Can anyone teach me how to solve this equation (x2+y2-1)3-x2y3=0 I don't know what to do when I have something like y2 + y3 = 0
The solution to the equation, (x^2+y^2 -1) - x^2y^3 =0 could be many pairs of solutions.
Some solutions are:
Let x = 0, y is solution of the equation(y^2-1)^3 = 0 forwhich one solution is (y^2-1)^3 = 0 or y^2 = 1 or y = 1 or y =-1. So the solution is (0, 1) and (0,-1).
Give x=1, (1+y^2-1)^3-1*y^3 = 0 or, y^6 -y^3 = 0 or
y^3(y^3-1) = 0 . So, y^3 = 0 <==> y=0 or y^3-1=0 gives: y = 1,w,and w^2 , where w is the cube root of unity.
So, the solution set is , (1,0), (1,1), (1,w), (1,w^2).
Like that you chose any desired value of x and find the paricular values of y. Obtain the solution set (x,y).
Thus for any given value of x we get many valued y, to solve which you have to solve the 6th degree equation in y. You may study the theory of equations which may help solving equations of order third ( cubic) or higher degrees.
Thre are 2 equations:
y2 + y3 = 0...................(2). To solve for x and y.
Given 2 equations are simultaneous nonlinear equations with 2 variables x and y. We have to eliminate one of the two variables and reduce the equations to one single variable and one single equation:
From the equation(2), y^2(y+1) = 0, we get y^2=0 or y=-1.
y^2=0 <==> y=0 and substituting y=0 in equation (1) :
(x^2+0-1)^3-x^2*0 = 0
(x^2-1)^3 = 0 or
x^2 -1 = 0 or
x^2 = 1 or x=1 or x = -1
When y = -1, equation (1)becomes:
(x^2+(-1)^2-1)^3-x^2*(-1)^3 = 0 or
(x^2)^3+x^2 = 0 or
x^6+x^2 = 0 or
x^2(x^4+1) = 0 or
x^2 = 0 or x^4+1=0 Or
x^2=0 gives x= 0 or
x^4+1=0 gives x^2 = +(-1)^(1/2) or x^2 = -(-1)^(1/2) .Or
x=(-1)^(1/4), x = -(-1)^(1/4) , x = (-1)^(3/4) , x= -(-1)^(3/4).
So ,summerising, the solution sets are:
(x:y) = (1 , 0) or (-1 ,0) or
(x,y) = (0 ,-1) or
((-1)^1/4 ,-1), or
(-(-1)^(1/4) , -1) or
((-1)^(3/4) , -1) or
Loved this question
and gave me a great insight
see the link below , hope you will also love it
oh my God... thanks a lot for your time Neela but I'm sorry to say you got the question wrong!
the equation which you have assumed to be the 2nd one was just an example!
This is the equation that makes a graph: (x2+y2-1)3-x2y3=0
now to map this equation, we need to choose any value of either x or y, and find corresponding y or x... to get 1 co-ordinate of the curve...