Solve for real y y^2+3+ 12/(y^2+3)=7
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to solve : y^2 +3 + 12/(y^2+3)=7
y^2 +3 + 12/(y^2+3)=7
=> (y^2 + 3)^2 + 12 = 7( y^2 + 3)
=> y^4 + 9 + 6y^2 + 12 = 7y^2 + 21
=> y^4 - y^2 = 0
=> y^2( y^2 - 1) = 0
y^2 = 0 or y = 0
y^2 - 1 = 0
=> y^2 = 1
=> y = 1 and y = -1
The required real values of y are 0 , -1 and 1
We'll note y^2 + 3 = t
We'll re-write the equation in t:
t + 12/t = 7
We'll multiply by t both sides, after that, bringing all terms to one side:
t^2 + 12 - 7t = 0
We'll re-arrange the terms:
t^2 - 7t + 12 = 0
We'll apply quadratic formula:
t1 = [7 + sqrt(49 - 48)]/2
t1 = (7+1)/2
t1 = 4
t2 = (7-1)/2
t2 = 3
But y^2 + 3 = t
For t = 4=> y^2 + 3 = 4
y^2 = 4 - 3
y^2 = 1
y1 = 1 and y2 = -1
For t = 3 => y^2 + 3 = 3
y^2 = 0
y3 = y4 = 0
The real roots of the equation are: {-1 ; 0 ; 1}.