Find the roots of the following equation $\displaystyle\,2y^2\,+\,\frac{15}{y^2}\,=\,12$, then
- $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6\,+\,\sqrt6}{4},\,y\,=\,\pm\,\sqrt\frac{6\,-\,\sqrt6}{2}$
- $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6\,+\,\sqrt6}{2},\,y\,=\,\pm\,\sqrt\frac{6\,-\,\sqrt5}{2}$
- $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6\,+\,\sqrt6}{2},\,y\,=\,\pm\,\sqrt\frac{6\,-\,\sqrt6}{2}$
- $\displaystyle\,y\,=\,\pm\,\sqrt\frac{6\,+\,\sqrt5}{2},\,y\,=\,\pm\,\sqrt\frac{6\,-\,\sqrt6}{2}$
Let u = y^2. 2u + 15/u = 12 => 2u^2 - 12u + 15 = 0. Using quadratic formula: u = [12 +/- sqrt(144 - 120)] / 4 = [12 +/- sqrt(24)] / 4 = [12 +/- 2*sqrt(6)] / 4 = (6 +/- sqrt(6)) / 2. Since y^2 = u, y = +/- sqrt((6 +/- sqrt(6))/2).
Multiply the equation 2y^2 + 15/y^2 = 12 by y^2 and divide by 2 to get the biquadratic y^4 - 6y^2 + 15/2 = 0. Using the quadratic formula for y^2, we get y^2 = (6 plus or minus the square root of (36 - 30))/2, which simplifies to y^2 = (6 plus or minus the square root of 6)/2. Taking the square root of both sides gives the roots y = plus or minus the square root of ((6 plus the square root of 6)/2) and y = plus or minus the square root of ((6 minus the square root of 6)/2).