Question 272595
you have 2 equations that I presume must be solved simultaneously.


Those equations are:


(2x^2-xy+y^2=8)and(xy=4)


first thing you should be able to do is replace xy in the first equation with 4.


your first equation becomes:


2x^2 - 4 + y^2 = 8


add 4 to both sides of this equation to get:


2x^2 + y^2 = 8 + 4 which becomes:


2x^2 + y^2 = 12


since your second equation states that xy = 4, you can solve for y to get:


y = 4/x


replace y in your first equation with 4/x to get:


2x^2 + (4/x)^2 = 12


since (4/x)^2 = 4^2/x^2 = 16/x^2, your equation becomes:


2x^2 + 16/x^2 = 12


multiply both sides of this equation by x^2 to get:


2x^4 + 16 = 12x^2


subtract 12x^2 from both sides of this equation to get:


2x^4 - 12x^2 + 16 = 0


divide both sides of this equation by 2 to get:


x^4 - 6x^2 + 8 = 0


let k = x^2


your equation becomes:


k^2 - 6k + 8 = 0


factor this quadratic equation to get:


(k-2) * (k-4) = 0


since either one of these factors can be equal to 0 to make this equation true, then you have:


k-2 = 0 or k-4 = 0


solve for k to get:


k = 2 or k = 4


plugging those values into the quadratic equation will make that equation true.


the quadratic equation is k^2 - 6k + 8 = 0


when k = 2, this equation becomes 4 - 12 + 8 = 0 which is true.
when k = 4, this equation becomes 16 - 24 + 8 = 0 which is also true.


the values for k are both good.


since k = x^2 ( you set it that way), then you can solve for x to get:


x = +/- square root of k


when k = 4, x = +/- 2


when k = 2, x = +/- square root of 2.


we also know that y = 4/x


this means that:


when x = 2, y = 4/2 = 2
when x = -2, y = 4/(-2) = -2
square root of 2 is the same as {{{sqrt(2)}}}
when x = {{{sqrt(2)}}}, y = {{{4/(sqrt(2))}}} = {{{(4*sqrt(2))/2}}} = {{{2*sqrt(2)}}}
when x = {{{-sqrt(2)}}}, y = {{{4/(-sqrt(2))}}} = {{{(4*(-sqrt(2)))/2}}} = {{{2*(-sqrt(2))}}} = {{{-2*sqrt(2)}}}


plug these values into your original equations and see whether those equations become true.


your original equations are:


(2x^2-xy+y^2=8)and(xy=4)


Since xy = 4, we changed the first equation to:


2x^2 - 4 + y^2 = 8


original equation is 2x^2 - 4 + y^2 = 8


when x = 2 and y = 2, this equation becomes:


2*2^2 - 4 + 2^2 = 8 which becomes:
2*4 - 4 + 4 = 8 which becomes:
8 - 4 + 4 = 8 which becomes:
8 = 8 which is true.


original equation is 2x^2 - 4 + y^2 = 8


when x = -2 and y = -2, this equation becomes:


2*(-2)^2 - 4 + (-2)^2 = 8 which becomes:
2*4 - 4 + 4 = 8 which becomes:
8 - 4 + 4 = 8 which becomes:
8 = 8 which is true.


original equation is 2x^2 - 4 + y^2 = 8


when x = {{{sqrt(2)}}} and y = {{{2*sqrt(2)}}}, this equation becomes:


{{{2*(sqrt(2))^2 - 4}}} + {{{(2*sqrt(2))^2}}} = 8 which becomes:
2*2 - 4 + 4*2 = 8 which becomes:
4 - 4 + 8 = 8 which becomes:
8 = 8 which is true.


original equation is 2x^2 - 4 + y^2 = 8


when x = {{{-sqrt(2)}}} and y = {{{-2*sqrt(2)}}}, this equation becomes:


{{{2*(-sqrt(2))^2 - 4}}} + {{{(-2*sqrt(2))^2}}} = 8 which becomes:
2*2 - 4 + 4*2 = 8 which becomes:
4 - 4 + 8 = 8 which becomes:
8 = 8 which is true.


your second equation is xy = 4


when x = 2 and y = 2, this equation becomes 4 = 4 which is true.


when x = -2 and y = -2, this equation becomes 4 = 4 which is true.


when x = {{{sqrt(2)}}} and y = {{{2*sqrt(2)}}}, this equation becomes:
{{{sqrt(2)*2*sqrt(2)}}} = 4 which becomes:
2*2 = 4 which becomes:
4 = 4 which is true.


when x = {{{-sqrt(2)}}} and y = {{{-2*sqrt(2)}}}, this equation becomes:
{{{-sqrt(2) * 2 * (-sqrt(2))}}} = 4 which becomes:
2 * 2 = 4 which becomes
4 = 4 which is true.


All equations are true so these values are good.


they are:


when x = 2, y = 2
when x = -2, y = -2
when x = {{{sqrt(2)}}}, y = {{{2*sqrt(2)}}}
when x = {{{-sqrt(2)}}}, y = {{{-2*sqrt(2)}}}


those are your 4 "sets" of answers.
each answer is a pair of values, 1 for x and 1 for y.