Polynomial Inequality ProblemDate: 2/7/96 at 12:44:6 From: Anonymous Subject: polynomial inequalities Solve: xsquared-5x-6>0 (x-6)(x+1)>0 x>6 x<-1 OR x<6 x>-1 How can I tell which one I'm suppose to use? 7ysquared<63 7ysquared-63<0 What do I need to do next? Can you please explain to me what I need to be doing with these kinds of problems? Date: 2/7/96 at 15:40:42 From: Doctor Elise Subject: Re: polynomial inequalities Hi! Let's look at the first one. You've done a great job factoring it. (x - 6)(x + 1) > 0 Okay. So we're multiplying two numbers together, and their product has to be a positive number. This means we're either multiplying a positive number by a positive number, or a negative number by a negative number, right? So EITHER: (x - 6) > 0 and (x + 1) > 0 (both numbers are positive) x > 6 and x > -1 which basically boils down to x > 6, since 6 > -1 OR (x - 6) < 0 and (x + 1) < 0 (both numbers are negative) x < 6 and x < -1 which boils down to x < -1 So your answer is either: x > 6, or x < -1 The second one, 7y^2 - 63 < 0 , needs to be factored. (I always write "y squared" as y^2). If possible, I try to factor out the constant in front of the y^2. We can do it here, so I'll show you what I mean. Notice that: 7y^2 - 63 = 7(y^2 - 9) That way, now we just have to factor the smaller term y^2 - 9. Remember though, that factoring out a constant isn't always possible. So now we need to factor: y^2 - 9 Because the first term is y^2, we know we at least have: (y )(y ) Since we have a -9, we know our signs must be like this: (y - )(y + ) Finally, since there's no "y" term, we know that the two middle terms will have to cancel out. This means that the constants we are looking for must be equal. In this case, they are both 3. Thus, 7y^2 - 63 factors to: 7(y - 3)(y + 3) So now we have: 7(y - 3)(y + 3) < 0 First, we see that we can divide by 7. It doesn't change the problem, and it makes the solution easier: (y - 3)(y + 3) < 0 In this case, we are multiplying two numbers together and getting a negative number, so one has to be positive and the other negative. negative * positive y - 3 < 0 and y + 3 > 0 y < 3 and y > -3 OR positive * negative y - 3 > 0 and y + 3 < 0 y > 3 and y < -3 This second solution is obviously impossible, because a number cannot simultaneously be greater than 3 and less than -3, so you just throw this answer out as "impossible" and go with y is between -3 and 3. Hope this helps! -Doctor Elise, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/