Answer: \(x=\pm1.932\) or \(x=\pm0.518\).

Answer: \(x=\pm1.932\) or \(x=\pm0.518\).

Substitute \(y=\frac{3}{2x}\) into

\(\frac{x^2}{4}+\frac{y^2}{9}=1.\)

This gives

so

\(\frac{x^2}{4}+\frac{1}{4x^2}=1.\)

Multiply by \(4x^2\):

\(x^4+1=4x^2.\)

Hence

\(x^4-4x^2+1=0.\)

Let \(u=x^2\). Then

\(u^2-4u+1=0.\)

Using the quadratic formula,

\(u=\frac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt3.\)

Therefore

\(x=\pm\sqrt{2+\sqrt3} \quad\text{or}\quad x=\pm\sqrt{2-\sqrt3}.\)

Correct to \(3\) decimal places,

\(x=\pm1.932 \quad\text{or}\quad x=\pm0.518.\)