# Determine all real roots of the equation x^2-9=-6/(x^2-4)?

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We have to determine the real roots of x^2 - 9 = -6/(x^2-4)

x^2 - 9 = -6/(x^2-4)

=> (x^2 - 9)(x^2 - 4) = -6

=> x^4 - 9x^2 - 4x^2 + 36 = -6

=> x^4 - 9x^2 - 4x^2 + 42 = 0

=> x^4 - 13x^3 + 42 = 0

Let x^2 = y

=> y^2 - 13y + 42 = 0

=> y^2 - 7y - 6y + 42 = 0

=> y(y - 7) - 6(y - 7) = 0

=> (y - 6)(y - 7) = 0

=> y = 6 and y = 7

As y = x^2

x = sqrt 6 and -sqtr 6

x = sqrt 7 and -sqrt 7

**The required solution for the equation is { sqrt 6, -sqtr 6, sqrt 7, -sqrt 7}**

First, we'll re-write the number 9, from the left side, as - 4 - 5 We'll re-write the equation:

x^2 - 4 - 5 = -6/(x^2-4)

We notice that we've created the structure x^2 - 4.

We'll note x^2 - 4 = t

t - 5 = -6/t

We'll multiply by t both sides:

t^2 - 5t + 6 = 0

We'll apply quadratic formula:

t1 = [5 + sqrt(25 - 24)]/2

t1= (5+1)/2

t1 =3

t2 = (5-1)/2

t2 = 2

We'll put x^2 - 4 = t1 => x^2 - 4 = 3

x^2 = 7

x1 = sqrt7 and x2 = -sqrt7

We'll put x^2 - 4 = t2 => x^2 - 4= 2

x^2 =6

x3 = sqrt6 and x4 = -sqrt6

**The all 4 real solutions of the equation are: {-sqrt7 ; -sqrt6 ; sqrt6 ; sqrt7}.**