What are the real solutions of the equation y^2+3=13/(y^2-9) ?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

We have the equation y^2 + 3 = 13/y^2 - 9 and we need to find the real solutions.

y^2 + 3 = 13/y^2 - 9

=> y^4 + 3y^2 = 13 - 9y^2

=> y^4 + 12y^2 - 13 = 0

let y^2 = x

=> x^2 + 12x - 13 = 0

=> x^2 + 13x - x - 13 = 0

=> x(x + 13) - 1(x + 13) = 0

=> ( x - 1)(x + 13) = 0

x = 1 and x = -13

As we want only the real solutions of y we can ignore x = -13 as y^2 = -13 and that will give complex values of y.

y^2 = 1

=> y = 1 and y = -1

The real solutions of y are y = 1 and y = -1

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial