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 
 A: Consider the first two equations, under this slightly different form:$$\left\{\begin{array}{l}(1+4\lambda)x+6y=0\\6x+(2+\lambda)y=0.\end{array}\right.$$Suppose that this homogeneous system has exatly one solution; then this solution is $(x,y)=(0,0)$, which is not a solution of the third one.
But\begin{align}\text{The system has more than one solution}&\iff\det\begin{bmatrix}1+4\lambda&6\\6&2+\lambda\end{bmatrix}=0\\&\iff\lambda=-\frac{17}4\vee\lambda=2.\end{align}So, deal only with the cases $\lambda=-\frac{17}4$ and $\lambda=2$. Can you do that?
A: From the first equation we get
$$\frac{y}{x}=-\frac{1+4\lambda}{6}$$
and from the second one
$$\frac{y}{x}=-\frac{6}{2+\lambda}$$
So you will get
$$\frac{1+4\lambda}{6}=\frac{6}{2+\lambda}$$
Can you finish?
A: Well, the first equation gives:
$$x+6y+4\lambda x=6y+x(1+4\lambda)=0\space\Longleftrightarrow\space x=-\frac{6y}{1+4\lambda}\tag1$$
Using the second equation, we get:
$$6x+2y+\lambda y=0\space\Longleftrightarrow\space x=-\frac{2y+\lambda y}{6}\tag2$$
So, we get:
$$-\frac{2y+\lambda y}{6}=-\frac{6y}{1+4\lambda}\space\Longleftrightarrow\space2+\lambda=\frac{36}{1+4\lambda}\space\Longleftrightarrow\space\lambda=-\frac{17}{2}\space\vee\lambda=2\tag3$$
A: 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?
