alpha, beta$ are the roots of the equation $x^{2} - 3 x + 1 = 0$ $x^2-3x+1=0$ then the equation whose roots are $\frac{1}{α - 2}, \frac{1}{β - 2}$ $1/(alpha-2) , 1/ (beta-2)$ will be?

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Answer is : $x^{2} - x - 1 = 0$ $x^2-x-1=0$

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Answer 1 · 2017-07-18T18:55:11

$x^{2} - x - 1 = 0 .$ $x^2-x-1=0.$

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

$a x^{2} + b x + c = 0; α + β = - \frac{b}{a}, and, α \cdot β = \frac{c}{a} .$ $ax^2+bx+c=0; alpha+beta=-b/a, and, alpha*beta=c/a.$

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.!