Can this particular method solve these quadratic equations?

In summary: Then either factor it into two quadratics (and solve those) or use a numerical method.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. Then either factor it into two quadratics (and solve those) or use a numerical method.In summary, this conversation discusses methods for solving two equations, S1 and S2, in terms of x and y. One method involves treating S1 as a quadratic equation in x and substituting the solution in S2. However, this does not lead to a solution for x and y. Other methods, such as eliminating one quadratic term, are suggested but may lead to a quartic
  • #1
JohnnyGui
796
51
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?
 
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  • #2
Why not eliminate one of the quadratic terms - either ##x^2## or ##y^2##?
 
  • #3
PeroK said:
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.
 
  • #4
JohnnyGui said:
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.
 
  • #5
PeroK said:
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.
 
  • #6
JohnnyGui said:
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.
 
  • #7
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.
 
  • #8
JohnnyGui said:
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.
 
  • #9
mfb said:
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?
 
  • #10
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.
 
  • #11
mathman said:
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##?
 
Last edited:
  • #12
JohnnyGui said:
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.
 

1. Can this method solve all types of quadratic equations?

Yes, this method can solve all types of quadratic equations, including those with real or imaginary solutions.

2. Is this method more efficient than other methods for solving quadratic equations?

It depends on the specific equation and the skill level of the person using the method. Some methods may be more efficient for certain types of equations, while others may be more efficient for different types.

3. Does this method require any special tools or software?

No, this method can be done using basic algebraic techniques and does not require any special tools or software.

4. Can this method be used to find all possible solutions to a quadratic equation?

Yes, this method can be used to find all possible solutions to a quadratic equation, including real and imaginary solutions.

5. Are there any limitations to this method for solving quadratic equations?

Like any method, there may be certain types of equations that are more difficult to solve using this method. It is always important to check for extraneous solutions and to use other methods as needed.

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