Question

$2y^2+\frac{15}{y^2}=12$

Answer 1

Given: $2y^2+\frac{15}{y^2}=12$
$\Rightarrow$2y⁴ +15 = 12y²
$\Rightarrow$ 2y⁴ – 12y² + 15 = 0
Let y² = m. Then, the equation can be written as
2m² – 12m + 15 = 0
On using the quadratic formula, we get:
$$\begin{gathered} m=\frac{-\left(-12\right)\pm \sqrt{{\left(-12\right)}^2-4\times 2\times 15}}{2\times 2} \\ =\frac{12\pm \sqrt{144-120}}{4} \\ =\frac{12\pm \sqrt{24}}{4} \\ =\frac{12\pm 2\sqrt{6}}{4} \\ =\frac{2(6\pm \sqrt{6})}{4} \\ =\frac{6\pm \sqrt{6}}{2} \\ \Rightarrow m=\frac{6+\sqrt{6}}{2},\frac{6-\sqrt{6}}{2} \end{gathered}$$
On substituting m = y², we get:
$$\begin{gathered} y^2=\frac{6+\sqrt{6}}{2} \\ \Rightarrow y=\pm \sqrt{\frac{6+\sqrt{6}}{2}} \\ \mathrm{or} \\ {\mathrm{y}}^2=\frac{6-\sqrt{6}}{2} \\ \Rightarrow \mathrm{y}=\pm \sqrt{\frac{6-\sqrt{6}}{2}} \end{gathered}$$
Therefore, the solutions of the given equation are $y=\pm \sqrt{\frac{6+\sqrt{6}}{2}},\pm \sqrt{\frac{6-\sqrt{6}}{2}}$