3.2*10^-10= (y)(2y)^2/ (0.140-2y)^2 or
3.2*10^-10= 2y^3/ (0.140-2y)^2
for all practical purposes, the left side of your equation is "zero"
(it has value 0.00000000032)
so we end up with 2y^3 = appr. 0
so y = appr. 0
How would i solve for Y:
3.2*10^-10= (y)(2y)^2/ (0.140-2y)^2
3 answers
(0.140-2y)^2 * (3.2*10^-10) = 4 y^5
(0.0196 -0.280y + 4y^2)*(3.2*10^-10)
= 4 y^5
6.282*10^-12 -8.96*10^-11 y +1.28*10^-9 y^2
= 4 y^5
This is a messy fifth order polynomial that is best solved by iteration. As a first approximation, I would assume 4 y^5 = 6.282*10^-12, since the y and Y^2 terms will be negligible compared to the constant.
y^5 = 1.57*10^-12
y = 4.36*10^-3
(0.0196 -0.280y + 4y^2)*(3.2*10^-10)
= 4 y^5
6.282*10^-12 -8.96*10^-11 y +1.28*10^-9 y^2
= 4 y^5
This is a messy fifth order polynomial that is best solved by iteration. As a first approximation, I would assume 4 y^5 = 6.282*10^-12, since the y and Y^2 terms will be negligible compared to the constant.
y^5 = 1.57*10^-12
y = 4.36*10^-3
Solve 7/18 using bar notation
1 answer