SOLUTION: Solve the inequality and graph the solution. s^2-5s-14<0

Algebra.Com
Question 201314: Solve the inequality and graph the solution.
s^2-5s-14<0

Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
Yours is a quadratic inequality with both second degree (s^2) and a first degree term (5s). Like quadratic equalities (equations), the key to solving these problems is to make one side of the sentence (equation or inequality) equal to zero. Then you either factor the other side or use the quadratic formula to find your solution.


Your inequality already has a zero on the right side so the next step is to try to factor the left side. Using trinomial factoring we find your inequality will factor:
(s - 7)(s + 2) < 0


Think about what this is saying. It says that multiplying two numbers, (s-7) and (s+2) is less than zero. In other words, (s-7) * (s+2) is negative. What does this tell you about the two numbers? How can multiplying two numbers result in a negative answer? The only way is if one of the numbers is positive and the other is negative.


Now it is very helpful if we can figure out, between (s-7) and (s+2), which is the positive one and which is the negative one. Since all positive numbers are larger than all negative numbers we only have to figure out which is the larger number, (s-7) or (s+2). Whatever number "s" is, which will be larger, (s-7) or (s+2)? Is should be obvious that if we add 2 to "s" we will get a bigger number than if we subtract 7 from "s" no matter what "s" happens to be. So (s+2) must be the positive factor and (s-7) will be the negative factor. (NOTE: If you you get this backwards (wrong) the "solution" you come up with will be impossible.)


Now we simply state what we have deducted in Mathematical "sentences":
(s-7)(s+2) < 0 means that
s+2 > 0 and s-7 < 0
Solving these separately we get:
s > -2 and s < 7
which can be rewritten as:
-2 < s < 7
(And since it is possible for numbers, "s", to be greater than 2 and less than 7 at the same time we know we correctly determined which factor was positive and which was negative!)


(Since I have not yet figured out how to use Algrebra.com's facilities to draw the graph of the solution I will have to describe it.) On a number line there will be open, empty circles (like an "O") on both -2 and 7 and between these two circles the number line should be shaded.

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