# Math Help - quadratic problem

2. Your notation is a bit strange. Do you mean $x^2- 2a_1x+ 1= 0$, $x^2- 4a_2x+ 2= 0$, and $x^2- 6a_3x+ 3= 0$?
By the quadratic formula, the solution to the first is $x= a_1\pm\sqrt{a_1^2- 1}$, to the second $x= 2a_2\pm\sqrt{4a_2^2- 2}$, and to the third $x= 3a_3\pm\sqrt{9a_3- 3}$. Compare those in pairs so see how you can get one common root.