I Solved A Nice Polynomial System

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[MarcelCox]

I used the approach that if you know the sum s and the product p of 2 numbers, then the 2 numbers are the solution of the quadratic equation x^2-sx+p=0
In this case, let's take u=x^2 and v=-y^2. In that case, you have the 2 equations u+v=9 and u*v=-9. By solving this, you can get u and v, and then deduce x and y by square roots and by choosing the correct signs using the second original equation.

[TheRhythmOfMathematics]

Nice method. We can also use the Vieta's formula
x^2 + (- y^2) = 9 and (x^2)(-y^2) = 9
=> K^2 - 9K + 9 = 0 with roots: x^2 and (-y^2)

[ianfowler]

This type of system is fairly common and where it shows up often is in finding the 2 square roots of complex numbers. Isolating x or y will always give you a quadratic in x^2 or y^2. As to other methods, "simpler" is a relative term. This method gives novices, or shall we say, "average" students a nice, clean and easy way to approach these types of systems - without having to know any identities. Keep up the good work.

[tarunmnair]

There's another method, consider y to be imaginäre (y=i.b) and square the 2nd equation.
Then the first eq is,
x²+b² = 9
and the second eq is
x².b²=-9
So sum of 2 numbers is 9, and their product is - 9.
Now form a quadratic eq, t²-9t-9=0 ; whose roots are x² and b².
Then solve for x and b, then find y.

[goldfing]

How about the following method:
x * y = 3
Square this equation:
x^2 * y^2 = 9
Multiply by (-1):
- (x^2 * y^2) = -9
x^2 * (-y^2) = -9
And we have the other equation
x^2 - y^2 = 9
x^2 + (-y^2) = 9
Now substitute
s = x^2
t = -y^2
And we have the sum and product of s and t:
s + t = 9
s * t = -9
According to Vieta's theorem, s and t are the solutions of the quadratic
z^2 - 9z - 9 = 0
z1, 2 = (9 +- sqrt(81 + 36))/2
s, t = (9 +- sqrt(117))/2
Either
s = x^2 = (9 + sqrt(117))/2
t = -y^2 = (9 - sqrt(117))/2
that means
x^2 = (9 + sqrt(117))/2
y^2 = (-9 + sqrt(117))/2
thus
x = +- sqrt((9 + sqrt*117))/2)
y = +- sqrt((-9 + sqrt(117))/2)
Or
s = x^2 = (9 - sqrt(117))/2
t = -y^2 = (9 + sqrt(117))/2
that means
x^2 = (9 - sqrt(117))/2
y^2 = (-9 - sqrt(117))/2
thus
x = +- sqrt((9 - sqrt(117)/2)
y = +- sqrt((-9 - sqrt(117))/2).
It seems that the latter solutions are complex, not real.

[Qermaq]

Here's what I did. From eq 1 y^2 = x^2-9. From eq 2 y^2 - 9/x^2. So we get the depressed quartic x^4-9x^2-9=0. Sub u for x^2, plug/chug, x = sqrt((9+sqrt117)/2) or sqrt((9-sqrt117)/2)). Corresponding y values just have a -9 in there. I'm too tired to be rigorous so I'ma call it that.

[mbmillermo]

If x²-y² = 2a² and xy = a, then the two hyperbolas are 45° rotations of each other. So if you had used 18 instead of 9 in the first equation, they would have been exact rotations.

[ghostingpeople]

3:49 can u use the quadratic method over here to get the two values ig?

[GeoffryGifari]

hey, looks like you can solve this one with hyperbolic trig!

[ガアラ-hh]

It easy multiply equation two by 2i then you do the ads make perfect square take the root then you can get value for one of them put in second equation and you done

[Ghost-txfs]

Check this method it very easy to find Real solutions with this method
Using complex number where x, y are real numbers
(x+iy)^2=(x^2 - y^2) + (2xy)i
(x+iy)^2=9+6i
Using polar form
x+iy=r(cos(u) +i sin(u)), where r is distance from origin in argand plane and u is the angle made line join origin and (x, y) then using euler's identity
cos(u) + isin(u)= e^(iu)
x+iy=re^(iu)
(x+iy)^2=(r^2)*(e^(i2u)

On comparison
r=(117)^(1/4)
u=(1/2)*arctan(2/3)
x=(+-)r*cos(u)
y=(+-)r*sin(u)
The plus and minus is because if we substitute -x and -y for x and y we get same equations implying if x=a is soln then x=-a is also a soln

[ghostingpeople]

I’m a student beginner trying to solve math questions daily, this questions seems long