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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
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7 years ago

147 Points
```										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,∞)
Please feel free to post as many doubts on our  discussion forum as you can.If you find any question Difficult to  understand - post it here and we will get you the answer and  detailed  solution very  quickly. We are all IITians and here to  help you in your IIT JEE preparation.All the best. Regards,Askiitians  ExpertsBadiuddin
```
7 years ago
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