How do you solve #x^2+2x-3<0#?

1 Answer
Oct 23, 2016

Answer:

#x^2+2x-3<0# for #-3 < x < 1#

Explanation:

#x^2+2x-3<0#
#rArr(x+3)(x-1)<0#

#color(blue)(x+3<0)rArrcolor(blue)(x<-3)#
#color(blue)(x+3>0)rArrcolor(blue)(x> -3)#

#color(purple)(x-1<0)rArrcolor(purple)(x<1)#
#color(purple)(x-1>0)rArrcolor(purple)(x>1)#

Then,
#x+3 < 0 and x-1 < 0 for -oo< x< -3 #
then,#(x+3)(x-1)>0# #color(red)(REJECTED)#

#x+3 > 0 and x-1 < 0 f or -3< x< 1 #
then,#(x+3)(x-1) < 0# #color(red)(AC CEPTED)#

#x+3 > 0 and x-1 > 0 for 1< x< +oo #
then,#(x+3)(x-1)>0# #color(red)(REJECTED)#

Therefore,

#x^2+2x-3<0# for #-3 < x < 1#