solve mod x^2 +4x +3=x+1.

x² + 4x + 3 = x +1 

x² + 3x +x + 3 = x + 1 (splitting the middle term)

x(x+3) 1(x+3) = x + 1

(x+1)(x+3) = x+1

x+3 = x+1/x+1

x+3 = 1

x= 1-3= -2

@sakthi meena  no your answer is wrong because it is mod before the term.

|x²+4x+3|=x+1,

Now, the LHS changes at different values of x.

x²+4x+3=(x+3)(x+1)=+ve for x -1  ...(i) while it is -ve for x between -3 and -1  ...(ii)

So, for (i) we have x²+4x+3=x+1 on solving which we get x= -2 or -1 but we cannot take -2 as it is inbetween -3 and -1 i.e. not satisfying range of x set in (i)      whereas for (ii) we have -(x²+4x+3)=x+1 on solving which we get x= -4 or -1 but we cannot take -4 as it doesnot lie in the range set in (ii) so our only solution is x=-1. 

correction!! x2+4x+3=(x+3)(x+1)=+ve for x -1. cheers!!

Something is wrong with the greater than and less than symbol.

I’m writing it in words: x2+4x+3=(x+3)(x+1)=+ve for when x is outside (-3,-1) i.e. less than -3 and greater than -1.