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\({3g \over g^3 - 9g}\) when trying to isolate g, what would you do? please show your steps because i'm lost :(

Guest Mar 14, 2018
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3+0 Answers

 #1
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0

What are you going to "isolate" g from, since every term has a "g'' in it?? All you can do is simplify the expression:

 

Simplify the following:

(3 g)/(g^3 - 9 g)

 

Factor g out of g^3 - 9 g:

(3 g)/(g (g^2 - 9))

 

(3 g)/(g (g^2 - 9)) = g/g×3/(g^2 - 9) = 3/(g^2 - 9):

3/(g^2 - 9)

 

g^2 - 9 = g^2 - 3^2:

3/(g^2 - 3^2)

 

Factor the difference of two squares. g^2 - 3^2 = (g - 3) (g + 3):

 

3 / ((g - 3) (g + 3))

Guest Mar 14, 2018
 #2
avatar+12248 
0

Nonpermissable values would result in the denominator = 0

 

so as guest found   3 and -3  are two possibilities   as is 0

ElectricPavlov  Mar 15, 2018
 #3
avatar+6954 
+1

\(\frac{3g}{g^3-9g}\)

 

To find the non-permissible value of  g , set the denominator equal to zero.

 

g3 - 9g   =  0    To solve this equation for  g , we need to factor the left side.

                        Start by factoring a  g  out of both terms.

g(g2 - 9)  =  0   Now we can factor  g2 - 9  as a difference of squares.   g2 - 9  =  (g + 3)(g - 3)

 

g(g + 3)(g - 3)  =  0      Set each factor equal to zero.

 

g  =  0     or     g + 3  =  0     or     g - 3  =  0

                          g  =  -3                 g  =  3

 

The non-permissible values of  g  are  0 ,  -3 , and  3

hectictar  Mar 15, 2018

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