15. (9th "Hope Cup" Invitational Training Question) Let the function $y=y_{1}+y_{2}$, where $y_{1}$ is directly proportional to $x^{2}$, and $y_{2}$ is inversely proportional to $x^{2}$. When $x=1$, $y=5$; when $x=\sqrt{3}$, $y=7$. Then when $y$ is at its minimum, $x=$ $\qquad$
15. Fill in the blank $\sqrt[1]{\frac{3}{2}}$. Reason: Since $y=k_{1} x^{2}+\frac{k_{2}}{x^{2}}$, and $5=k_{1}+k_{2}, 7=3 k_{1}+\frac{k_{2}}{3}$. Then $k_{1}=2, k_{2}=3$. Therefore, $y=$ $2 x^{2}+\frac{3}{x^{2}}=\left(\sqrt{2}|x|-\frac{\sqrt{3}}{|x|}\right)^{2}+2 \sqrt{6} \geqslant 2 \sqrt{6}$. Hence, when $\sqrt{2}|x|=\frac{\sqrt{3}}{|x|}$, i.e., $x= \pm \sqrt[4]{\frac{3}{2}}$, $y$ reaches its minimum value.