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