SOLUTION: Example 4 on page 352 shows how to do this problem, but I am having problems factoring once I get to the point of setting the 2-x-x^2>0 and then solving for x (factoring is one of

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Example 4 on page 352 shows how to do this problem, but I am having problems factoring once I get to the point of setting the 2-x-x^2>0 and then solving for x (factoring is one of       Log On


   

Question 151505This question is from textbook precalculus
: Example 4 on page 352 shows how to do this problem, but I am having problems factoring once I get to the point of setting the 2-x-x^2>0 and then solving for x (factoring is one of my major weaknesses). I seem to have trouble with factoring whenever I am faced with needing to combine the same variable in two different forms (e.g. the -x-x^2). Thanks. This question is from textbook precalculus

Found 2 solutions by checkley77, jim_thompson5910:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
2-x-x^2>0
-X^2-X+2>0
(-X+1)(X+2)=0
-X+1=0
-X=-1
X=1 ANSWER.
X+2=0
X=-2 ANSWER.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
2-x-x%5E2%3E0 Start with the given inequality.


-x%5E2-x%2B2%3E0 Rearrange the terms.


-x%5E2-x%2B2%3E0 Rearrange the terms.


-%28x%2B2%29%28x-1%29%3E0 Factor the left side (note: if you need help with factoring, check out this solver)



-%28x%2B2%29%28x-1%29=0 Set the left side equal to zero


Set each individual factor equal to zero:

x%2B2=0 or x-1=0

Solve for x in each case:

x=-2 or x=1


So our critical values are x=-2 and x=1

Now set up a number line and plot the critical values on the number line

number_line%28+600%2C+-10%2C+10%2C-2%2C1%29



So let's pick some test points that are near the critical values and evaluate them.


Let's pick a test value that is less than -2 (notice how it's to the left of the leftmost endpoint):

So the interval that we are testing is ()


So let's pick x=-3 to test the interval

-x%5E2-x%2B2%3E0 Start with the given inequality


-%28-3%2B2%29%28-3-1%29%3E+0 Plug in x=-3


-4%3E+0 Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not in our solution set and we can ignore it.


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Let's pick a test value that is in between -2 and 1:


So the interval that we are testing is ()

So let's pick x=-0 to test the interval

-x%5E2-x%2B2%3E0 Start with the given inequality


-%28-0%2B2%29%28-0-1%29%3E+0 Plug in x=-0


2%3E+0 Evaluate and simplify the left side

Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.


So part our solution in interval notation is ()





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Let's pick a test value that is greater than 1 (notice how it's to the right of the rightmost endpoint):


So the interval that we are testing is ()


So let's pick x=2

-x%5E2-x%2B2%3E0 Start with the given inequality


-%282%2B2%29%282-1%29%3E+0 Plug in x=2


-4%3E+0 Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not in our solution set and we can ignore it.


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Summary:

So the solution in interval notation is:


()





Here's a graph to visually verify the answer:

+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+-%28x%2B2%29%28x-1%29%29+