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

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

Grade:12

5 Answers

sanju Ramanathan
47 Points
8 years ago
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
Akshita Goyal
17 Points
8 years ago
@sakthi meena  no your answer is wrong because it is mod before the term.
Shubham Swastik Behera
43 Points
8 years ago
|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. 
Shubham Swastik Behera
43 Points
8 years ago
correction!! x2+4x+3=(x+3)(x+1)=+ve for x -1. cheers!!
Shubham Swastik Behera
43 Points
8 years ago
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.
Cheers!!

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