Find $\left \lfloor{(x+y)^4}\right \rfloor$ if $3x^2-2xy+3y^2-8y+6=0$.

Question

Find $\left \lfloor{(x+y)^4}\right \rfloor$ if $3x^2-2xy+3y^2-8y+6=0$.

I'm pretty bad at algebra so I struggle with questions like these, and I have literally no idea how to get $\left \lfloor{(x+y)^4}\right \rfloor$ out of $3x^2-2xy+3y^2-8y+6=0$. I considered expressing $x$ in terms of $y$ (since the equation is a quadratic in $x$) but I don't see how that could help me. Maybe there's a clever factorization I'm missing?

Answer 1

Hint: The solutions to the equation is actually just a single point.
We can determine this by calculating the various determinants:
$ AC - \frac{B^2}{4} = 3 \times 3 -\frac{(-2)^2}/4 > 0 $ so we have an ellipse.
$\Delta = (AC - \frac{B^2}{4} ) F + \frac{BED}{4} - \frac{CD^2}{4} - \frac{AE^2}{4} = 0 $ so it's a degenerate ellipse, namely a single point. (Note that $\Delta$ is the discriminant of Will's matrix.)

$3x^2-2xy+3y^2-8y+6 = 2(x-y+1)^2 + (x+y-2)^2 \quad (1) $
Hence, the expression = 0 iff $x-y+1 = 0, x + y - 2 = 0$, which is the point $ (1/2, 3/2)$.
Thus $(x+y)^4 = 16.$

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Algorithm for calculating (1): Convert a conic section into standard form. This is shown in Steven's solution.

Answer 2

Your polynomial is $$ 3 \left(x - \frac{y}{3} \right)^2 + \frac{8}{3} \left(y - \frac{3}{2} \right)^2 = 0 $$

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 3 } & 1 & 0 \\ \frac{ 1 }{ 2 } & \frac{ 3 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & - 1 & 0 \\ - 1 & 3 & - 4 \\ 0 & - 4 & 6 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 3 } & \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 3 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 8 }{ 3 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 3 } & 1 & 0 \\ 0 & - \frac{ 3 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{ 8 }{ 3 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & - \frac{ 3 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 3 & - 1 & 0 \\ - 1 & 3 & - 4 \\ 0 & - 4 & 6 \\ \end{array} \right) $$

Answer 3

$$3x^2-2xy+3y^2-8y+6=0$$

We can eliminate the $xy$ term by rotating the axis through an angle of $\theta$ where $$\cot(2\theta) = \dfrac{a-c}{b} = \dfrac{3-3}{-2}= 0$$

The simplest solution is $2\theta = 90^\circ$. So $\theta = 45^\circ$.

So $$\left( \begin{array}{c} x \\ y \end{array} \right) = \left(\begin{array}{c} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right) \left( \begin{array}{c} u \\ v \end{array} \right) = \dfrac{1}{\sqrt 2}\left(\begin{array}{c} 1 & -1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{c} u \\ v \end{array} \right)= \left( \begin{array}{c} \dfrac{u- v}{\sqrt 2} \\ \dfrac{u + v}{\sqrt 2} \end{array} \right)$$

substituting and simplifying, we get

$$\begin{align} 2 u^2 - 4 \sqrt 2 u + 4 v^2 - 4 \sqrt 2 v + 6 &= 0 \\ 2(u^2 - 2 \sqrt 2 u) + 4 v^2 - 4 \sqrt 2 v + 6 &= 0 \\ 2(u^2 - 2 \sqrt 2 u + 2) + (4 v^2 - 4 \sqrt 2 v + 2) &= 0 \\ 2(u - \sqrt 2)^2 +(2v-\sqrt 2)^2 = 0 \end{align}$$

So $u = \sqrt 2$ and $v = \dfrac{1}{\sqrt 2}$.

It follows that $x = \dfrac 12$ and $y = \dfrac 32$

$$(x+y)^4 = 2^4=16$$