$$4x^2+25y^2+9z^2-10xy-15yz-6zx=0$$
$$x+y+z=5$$
I tried two approaches
1) Substituting $x$ as $5-y-z$ in the first equation but didn't work out, I was getting $39y^2+19z^2-31yz-90y-70z+100=0$ which can't be factorized
2) First equation corresponds to $a^2+b^2+c^2-ab-bc-ca=0$, which means $a^3+b^3+c^3=3abc$, but didn't get a breakthrough.
I am stuck here, please help me.
On
Just wanted to see. Here is how it looks if you make the Hessian matrix congruent to a diagonal matrix, the one that gives $$ \frac{1}{4} \left(4x - 5 y - 3 z \right)^2 + \frac{3}{4} \left( 5y - 3 z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy $$
is $ Q^T D Q = H $
The theorem is that the form does factor, allowing complex coefficients, if and only if the discriminant is zero. In this case, we see that $\det H = 0.$
As you saw, setting to zero gives us $5y=3z,$ after which $4x = 5y + 5y = 10y,$ giving $2x=5y=3z$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 5 }{ 4 } & 1 & 0 \\ \frac{ 3 }{ 2 } & \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 8 & - 10 & - 6 \\ - 10 & 50 & - 15 \\ - 6 & - 15 & 18 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 5 }{ 4 } & \frac{ 3 }{ 2 } \\ 0 & 1 & \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & \frac{ 75 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 5 }{ 4 } & 1 & 0 \\ - \frac{ 3 }{ 4 } & - \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 8 & 0 & 0 \\ 0 & \frac{ 75 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 5 }{ 4 } & - \frac{ 3 }{ 4 } \\ 0 & 1 & - \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 8 & - 10 & - 6 \\ - 10 & 50 & - 15 \\ - 6 & - 15 & 18 \\ \end{array} \right) $$
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
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$$ 4 \left(x - \frac{5}{4} y - \frac{3}{4} z \right)^2 + \frac{75}{4} \left( y - \frac{3}{5} z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy $$
$$ \frac{1}{4} \left(4x - 5 y - 3 z \right)^2 + \frac{3}{4} \left( 5y - 3 z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy $$
On
Completing the squares, the quadric equation can be rewritten as $$\frac14\left(4x-5y-3z\right)^2+\frac34(5y-3z)^2=0,$$ so its spectrum is $(+,+,0,0)$, which makes it a collapsed elliptic cylinder, i.e., a line. Examining the above equation, we can see that this line is the intersection of the planes $4x-5y-3z=0$ and $5y-3z=0$, and adding the equation of the other given plane results in a straightforward system of linear equations to solve.
On
Continuing your second method.
Denote: $a=2x,b=5y,c=3z$, then: $$\begin{cases}4x^2+25y^2+9z^2-10xy-15yz-6zx=0\\ x+y+z=5\end{cases} \Rightarrow \\ \begin{cases}a^2+b^2+c^2-ab-bc-ca=0\\ \frac a2+\frac b5+\frac c3=5\end{cases} \Rightarrow \\ 2a^2+2b^2+2c^2-2ab-2bc-2ca=0 \Rightarrow \\ (a-b)^2+(b-c)^2+(c-a)^2=0 \Rightarrow \\ a=b=c=\frac{150}{31} \Rightarrow \\ (x,y,z)=\left(\frac{75}{31},\frac{30}{31},\frac{50}{31}\right)$$ WA answer.
$$(a-b)^2+(b-c)^2+(c-a)^2=2(?)=0$$
Now for real $d,d^2\ge0$
If $a,b,c$ are real, we can establish $$a-b=b-c=c-a=0$$