Solving inequalities, need some confirmation

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  • #1
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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
 

Answers and Replies

  • #2
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Your first one looks fine....

ability said:
B. x^2+8x > 0
my work:
soutions I got were 0, -8
so my answer is:
-8 < x < 0
[tex]y = x^2+8x[/tex]

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.
[tex]
x^2 - x +1 = (x - 1/2)^2 + 3/4
[/tex]

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?
 
  • #3
arildno
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No, no no!
I'll take B for you:
[tex]x^{2}+8x>0[/tex]
This can be rewritten as:
[tex]x(x+8)>0[/tex]

What you call "solutions", are the solutions to the equation [tex]x^{2}+8x=0[/tex]
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:
[tex]x(x+8)>0[/tex]
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!
 

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