anonymous
  • anonymous
Solve the inequality in terms of intervals. 2x^2 + x <= 1
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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phi
  • phi
this one is tricker. you have to move the 1 to the left side, and factor the quadratic
anonymous
  • anonymous
|dw:1377977958266:dw|
anonymous
  • anonymous
OH, so I need the quadratic formula for this one?

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phi
  • phi
true, but that is not how to factor it.
phi
  • phi
I think 2 x^2 + x -1 ≤ 0 factors nicely (2x ) (x ) ≤ 0 it looks like 2x - 1 and x+1
phi
  • phi
once you get A * B ≤ 0 you have to test two cases A*B is negative (less than 0) if either 1) A < 0 and *at the same time* B > 0 (minus times a plus is minus) 2) A> 0 and at the same time B <0 same idea
phi
  • phi
** in this case use ≤ or ≥ rather than < or > in the above cases
phi
  • phi
in other words 2 x^2 + x -1 ≤ 0 is the same as (2x -1)(x+1) ≤ 0 this means the stuff on the left side is either 0 or a negative number. you need to find when the left side is negative. It will be negative when the two terms have opposite signs.
anonymous
  • anonymous
Alright, so I follow now with the idea that 2x^2 +x -1 is factorable and that x = 1/2 and -1, and using the test interval method I found |dw:1377978522870:dw| How do I know what the answer is from this? Is it the center section or the outside sections?
phi
  • phi
you want the interval that is labeled with a -, because that is where the expression 2 x^2 + x -1 ≤ 0 is true
phi
  • phi
if the problem were 2 x^2 + x -1 ≥ 0 you would pick the intervals labeled +
anonymous
  • anonymous
So, if the less than or equal sign were greater than or equal to sign that I'd be looking for the positives?
anonymous
  • anonymous
YAY!!!!!!
anonymous
  • anonymous
Now I understand the concept of the positives and the negatives. :)

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