HELP...!!! Quadratic equation?

Question

Transform the equation in the Quadratic Equation form and solve the equation for the indicated variable.

For "r"

1/r+ 2/1-r=4/r^2

Show me how transform the equation in the Quadratic Equation form .
Show me the method by steps please

Answer 1

make common donimenator:
(1-r+2r)/(r*(1-r)) = (4/r^2)
(1+r)/(r*(1-r)) = (4/r^2)
multiply the numerator of the RHS to denominator of LHS and the vice versa
4r(1-r)=(1+r)r^2
4r-4r^2=r^2+r^3
r^3+5r^2-4r=0
r(r^2+5r-4)=0
so r=0 or r^2+5r-4=0 which is quadratic equation... use the rule to get its two roots... r= -5+sqrt(5^2-4*1*-4)/2*1
and r=-5-sqrt(5^2-4*1*-4)/2*1
so the first root is r=0
the second is r=0.7015
the third is r=-5.7015

regards

Answer 2

I assume you mean 1/r + 2/(1 - r) = 4/r^2
Multiplying both sides by r^2(1 - r),
=> r - r^2 + 2r^2 = 4
=> r^2 + r - 4 = 0
=> r = (1/2)[ -1 +or- sqrt (17) ]

Answer 3

r is
r1= (-1/4+33^1/2 /4)^1/3
r2= (-1/4-33^1/2 /4)^1/3

Answer 4

eqn. can be written as

r^2 + 5r - 4 = 0

r = -6 &1.07

Answer 5

Hi,
-----------------------------------------------------------
(1/r)+2/(1-r)=4/r^2
or (1-r+2r)/{r(r-1)}=4/r^2
or (1+r)/{r(r-1)}=4/r^2
or r^2 + r^3=4r-4r^2
or r^3=4r - 4r^2 - r^2
or r^3=4r-5r^2
or r^2=4 - 5r
or r^2 + 5r -4=0
------------------------------------------------------------