1and obviously (0,0) also
\hspace{-16}$\textbf{Solve System of Equations.}\\\\ $\begin{matrix} \bold{x^4+y^2-xy^3-\frac{9}{8}x=0} & \\\\ \bold{y^4+x^2-x^3y-\frac{9}{8}y=0} & \end{matrix}\right.
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7 Answers
1the second equation is obtained by interchanging x and y.
so the sloution shud be of the form (α,α)
=> α = 9/8
.:. answer shud be (9/8,9/8)
62rishabh, there could be more solutions. you have given some good solutions. but see the polynomial is of degree 4 so there is apossiblitly of more solutions.
66The trivial solution is (0,0).
Next we assume x≠0, y≠0.
The two equations can be rewritten as
x(x^3-y^3) + y^2 -\frac{9}{8}x=0
y(y^3-x^3) + x^2 -\frac{9}{8}y=0
That is
x^3-y^3 + \frac{y^2}{x} -\frac{9}{8}=0
y^3-x^3 + \frac{x^2}{y} -\frac{9}{8}=0
Subtracting gives
2(x^3-y^3)+\frac{y^2}{x}-\frac{x^2}{y}=0
that is
(x^3-y^3)\left(2-\frac{1}{xy}\right)=0
Therefore either x3 = y3 or xy = 1/2
If y3 =x3, we get
y = x,\ \omega x,\ \omega^2 x
where \omega is a complex cube root of unity. With this condition, the first of the original equations gives
y^2=\frac{9}{8}x
The non-trivial solutions are:
• for y = x, (9/8, 9/8)
• for y = ωx, (98ω, 98ω2)
• for y = ω2x, (98ω2, 98ω)
On the other hand, if xy = 1/2, the first of the original equations gives after simplification
8x^6-9x^3+1=0
which gives x3 = 1, and x3 = 1/8 whence it follows that
x= 1, ω, ω2 or x=12, 12ω, 12ω2
Thus in total we get the following 10 solutions
• (0,0)
• (9/8, 9/8)
• (98ω, 98ω2)
• (98ω2, 98ω)
• (1,12)
• (ω, ω22)
• (ω2, ω2)
• (12,1)
• (ω2, ω2)
• (ω22, ω)
1@nishant sir,
can you please mail me the pdf which you were talking about here , http://targetiit.com/iit-jee-forum/posts/rmo-2009-12793.html
my id : rishabh.bohara20@gmail.com
11Adding the two given equations gives (x+y)^4+4x^2y^2-5xy(x+y)^2+(x+y)^2-2xy=\frac{9}{8}(x+y)
Similalrly subtracting gives -(x+y)^3+xy(x+y)+(x+y)+\frac{9}{8}=0.........(\tex{*}) provided x≠y.
Now putting the value of 98 from (*) in the 1st eqn, we have \boxed{x^2+y^2=-xy}
From the boxed relation we get that if (x,y)\ne (0,0), then either of x or y is negative - which means we can have (x,y)=(x,-y') WLOG
Putting this in the second given relation gives y'=0.
The conclusion follows.
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