SOLUTION: {{{ x^2+4x+3>0 }}}. {{{ x^2-2x-3<0 }}}. {{{ x^2+3x+2_>0 }}}. {{{ x^2+5x+4_<0 }}}.

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Question 151578: .
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Found 2 solutions by checkley77, jim_thompson5910:
Answer by checkley77(12844)   (Show Source): You can put this solution on YOUR website!
.(x+3)(x+1)>0). x>-3, x>-1
.(x-3)(x+1)<0. x<3, x<-1
. (x+2)(x+1)_>0. x_>-2, x_>-1
. (x+4)(x+1)_<0. x_<-4, x_<-1
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
I'll do the first two to get you started

# 1




Start with the given inequality



Factor the left side


Set the left side equal to zero


Set each individual factor equal to zero:

or

Solve for x in each case:

or


So our critical values are and

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





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 (notice how it's to the left of the leftmost endpoint):

So let's pick

Start with the given inequality


Plug in


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 ()





---------------------------------------------------------------------------------------------



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

So let's pick

Start with the given inequality


Plug in


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.


---------------------------------------------------------------------------------------------



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

So let's pick

Start with the given inequality


Plug in


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 ()





---------------------------------------------------------------------------------------------





Summary:

So the solution in interval notation is:


() ()





Here's a graph to visually verify the answer







# 2





Start with the given inequality



Factor the left side


Set the left side equal to zero


Set each individual factor equal to zero:

or

Solve for x in each case:

or


So our critical values are and

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





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 (notice how it's to the left of the leftmost endpoint):

So let's pick

Start with the given inequality


Plug in


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.


---------------------------------------------------------------------------------------------



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

So let's pick

Start with the given inequality


Plug in


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 ()





---------------------------------------------------------------------------------------------



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

So let's pick

Start with the given inequality


Plug in


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.


---------------------------------------------------------------------------------------------





Summary:

So the solution in interval notation is:


()





Here's a graph to visually verify the answer



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