Vikas TU
Last Activity: 8 Years ago
Dear Mudit,
∣x²+4x+3∣=3
mod always open with + and – sign cases.
Therfore,
case 1 with + sign,
x²+4x+3= + 3
x²+4x = 0
x(x + 4) = 0
x = 0 and x = – 4
Case 2
x²+4x+3= – 3
x²+4x+6 = 0
D
x²+4x+6 > 0
never zero therefore x belongs to null.
So the solutions are only x = 0 and x = – 4.