Answer:
See below
All the inequalities can be solved by making a table, but once they are simple factored quadratic expressions, we already know what interval it will have.
Step-by-step explanation:
1)
[tex]x^2+9x+14>0[/tex]
Factoring the expression, we have
[tex](x+2)(x+7)>0[/tex]
Considering that
[tex](x+2)(x+7)=0[/tex]
for [tex]x = -2[/tex] and [tex]x=-7[/tex] you can see that for
[tex](x+2)(x+7)>0[/tex]
[tex]x \not\in (-7, -2)[/tex]
Therefore,
[tex]S = \{x\in\mathbb{R} : x<-7 \text{ or } x>-2 \}[/tex]
[tex]S = (-\infty,-7)\cup (-2,\infty)[/tex]
2)
[tex]x^2-10x+16<0[/tex]
You can factor it again.
[tex](x-2)(x-8)<0[/tex]
You can make the table to solve all the inequalities, but once we have the expression [tex]< 0[/tex], we know that
[tex]2<x<8[/tex], therefore,
[tex]S = (2, 8)[/tex]
3)
[tex]2x^2+11x+12<0[/tex]
Factoring we have
[tex](2x+3)(x+4)<0[/tex]
Find the solutions for
[tex](2x+3)(x+4)=0[/tex]
And we know that
[tex]-4<x<-\dfrac{3}{2}[/tex]
[tex]S = \left(-4,-\dfrac{3}{2}\right)[/tex]
