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

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Answer 1 · 2015-06-11T20:47:09

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#