Two roots of quadratic equations are givne, frame the equation :
`1-3sqrt5and1+3sqrt5`

Answer

`x^(2)-2x-44=0` is the required quadratic equation.

Explanation

Let `alpha=1-3sqrt5andbeta=1+3sqrt5`
Then `alpha+beta=1-3sqrt5+1+3sqrt5=2`
`alphabeta=(1-3sqrt5)(1+3sqrt5)`
`=(1)^(2)-(3sqrt5)^(2)=1-45=-44`
The required quadratic equation is
`x^(2)-(alpha+beta)x+alphabeta=0`
i.e. `x^(2)-(2)x+(-44)=0`
i.e. `x^(2)-2x-44=0`