<s>[INST] solve the Lagrange multiplier equations the equations :
$$\left\{\begin{array}{l}{x+6 y+4 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}\right.$$
I've done many transformations,but I still can't get the answer.
the last I did this:
$$\lambda=\left(6 \frac{x}{y}+2\right)=\left(\frac{1}{4}+\frac{3}{2} \frac{y}{x}\right),y=tx \\=\left(6 \cdot \frac{1}{t}+2\right)=\left(\frac{1}{4}+\frac{3}{2} t\right)\\t^{2}-\frac{7}{6} t-4=0$$
the  quadratic formula's roots have square number , but the answer doesn't have,so I think I might be wrong.
I don't know how to solve it 
 [/INST] Restoring the objective function from the constraint (third) equation and partial derivatives:
$$Optimize \ \ z(x,y)=\frac12x^2+6xy+y^2 \ \ s.t. \ \ 4x^2+y^2-25=0.$$
If you take partial derivatives now:
$$\left\{\begin{array}{l}{x+6 y+\color{red}8 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}\right.$$
Note that if $x=0$, then from the first equation $y=0$, but it contradicts the last equation. So, $x\ne 0$. Similarly, if $y=0$, then from the second equation $x=0$, but it contradicts the last equation. So, $y\ne 0$ too. Thus:
$$\begin{cases}y=\frac{-(1+8\lambda)x}{6}\\ 6x+(2+\lambda)\frac{-(1+8\lambda)x}{6}=0 \Rightarrow x(36-(2+\lambda)(1+8\lambda))=0 \Rightarrow 36-(2+\lambda)(1+8\lambda)=0\end{cases} \Rightarrow \\
8\lambda^2+17\lambda-34=0 \Rightarrow \lambda_{1,2}=\frac{-17\pm 9\sqrt{17}}{16}.
$$
Can you do the rest?
</s>