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:

$y = (-b +- sqrt(Delta))/(2a) = (-5+-sqrt(21))/2$

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