+0  
 
0
258
1
avatar+564 

 

Could someone please help me with this? 

chilledz3non  Feb 9, 2016
 #1
avatar+87604 
0

Let's do this by process of elimination......

 

Notice that  the graph indicates that 0 is in the solution set....  but...if I  put 0  into A,   6/4 >= 4  is not true....so that one is no good

 

And the graph indicates that a "large" negative value will satisy some inequality.......but.....putting a "large" negative into D  will cause it to become false

 

Look at C.....the graph indicates that -4 can't be part of the solution set.....but.... -3 / [-4 + 2.5]  = 2

And this certainly satisfies C

 

Look at B....we can find the possible intervals that satisfy this as follows:

 

3/[x + 4]  <= 2        get a common denominator on the right

 

3/[x + 4]  <=  2[x + 4] / [x + 4]      turn this into an equality

 

3/ [x + 4] = [2x + 8] / [x + 4]      cross-multiply

 

3[x + 4] = [x + 4][2x + 8]

 

3x + 12  = 2x^2 + 16x + 32    re-arrange

 

2x^2 + 13x + 20  = 0   factor

 

(2x + 5) (x + 4)  = 0

 

Setting each of the factors to 0, we have that x = -5/2  = -2.5   or x = -4

 

So........the solution will come from these possible intervals

 

(-inf, -4) , (-4, -2.5]  or [-2.5, inf )      [note that -4 can't be included, because it makes the denominator of the inequality = 0 ]

 

Notice that if we put -3 into   3/[x + 4]  we get 3....and this does not satisfy B

 

So....the solution comes from the other two intervals  (-inf, -4) U [-2.5, inf)    which is the same as the graphed solution  ......so, B is correct

 

 

 

cool cool cool

CPhill  Feb 9, 2016

13 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.