How do you solve #y + 2/y = 1/y - 5#?

Question

Abigail Allen · Accepted Answer

Multiply through by #y#, rearrange into standard quadratic form and solve using the quadratic formula to give:

#y = (-5+-sqrt(21))/2# Given: #y+2/y=1/y-5# First multiply both sides by #y# to get: #y^2+2 = 1-5y# (Note that in general multiplying both sides of an equation by #y# could introduce a spurious solution of #y=0#, but that won't happen in this case) Add #5y-1# to both sides to get: #y^2+5y+1 = 0# By the rational roots theorem we can immediately see that the only possible rational roots would be #y = +-1#, but neither of those works. Let's use the quadratic formula. Our quadratic equation is of the form #ay^2+by+c = 0#, with #a=1#, #b=5# and #c=1#. The discriminant is given by the formula: #Delta = b^2-4ac = 5^2 - (4xx1xx1) = 25 - 4 = 21# Being positive but not a perfect square we can tell that our equations has two distinct irrational real roots. The solutions are given by the formula: #y = (-b +- sqrt(Delta))/(2a) = (-5+-sqrt(21))/2#

Jameson Allen · Answer

undefined To solve the equation y + 2/y = 1/y - 5, we can start by multiplying both sides of the equation by y to eliminate the denominators. This gives us y^2 + 2 = 1 - 5y. Rearranging the equation, we have y^2 + 5y + 3 = 0. To solve this quadratic equation, we can factor it as (y + 3)(y + 1) = 0. Therefore, the solutions are y = -3 and y = -1.