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

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### 2 Answers

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**

First, we'll re-write the number 3, fro the left side, as 12 - 9. We'll re-write the equation:

y^2 - 9 + 12 = 13/(y^2 - 9)

We've made this change, to create the structure y^2 - 9.

We'll note y^2 - 9 = t

t + 12 = 13/t

We'll multiply by t both sides:

t^2 + 12t - 13 = 0

We'll apply quadratic formula:

t1 = [-12 + sqrt(144 + 52)]/2

t1= (-12 + sqrt196)/2

t1 = (-12+14)/2

t1 =1

t2 = (-12-14)/2

t2 = -13

We'll put y^2 - 9 = t1 => y^2 - 9 = 1

y^2 = 10

y1 = sqrt10 and y2 = -sqrt10

We'll put y^2 - 9 = t2 => y^2 - 9 = -13

y^2 = -13+9

y^2 = -4

**Since there is no real value for y, that raised to square to give -4, we'll conclude that the real solutions of the equation are: {-sqrt10 ; sqrt10}.**