The real number satisfies \($x^2 - 5x + 6 < 0.$\) Find all possible values of \($x^2 + 5x + 6.$\)
+171 we are given $ x^2+5x+6<0 $ right? turn the $<$ into $=$:
$ x^2+5x+6=0 $
again, a binomial in which $ a=1$ ; $b=5$ ; $c=6 $
plugging that into $ _1x_2=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $ we get:
$ _1x_2= \frac{-5\pm \sqrt{5^2-4\cdot \:1\cdot \:6}}{2\cdot \:1} $
$ _1x_2= \frac{-5\pm \:\sqrt{25-24} }{2\cdot \:1} $
$ _1x_2= \frac{-5\pm \:1}{2\cdot \:1} $
that gives us:
$ x_1=-2$ and $x=-3 $
