The sum of squares of all the real solutions of the equation (x+1) (2x2+5x+3 )=4- (2x+3) (x2+2x+1 ) is equal to .
Answer
2
Solution
Conditions: x+1>0, x+1 1, 2x+3>0, 2x+3 1 x>-1, x 0
2x2+5x+3=(2x+3)(x+1), x2+2x+1=(x+1)2
Let t= (x+1)(2x+3) (t 0)
LHS =1+t; RHS =4-2 (2x+3)(x+1)=4- 2t
1+t=4- 2t t2-3t+2=0 t=1 or 2
t=1: 2x+3=x+1 x=-2 (rejected, x>-1)
t=2: 2x+3=(x+1)2 x2=2 x= 2 (x=- 2<-1 rejected)
sum of squares of real solutions = ( 2 )2=2
Answer: 2