The equation goes: 2G(x)-1/(G(x)+1=2√x^{2}+1 -1/(√x^{2}+1 +1)
I need to know how would you go about solving for the G(x) because i'm so stuck!
Note: The "x^{2}+1" is under the square root. It jus didn't look right so I had to make a note.
The equation goes: 2G(x)-1/(G(x)+1=2√x^{2}+1 -1/(√x^{2}+1 +1)
I need to know how would you go about solving for the G(x) because i'm so stuck!
Note: The "x^{2}+1" is under the square root. It jus didn't look right so I had to make a note.
Because if it's (2G(x)-1)/(G(x)) then just split up the numerator and you'll get 2-1/G(x)= 2√x 2+1 -1/(√x2+1 +1) and that's the equivalent of 2-[2√x2+1 -1/(√x2+1 +1) ]=1/G(x)
So that'll give you G(x)=1/(2-[2√x2+1 -1/(√x2+1 +1)])
Hello, EJdive43!
How is this any different from any other "solve for G" problem?
$\displaystyle \text{Solve for G: }\:\frac{2G-1}{G+1} \:=\:\frac{2\sqrt{x^2+1} - 1}{\sqrt{x^2+1} + 1}$
We have: .$\displaystyle (2G-1)(\sqrt{x^2+1} + 1) \:=\:(G+1)(2\sqrt{x^2+1} - 1)$
. . $\displaystyle 2G\sqrt{x^2+1} + 2G - \sqrt{x^2+1} - 1 \:=\:2G\sqrt{x^2+1} - G + 2\sqrt{x^2+1} - 1$
x . . . . . . . . . . . . . . . . . . . . $\displaystyle 3G \:=\:3\sqrt{x^2+1}$
. . . . . . . . . . . . . . . . . . . . . . $\displaystyle G \:=\:\sqrt{x^2+1}$