Let The roots of the equation #x^2 + ax + b = 0# be #alpha# and #beta#.
Similarly, Let the roots of the equation #x^2 + bx + a = 0# be #alpha# and #gamma#. [There is a common root.]
So, #alpha + beta# = #-a rArr alpha = -(a + beta)#
And #alpha + gamma# = #-b rArr alpha = -(b + gamma)#
So, #-(a + beta) = -(b + gamma)#
#rArr a + beta = b + gamma#
#rArr beta - gamma = b -a#
#rArr (beta - gamma)^2 = b^2 - 2ab + a^2#...............(i)
Again, #alphabeta = b rArr beta = b/alpha#
And, #alphagamma = a rArr gamma = a/alpha#
Then, #beta - gamma = (b -a)/alpha# and #betagamma = (ab)/alpha^2#
Putting this in eq (i)
#((b-a)/alpha)^2 = (b-a)^2#
#rArr 1/alpha^2 = 1#
#rArr alpha = +-1#
So, #betagamma = +-ab# and #beta + gamma = (b + a)/alpha = +-(a + b)#
So, The Required Equation, which has #beta and gamma# as roots is
#x^2 +- (a + b)x +- ab = 0#
But #(a + b) = +- 1# [As, #alpha = +-1, beta = b# and #gamma = a#]
So, The equation becomes #x^2 +-x +-ab = 0#
Hence Proved.
