Advanced System Of Equations
How many ordered pairs of real numbers ( x , y ) are there such that
{ x 2 − y 2 2 x y + x 2 + y 2 π x + ϕ y + x 2 + y 2 ϕ x − π y = 2 , = 0 .
( ϕ is the golden ratio 2 1 + 5 . π is pi, which is approximately 3 . 1 4 1 5 9 .)
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4 solutions
cool! Dude
nice solution
wow...that was real mathematical thinking!
Write z = x + y i , w = π + ϕ i . Then the equations are equivalent to z 2 + z w = 2 (the first one is the real part and the second is the imaginary part). This simplifies to a cubic in z , and it's easy to check (e.g. by taking the derivative) that there are no repeated roots. So it has three solutions.
Nice solution, but one thing. Well z is a complex number, do the usual rules of derivative apply, I mean the laws of calculus apply to real numbers right?
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The property I'm referring to is the property that a polynomial has a repeated factor if and only if the gcd of it and its derivative is nonconstant.
This works for a polynomial with coefficients in any field. You don't even need to know calculus; for our purposes, the derivative of a polynomial is just the thing that you get by mechanically applying the power rule. (I believe that all you need to know to prove the property is that this satisfies the Product Rule, which shouldn't require any calculus to prove.)
It is not necessary to invoke complex numbers.The above two eqations can be reduced to a polynomial of degree three in x and y.ie fi.x^3+pi.y^3-fi.x.y^2-3.pi.y.x^2-r00t(2)fi.x-root(2)pi.y=0 so it will have three roots.
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Why must a polynomial of degree 3 only have 3 roots? This is not true of conic sections, which are polynomials of degree 2 but have infinitely many roots.
If all the terms had degree 3, then we could say that y x has 3 possible values, which follows from dividing by y 3 and applying the Fundamental Theorem of Algebra.
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You are confusing with the polynomial in single variable.As per Fundamental Theorem of algebra of Gauss,any polynomial in single variable will of degree n will have only n solutions.Now your polynomial in two variables,so it represents a Cuve in 2-D.So all pairs of (X,Y) lying on the curve are solutions and they are infinite.So if you hold one variable then threr will be only n solutions.So a conic is a 3-D curve so plot y v/s x means Y is dependent variable so when you cahange y you will get another solution actually two roots for a given y.Thereare infite nuber of choice of Y so you get an infinte set of doublets.
Can you explain the following: "This is not true of conic sections, which are polynomials of degree 2 but have infinitely many roots."
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@Muhammad Shariq – As an explicit example, the equation x 2 − y 2 = 0 has infinitely many real solutions (of the form x = y and x = − y ).
good
We see x 2 − y 2 and 2 x y , and we remember that ( x + y i ) 2 = ( x 2 − y 2 ) + ( 2 x y ) i . Also, we notice that ∣ x + y i ∣ 2 = x 2 + y 2 and we suspect complex numbers. So we multiply the second equation by i and add it to the first. We get ( x 2 + 2 x y i − y 2 ) + x 2 + y 2 π x + ϕ i x + ϕ y − π i y = 2 . Now we can factor, so we have ( x + y i ) 2 + ( x + y i ) ( x − y i ) ( π + ϕ i ) x − ( π + ϕ i ) i y = 2 Let z = x + y i . Substituting gives z 2 + z ⋅ z ˉ ( π + ϕ i ) z ˉ = 2 Then z 3 − 2 z + ( π + ϕ i ) = 0 There is a maximum of 3 solutions and there cannot be 1 (since this is not a perfect cube polynomial), we go with the answer choice 3.
(x, y):
(-1.80707075451828, -0.193899517098122)
(1.15744570604683, -0.938032202414793)
(0.649625048471628, 1.13193171951272)
Let z = x + i y , multiply the second equation by i and add it to the first equation we have: z 2 + z z ˉ ( π + i ϕ ) z ˉ = 2 which is equivalent to z 3 − 2 z + ( π + i ϕ ) = 0 This should be enough to conclude that the answer is 3 .