How do you solve #x^2+4x+3=0#?

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Answer 1 · 2016-03-11T12:06:34

#x=-1and x=-3#

#=>x^2+3x+x+3#
#=>x(x+3)+1(x+3)#
#=>(x+3)(x+1)#
#x=-1and x=-3#

Answer 2 · 2016-03-11T13:03:28

x = -1 ; x = -3

Since (a - b + c) = 0, use shortcut --> x = -1 and #x = -c/a = -3#.

Reminder of Shortcut.
- When a + b + c = 0 --> x = 1 and #x = c/a#
- When a - b + c = 0 --> x = - 1 and #x = - c/a#