Describe the solution set of each of the following quadratic inequality or not​

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]

Multiplying intergers practice look at pic

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