Solving inequalities, need some confirmation

[ability]

I have these three inequalities that I am supposed to solve, I think I came up with the right answer but I'm not even 100% sure it's in the correct format.

A. 6x^2 < 6+5x
my work:
6x^2-5x-6 < 0
solutions are then 3/2 and -2/3
so the answer I got is:
-2/3 < x < 3/2

B. x^2+8x > 0
my work:
soutions I got were 0, -8
so my answer is:
-8 < x < 0

C. (x+2)(x^2-x+1) > 0
my work:
x+2 > 0 and x^2-x+1 > 0
solutions are then -.414 and 2.414
so my answer is:
-4.14 < x < 2.414

 

[Townsend]

Your first one looks fine...

B. x^2+8x > 0
my work:
soutions I got were 0, -8
so my answer is:
-8 < x < 0

y = x^2+8x

Opens up...the vertex is at (-4, -16)...you found the zeros at 0 and -8.
Graph this...Now find the parts of the graph that are BIGGER than 0...in other words, what parts of the graph are above the x-axis?

C. (x+2)(x^2-x+1) > 0
my work:
x+2 > 0 and x^2-x+1 > 0
solutions are then -.414 and 2.414
so my answer is:
-4.14 < x < 2.414

This has only one zero...x = -2.
<br /> x^2 - x +1 = (x - 1/2)^2 + 3/4 <br />

Which is clearly always above the x-axis and so you have no real roots...

So to the left of ( -2, 0)...for example x = -3, what would your expression evaluate to? A positive or negative number?

 

[arildno]

No, no no!
I'll take B for you:
x^{2}+8x&gt;0
This can be rewritten as:
x(x+8)&gt;0

What you call "solutions", are the solutions to the equation x^{2}+8x=0
These values are important in determining the regions of x-values where the INEQUALITY holds, but they are by no means indicative of these regions in the manner you think.

Let us go back to:
x(x+8)&gt;0
The left-hand side has two factors.
The product of two numbers are positive if
a) each factor is positive (that is, x>0 AND, x+8>0)
OR
b) each factor is negative (that is, x<0 AND x+8<0)

Try now to identify the regions on the x-axis where the inequality holds!