Question 50458This question is from textbook Pre-Calculus with Limits
: I tried the following problem and came up with a different answer than the textbook.
x^2+4x+4>9
x^2+4x-5>0
(x+5)(x-1)>0
x+5>0, x-1>0
x>-5, x>1
The book has the answer to be x<-5 and x>1. I checked my work a few times and I still can't find what I did wrong. Please help me find my error. Thank you in advance.
This question is from textbook Pre-Calculus with Limits
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! x^2+4x+4>9
x^2+4x-5>0
(x+5)(x-1)>0
At this point you need to mark -5 and +1 on a number line.
Those two points create three segments on the line.
You need to check a value in each segment to see where your
solutions lie.
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For example choose values -10,0,and 10
If x=-10 the inequality becomes (-5)(-11) which is greater zero
This means there are solutions in the interval (-infinity,-5)
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If x=0 the inequality becomes (5)(-1) which is less than zero
So, no solutions in the interval (-5,1)
Cheers,
Stan H.
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If x=10 the inequality becomes (15)(9) which is greater than zero.
This means the interval (1,+infinity) is part of the solution set of the inequality.
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Conclusion:
x<-5 or x>1 is the solution set.
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