quadratic problem

**prasum** · March 12th 2011, 06:22 AM

if x^2-2ra[r]x+r=0,r=1,2,3 are three quadratic equations of which each pair has exactly one root common then the number of solutions of the triplet (a1,a2,a3) is

**HallsofIvy** · March 12th 2011, 07:48 AM

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.

**prasum** · March 12th 2011, 09:37 AM

how do we compare these two in pairs

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