\({3g \over g^3 - 9g}\) when trying to isolate g, what would you do? please show your steps because i'm lost :(
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))
+36916 Nonpermissable values would result in the denominator = 0
so as guest found 3 and -3 are two possibilities as is 0
+9466 \(\frac{3g}{g^3-9g}\)
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
