Question

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?

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

Answer 1

`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.!