Respuesta :
im assuming the equation is:-
3 / (2g + 8) = (g + 2) / (g^2 - 16)
3(g^2 - 16) = (2g + 8)(g + 2)
3g^2 - 48 = 2g^2 + 4g + 8g + 16
Answer is A x = -4
3g^2 - 2g^2 - 12g - 64 = 0
g^2 - 12g - 64 = 0
( g + 4)(g - 16) = 0
g = - 4, 16.
Test these solutions:-
g = -4
left side = 3/0 which is indeterminate
right side = -2/0 indeterminate
So x = -4 is extraneous.
x = 16:- LHS = 0.075 RHS = 0.075 so x = 16 is a root.
Answer:
g = -4 is the extraneous solution.
Step-by-step explanation:
The given equation :
[tex]\frac{3}{(2g+8)}=\frac{(g+2)}{(g^{2}-16)}[/tex]
3 (g²-16) = ( g + 2 ) ( 2g + 8 )
3g² - 48 = 2g² + 8g + 4g + 16
( 3g² - 2g² ) = ( 8g + 4g ) + ( 16 + 48 )
g² = 12g + 64
g² - 12g - 64 = 0
g² - 16g + 4g - 64 = 0
g ( g- 16 ) + 4 ( g - 16 ) = 0
( g + 4 ) ( g - 16 ) = 0
Since ( g + 4 ) & ( g - 16 ) are zero factors
So, g = -4, 16
Now we test these solutions. for (g = -4)
L.H.S (Left Hand Side)= [tex]\frac{3}{2(-4)+16}=\frac{3}{(-8+16)}=-(\frac{3}{8})[/tex]
R.H.S. (Right Hand Side) = [tex]\frac{(g+2)}{(g^{2}-16) }=\frac{-4+2}{16-16}=\frac{2}{0}[/tex]
L.H.S. ≠ R.H.S. So g = ( -4) is not the answer, so g = -4 is the extraneous solution.
Now for g = 16
L.H.S. [tex](\frac{3}{32+8})=\frac{3}{40}[/tex]
and R.H.S. [tex]\frac{16+2}{256-16}=\frac{18}{240}=\frac{3}{40}[/tex]
so LHS = RHS
Therefore. g = 16 is not the extraneous solution.
