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piyush shukla Grade: 11
        

 

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


7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
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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Badiuddin

7 years ago
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