find set of a which equation x(x+1)(x+a(x+a+1)=a² has four real roots answer is (-∞,-2-√5]U[2-√5,-2+√5]U[2+√5,∞) but plz give me whole solution of this question

Dear piyush

x(x+1)(x+a)(x+a+1)=a²

x(x+1)(x+a){(x+1)+a}=a²

x(x+1){x(x+1)+2(xa)+a +a²}=a²

{x(x+1)}² + 2(xa){x(x+1)} + (a+a²) x(x+1) =a²

{x(x+1)}² + 2(xa){x(x+1)} + (ax)²  +ax(x+1+a)  =a²

 {x(x+1) +ax}² + a{x(x+1) +ax} =a²

 {x(x+1) +ax +a/2}²   =(√5a/2)²

{x(x+1) +ax +a/2 -√5a/2}{x(x+1) +ax +a/2 +√5a/2} =0

{x² +x(a+1) +a/2 -√5a/2}{x² +x(a+1) +a/2 +√5a/2} =0

x² +x(a+1) +a/2 +√5a/2 =0   and  x² +x(a+1) +a/2 -√5a/2 =0

for 4 real value discrimnent of both the equation must be grater than or equal to zero

x² +x(a+1) +a/2 +√5a/2 =0            and       x² +x(a+1) +a/2 -√5a/2 =0

D≥0                                                                      D≥0

(a+1)² -4(a/2 +√5a/2)≥0                      (a+1)² -4(a/2 -√5a/2)≥0                 

  (a-(√5+2))(a-(√5-2))≥0                      (a-(-√5+2))(a-(-√5-2))≥0

 so solution is

(-∞,-2-√5]U[2-√5,-2+√5]U[2+√5,∞)

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