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`        find set of a which equation x(x+1)(x+a(x+a+1)=a² has four real roots `
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,∞)
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```
7 years ago
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