Solve for y: Explicit Function of y^2 + xy = x^2 | Tips and Examples

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The discussion focuses on solving the equation y^2 + xy = x^2 for y as an explicit function of x. Participants explore methods to express y, including factoring and using the quadratic formula. The quadratic equation is reformulated as y^2 + xy - x^2 = 0, leading to the solutions y(x) = (1/2)(-x ± x√5). There is some confusion about the steps involved in solving the quadratic, but ultimately, two valid expressions for y in terms of x are derived. The conversation emphasizes the importance of checking solutions against the original equation.
chwala
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Express y as an explicit function
y^2 +xy = x^2

I was unable to get this this are my steps:
Y(x + y) = x^2
X + y = x^2/y
X=x^2/y – y………..kindly assist
 
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It appears that you have to solve a quadratic equation, right ? Remeber you need y=y(x).
 
so will i be right to say y=0 and y=x^2-x.
 
I don;t think so. Just check the 2 solutions you found/propose by plugging in the initial implicit equation. You should normally find tautological relations, but you don't.

So one more time

y^2 + xy - x^2 = 0 what is y in terms of x equal to ?
 
dextercioby said:
I don;t think so. Just check the 2 solutions you found/propose by plugging in the initial implicit equation. You should normally find tautological relations, but you don't.

So one more time

y^2 + xy - x^2 = 0 what is y in terms of x equal to ?

ok am bit confused here make it a bit simple.we need to exprex y as a function of x.and that is the information i need.simplify the quadratic for me
 
dextercioby said:
So you don't know how to solve the quadratic equation ? This is quite basic stuff. http://en.wikipedia.org/wiki/Quadratic_equation

i still insist show me how you gat the factors of the quadratic equation i have followed examples and not any close my study has been on the following procedure:,
Quadratic Equation in Two Variables

Introduction about quadratic equation:

An equation which contains more than one terms are squared but no higher power in terms, having the syntax, ax2+bxy+cy2 =0 where a represents the numerical coefficient of x2, b represents the numerical coefficient of xy, and c represents the numerical coefficient of y2

Example: x2+2xy+y2

In this article we shall discuss about the how to solve quadratic equation in two variables with suitable example problem.
Example Problem of Quadratic Equation in Two Variables:

Solve the quadratic equation by factoring method and find the factor?

X2+7xy+10y2

Solution:

Step 1: Multiply the coefficient of x2 and the coefficient of y2,

1*10 =10 (product term)

Step 2: Find the factors for the product term

10--- > 5*2= 10 (factors are 5 and 2)

10 --- > 5+2 = 7(7 is equal to the coefficient of xy)

Step 3: Separate the coefficient of xy

X2+7xy+10y2

x2+5xy+2xy+10y2

Step 4: Taking the common term x for the first two terms and 2y for the next two terms

x(x+5y) +2y(x+5y)

(x+2y) (x+5y)

Factors are (x+2y) (x+5y)
 
y^2 + xy - x^2 = 0
How about this? Solve for y by using the quadratic formula, with
a = 1
b = x
c = -x2
 
eumyang said:
y^2 + xy - x^2 = 0
how about this? Solve for y by using the quadratic formula, with
a = 1
b = x
c = -x2

you get f=( -x + or - x√5)/2 so?
 
  • #10
So that means that you've got several solutions.

y(x) = \frac{1}{2} \left(-x\pm |x|\sqrt 5\right)

Any of the two functions can be considered an explicitation of y wrt x.
 
  • #11
dextercioby said:
So that means that you've got several solutions.

y(x) = \frac{1}{2} \left(-x\pm |x|\sqrt 5\right)

Any of the two functions can be considered an explicitation of y wrt x.[/QUOTE

Thanks a lot long live physics forum man.I love Mathematics.
 
  • #12
dextercioby said:
y(x) = \frac{1}{2} \left(-x\pm |x|\sqrt 5\right)

You don't need the absolute value sign since we have a \pm just before it :wink:
 
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