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

Expert Answers

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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

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