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

Guest Mar 14, 2018

#1**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**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**+1 **

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

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

g^{3} - 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(g^{2} - 9) = 0 Now we can factor g^{2} - 9 as a difference of squares. g^{2} - 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