For the record, 2 and -3 are two roots. The last two roots can easily be found from the remaining quadratic equation. -3 can be discarded due to being a negative number whereas a root is always positive. As for the two roots from the quadratic equation, one is negative, whereas the other one is larger than \sqrt{7} which is in contradiction with the initial root problem, so both can be discarded too. Only 2 is a valid solution to the initial problem.
2and-3are two roots. The last two roots can easily be found from the remaining quadratic equation.-3can be discarded due to being a negative number whereas a root is always positive. As for the two roots from the quadratic equation, one is negative, whereas the other one is larger than\sqrt{7}which is in contradiction with the initial root problem, so both can be discarded too. Only2is a valid solution to the initial problem.kshows up on the right hand side of the equation too...