2
$\begingroup$

Based on the previous question, I tried to find the roots of x with variable y includes in the equation like this: $2x^2-x(3y-7)-(2y^2-6y+4)=0$

Then, I can apply the Quadratic formula like here: https://en.wikipedia.org/wiki/Quadratic_formula $x= \frac {(3y-7)\pm \sqrt{(3y-7)^2-4(-2(2y^2-6y+4))}}{4}$

$x=\frac{(3y-7)\pm \sqrt{(25y^2-90y+81)}}{4}$

Until then, I cannot go further because the equation inside the square root is cannot factorized.

Then, I'm trying to use the other way like this:

$2x^2-x(3x-7)-(2y^2-6y+4)= (ax+by+c)(dx+ey+f)$

$2x^2-x(3x-7)-(2y^2-6y+4)=(adx^2+aexy+afx+bdxy+bey^2+bfy+cdx+cey+cf)$

$2x^2-x(3x-7)-(2y^2-6y+4)=(adx^2+x(aey+bdy+af+cd)+bey^2+bfy+cey+cf)$

But, I think this equation still hard to solve and need some effort to find the answer.

How can I get the easiest way to factorize this equation and also without software too?

$\endgroup$
3
  • 1
    $\begingroup$ You have done it note that inside the radical you have perfect square, it is (5x-9). It is over. $\endgroup$
    – Z Ahmed
    Oct 4, 2021 at 9:28
  • 2
    $\begingroup$ Try factoring $25y^2-90y+81$ and you're in for a nice surprise. $\endgroup$ Oct 4, 2021 at 9:30
  • $\begingroup$ thanks for the help, I think this form will produce a bad ones in the first thinking. Thanks, I'll solve it with my ways $\endgroup$
    – akusaja
    Oct 4, 2021 at 9:35

4 Answers 4

2
$\begingroup$

Since the polynomial is quadratic in both $x$ and $y$, you can write it in the form: $$f(x,y) =2 x^2 - x (3 y - 7) - (2 y^2 - 6 y + 4) $$ $$= p^i F_{ij} p^j=(x-x_0,y-y_0)^i F_{ij}(x-x_0,y-y_0)^j$$ where $p^i=(x-x_0,y-y_0)$, while $(x_0,y_0)$ is the value at the minimum/maximum and $F_{ij}$ is the Fisher matrix defined as $F_{ij} =\frac12 \partial_i \partial_j f(x,y)|_{(x,y)->(x_0,y_0)}$.

Simple algebra gives: $$ F_{ij}=\left( \begin{matrix} 2 & -3/2 \\ -3/2 & -2 \\ \end{matrix} \right), $$ and $(x_0,y_0)=(-2/5,9/5)$.

$\endgroup$
1
$\begingroup$

If you factorise the quadratic terms first, leaving unknowns for the constant terms: $$(2x+y+a)(x-2y+b)$$

Now compare coefficients of $x$ and $y$, and solve $$2b+a=7, b-2a=6\implies a=-1,b=4$$

and this agrees with $ab=-4$

$\endgroup$
1
$\begingroup$

Let graph the quadratic form below : https://www.desmos.com/calculator/ezwhow2rmf

$$f(x,y)=2x^2-3xy-2y^2+7x+6y-4$$

Notice that with conics (red curve) you can always find a translation to center it at the origin (green curve), which is equivalent to eliminate terms in $x$ and $y$ and reduce to terms in $x^2,y^2,xy$ only

Just set $\begin{cases}X=x-a\\Y=y-b\end{cases}$ and substitute $f(x,y)=f(X+a,Y+b)$ and try to remove terms in $X^1$ and $Y^1$.

In our case this gives:

$$\underbrace{(2a^2-3ab-2b^2+7a+6b-4)}_{f(a,b)}+\underbrace{(4a-3b+7)}_{f_x(a,b)}X+\underbrace{(-3a-4b+6)}_{f_y(a,b)}Y+2X^2-3XY-2Y^2$$

Annulating the partial derivatives $f_x,f_y$ gives you a system to find $a,b$ directly.

This leads to $(a,b)=(-\frac 25,\frac 95)$.

The simplified translated conic becomes: $$2X^2-3XY-2Y^2$$

And this equation is way easier to factorize into $(2X+Y)(X-2Y)$

$\endgroup$
1
  • $\begingroup$ Note that $f(a,b)$ is not necessarily $0$, but here it is the case because the conic is degenerated into two straight lines (this is why it factorizes). $\endgroup$
    – zwim
    Oct 4, 2021 at 10:37
1
$\begingroup$

Your method is good and correct.

If your all polynomial factors are also polynomials, then $\Delta\left[\Delta_x\right]=0$ or $\Delta\left[\Delta_y\right]=0$.

Indeed, we have

$\Delta\left[\Delta_x\right]=45^2-25\times 81=0$.

This implies,

$$y_1=y_2=\frac 95$$

You can conclude that

$$\begin{align}25y^2-90y+81&=25\left(y-\frac 95\right)^2\\ &=(5y-9)^2\end{align}$$

Thus you get,

$$x_{1,2}=\frac{3y-7±5y-9}{4}$$


The advantage of this method is:

If factoring isn't possible, you'll see it right away. For example, if $\Delta=25y^2-90y+85$, then we could quickly conclude that polynomial factorization of $f(x,y)$ is impossible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.