SOLUTION: Provide answers in interval notation . s^2 - 2s - 35 < 0

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Question 363215: Provide answers in interval notation .
s^2 - 2s - 35 < 0
Found 2 solutions by Fombitz, stanbon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
s%5E2+-+2s+-+35+%3C+0
%28s-7%29%28s%2B5%29+%3C+0
Break up the number line into 3 regions using the critical points of the function.
Region 1:(-infinity,-5)
Region 2:(-5,7)
Region 3:(7,infinity)
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For each region, choose a point in the region (not an endpoint).
Test the inequality.
If the inequality is satisfied, the region is part of the solution region.
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Region 1:s=-6
%28s-7%29%28s%2B5%29+%3C+0
%28-6-7%29%28-6%2B5%29+%3C+0
%28-13%29%28-1%29%3C0+
13%3C0
False, Region 1 is not part of the solution region.
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Region 2:s=0
%28s-7%29%28s%2B5%29+%3C+0
%28-7%29%285%29+%3C+0
-35%3C0
True, Region 2 is part of the solution region.
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Region 3:s=8
%28s-7%29%28s%2B5%29+%3C+0
%288-7%29%288%2B5%29+%3C+0
%281%29%2813%29%3C0+
13%3C0
False, Region 3 is not part of the solution region.
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Solution Region:(-5,7)
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Graphical verification: Look for regions where the function is above the x-axis (y%3C0)
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Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Provide answers in interval notation .
s^2 - 2s - 35 < 0
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Find the boundary value for "s":
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s^2 - 2s - 35 = 0
(x-7)(x+5) = 0
x = -5 or x = 7
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Plot x=-5 and x=7 on a number line.
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Check a value from each of the resulting intervals
in the INEQUALITY:
(x-7)(x+5) < 0
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Let x = -10 to get: -*- < 0 ; false
Let x = 0 to get: -*+ < 0 ; true, so solutions in (-5,7)
Let x = 10 to get: +*+ < 0 ; false
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Solution interval: (-5,7)
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Cheers,
Stan H.