How do you solve and write the following in interval notation: #x^2+4x-12>0#?

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Answer 1 · 2017-06-01T04:55:26

#x in RR: (-oo,-6)uu(+2,+oo)#

#x^2+4x-12>0#

To solve this inequality and write in interval notation we must find the values of #x in RR# for which #x^2+4x-12>0#

Let's first consider the critical points for which #x^2+4x-12=0#

#x^2+4x-12 = (x+6)(x-2)#
which #= 0# for #x=-6 or x=2#

We know #x^2+4x-12# is a parabola with a minimum value (since the coefficient of #x^2 >0#

Hence, #x^2+4x-12>0# for all real #x# from #-oo# up to #-6# exclusive and from #+2# exclusive up to #+oo#

This can be written as: #x in RR: (-oo,-6)uu(+2,+oo)#

We can see this from the graph of #x^2+4x-12# below:

graph{x^2+4x-12 [-26.85, 24.5, -17.23, 8.45]}