Can this particular method solve these quadratic equations?

[JohnnyGui]

Given are two equations:
$$S_1 = ax^2+2hxy+by^2 + c=0$$
$$S_2 = a'x^2+2h'xy+b'y^2 + c'=0$$
This source states that there are several methods to solve for $x$ and $y$. One of them is the following quote:"Treat equation $S_1$ as a quadratic equation in $x$ and solve it for $x$ in terms of $y$. Then substitute this in equation $S_2$."

If $A=a$, $B=2hy$ and $C=by^2+c$, this means that:
$$x=\frac{-2hy\pm \sqrt{4h^2y^2-4a(by^2+c)}}{2a}$$
However, substituting this in $S_2$ does not make me able to solve for $x$ and $y$. Expressing $x$ also in terms of the constants of $S_2$ and substituting that in $S_1$ also doesn't help me.

Am I missing something?

 

[PeroK]

Why not eliminate one of the quadratic terms - either $x^2$ or $y^2$?

 

[JohnnyGui]

Why not eliminate one of the quadratic terms - either $x^2$ or $y^2$?

You mean as in dividing $S_1$ and $S_2$ by either $x^2$ or $y^2$? I have tried that but I'm not sure how to continue to get $x$ or $y$ on one side.

 

[PeroK]

You mean as in dividing $S_1$ and $S_2$ by either $x^2$ or $y^2$? I have tried that but I'm not sure how to continue to get $x$ or $y$ on one side.

That won't work. Multiply $S_1$ by $a'$ and $S_2$ by $a$ and subtract one from the other.

 

[JohnnyGui]

That won't work. Multiply S1 by a′ and S2 by a and subtract one from the other.

Thanks, but wouldn't this lead to another method to solve these equations? I was wondering whether the quoted method can actually be used to solve this.

 

[PeroK]

Thanks, but wouldn't this lead to another method to solve these equations? I was wondering whether the quoted method can actually be used to solve this.

It doesn't look very promising to me.

 

[mfb]

Solving S1 for x gives you an expression of the general form $x=my\pm\sqrt{ny^2+p}$ where m,n,p are some constants.. That means $x^2 = m^2 y^2\pm 2ym\sqrt{ny^2+p}+ny^2+p$. I'll ignore coefficients now, imagine prefactors for every term. Your full second equation then reads:

$0 = y^2 + y + 1 \pm y\sqrt{y^2+1}+y\sqrt{y^2+1}$.
The latter two terms can be combined and isolated:
$y\sqrt{y^2+1} = y^2 + y + 1$
Now square both sides:
$y^2(y^2+1) = (y^2 + y + 1)^2$
That's a quartic equation that is in principle solvable, but it doesn't sound like something you want to do. There might be combinations of coefficients that make this easier to handle.

 

[mathman]

Given are two equations:
$$S_1 = ax^2+2hxy+by^2 + c=0$$
$$S_2 = a'x^2+2h'xy+b'y^2 + c'=0$$
This source states that there are several methods to solve for $x$ and $y$. One of them is the following quote:"Treat equation $S_1$ as a quadratic equation in $x$ and solve it for $x$ in terms of $y$. Then substitute this in equation $S_2$."

If $A=a$, $B=2hy$ and $C=by^2+c$, this means that:
$$x=\frac{-2hy\pm \sqrt{4h^2y^2-4a(by^2+c)}}{2a}$$
However, substituting this in $S_2$ does not make me able to solve for $x$ and $y$. Expressing $x$ also in terms of the constants of $S_2$ and substituting that in $S_1$ also doesn't help me.

Am I missing something?

You end up in a quartic in y, which is solvable.

 

[JohnnyGui]

Solving S1 for x gives you an expression of the general form $x=my\pm\sqrt{ny^2+p}$ where m,n,p are some constants.

Why does this solution for $x$ not have the form of a fraction? It's missing a factor of $\frac{1}{2a}$ as mentioned in my opening post, isn't that how the solution of a quadratic function is defined according to the abc-formula?

 

[mfb]

1/(2a) or its square is part of the constants m,n,p, I didn't write everything in terms of a,b,c because that would have made the equations unhandy.

 

[JohnnyGui]

You end up in a quartic in y, which is solvable.

When substituting the solution for $x$ in the other equation, I get:
$$y^2 \bigg(a'\cdot \frac{(-2h\pm s)^2}{4a^2}+2h'\cdot \frac{-2h\pm s}{2a}+b'\bigg)+c'=0$$
Where $s=\sqrt{4h^2-4ab+\frac{4ac}{y^2}}$. Not sure how such an equation can be solved.

Also, shouldn't a quartic equation have a variable to the power of $>2$?

 

[mathman]

When substituting the solution for $x$ in the other equation, I get:
$$y^2 \bigg(a'\cdot \frac{(-2h\pm s)^2}{4a^2}+2h'\cdot \frac{-2h\pm s}{2a}+b'\bigg)+c'=0$$
Where $s=\sqrt{4h^2-4ab+\frac{4ac}{y^2}}$. Not sure how such an equation can be solved.

Also, shouldn't a quartic equation have a variable to the power of $>2$?

You need to rearrange terms so that the square root expression is alone on one side of the equation. Square both sides to get a quartic polynomial.