Question

Find all pairs of real numbers (x; y), that are solutions of equality: 5x^2 + 5y^2 + 8xy + 2y - 2x + 2 = 0

Answer 1

@ganeshie8 @welshfella @Kainui

Answer 2

Partially differentiate thr function with respect to x and y separately.

Then u will get two linear equation in terms of x and y.

Answer 3

Solve those two equation

Answer 4

@jiteshmeghwai9 I would be glad to do what you said, however, the question is supposed to be solvable by 10th graders, not university students (or where would I learn about that?), so I think that there should be another way to solve this.

Answer 5

I think that it should be possible to make whole LHS into a multiplication thing and then use the x1 = 0 or x2 = 0. However, I am unable to notice how to construct (x1)(x2) = 0 in this case...

Answer 6

5x^2+5y^2+8xy+2y-2x+2=0
This may represent two straight line equation
(Ax+by+c)(lx+my+n)=0

Comparing we get
Al=5
bm=5
Am+bl=8
An=-2
bn=2
Cn=2

Answer 7

From this information u may find two linear equation and solve them

Answer 8

@jiteshmeghwal9 thank you very much! Now I am trying to solve for all the letters. Seems to be easier said than done :)

Answer 9

Yw

Answer 10

The result will be all integers number

Answer 11

@Kevin thank you for the hint! :)

Answer 12

@jiteshmeghwal9, I think that you had made a mistake with those letters. Expanding (Ax+by+c)(lx+my+n)=0 we get:
Alx^2 + Amxy + Anx + blxy + bmy^2 + bny + clx + cmy + cn, so it boils down to a different system of equations:
Al = 5
bm = 5
Am + bl = 8
An + cl = -2
bn + cm = 2
cn = 2

It seems that your system of equations has no solutions, so I think that mine is the right one(?)

Answer 13

Ooh yup sorry for the inconvenience

Answer 14

Ur system is 100%correct

Answer 15

I wonder how to solve it tho. Have been meddling with it for quite a long while with no results.

Answer 16

There is a reason behind it.

For an equation of second degree ax^2+2hxy+by^2+2gx+2fy+c=0 to represent a pair of linear equation it should satisfy the condition h^2-ab greater than equal to zero.

Answer 17

This equation ain't satisfying this condition

Answer 18

Does that mean that the equation has no solutions? @jiteshmeghwal9

Answer 19

Yeah

Answer 20

More clearly this equation has no real solution

Answer 21

What is the name of h^2 - ab identity?
Thank you for your help again, @jiteshmeghwal9 ! :)

Answer 22

And one extra question: the solutions to this equality would have to involve imaginary numbers, yes?

Answer 23

It is not an identity it is just a condition for an equation of second degree to represent real lines

Answer 24

And the solution involve imaginary number.

Probably, not so sure about it.

Answer 25

@jiteshmeghwal9 it is solvable. I had written a post in MSE and they had solved it: http://math.stackexchange.com/questions/1930068/quadratic-with-two-unknowns#1930070