How do you solve # 14-5x-x^2<0#?

1 Answer
NickTheTurtle
Mar 29, 2017

#x<-7 uu x>2#

Explanation:

Just to make things easier, we multiply both sides by #-1# and flip the inequality sign: #x^2+5x-14>0#. This way the coefficient before #x^2# is positive. (Note: this is not a necessary step.)

We factorize the expression: #x^2+5x-14=(x+7)(x-2)>0# to find the roots: #x=-7# or #2#.

These are the only two possible points where the expression can change signs. Thus, there are three regions that have the same sign:

  • #x<-7#
  • #-7 < x < 2#
  • #x>2#

Since there are no repeated roots, the signs of the region alternate. We substitute an arbitrary number in the second region to find the sign: #x=0#. Then #x^2+5x-14=0^2+5*0-14=-14<0#. Thus, the second region is negative.

Therefore, the first and third regions are positive (you can use substitution to be sure). Looking at the inequality, #x^2+5x-14>0#, we notice that we need to find regions where the value is positive.

The solution is the first and third regions: #x<-7 uu x>2#.

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