For all real $x, y$ that satisfy $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ I started with $x+y=a$ and $xy=b$ and I rewrote the equations with a and b.
I got
$$b=a^2+a-4$$
$$x^3+y^3=2a^3+3a^2-12a^2=7$$
$$f(a)=-2a^3+3a^2-12a^2-7=0$$
I factorised it to get
$$f(a)=-(a-1)(a-1)(2a+7)=0$$
So $a=1,b=-2$ or $a=\frac{-7}{2},b=\frac{19}{4}$
Now I know that values of a and b but how do I get x and y?
Note:
The textbook solution says that if we consider a=1 and b=-2 then x and y are the roots of
$$t^2+t-2=0$$
But shouldn't it be
$$t^2-t-2=0$$
from Vieta's formulas?
I would like a clarification for this or any other solutions would be fine too.
 A: $$ax^2+bx+c=0$$ so $$x_1=\frac{-b + \sqrt{b^2-4ac}}{2a} \; and \; x_2=\frac{-b - \sqrt{b^2-4ac}}{2a}$$
From this we have $x_1x_2=\frac{c}{a}$ and $x_1+x_2=-\frac{b}{a}$
A: Let $t^2+u t+v$ be a quadratic polynomial with roots $a$ and $b$. Then
$$t^2+u t+v=(t-a)(t-b)=t^2-(a+b)t+ab,$$
and hence $v=ab$ and $a=-(a+b)$. Note the $-$-sign. In you particular case with $a=1$ and $b=-2$
$$u=-(a+b)=1\qquad\text{ and }\qquad v=ab=-2,$$
corresponding to the polynomial $t^2+t-2$.
A: I like the following way.
Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Thus, $$8u^3-6uv^2=7$$ and $$4u^2-2v^2+2u+v^2=4,$$
which gives $$v^2=4u^2+2u-4.$$
But since $$v^2\leq u^2,$$ we obtain:
$$4u^2+2u-4\leq u^2$$ or
$$3u^2+2u-4\leq0$$ or
$$\frac{-1-\sqrt{13}}{3}\leq u\leq\frac{-1+\sqrt{13}}{3}.$$
Thus, $$8u^3-6u(4u^2+2u-4)=7$$ or
$$16u^3+12u^2-24u+7=0$$ or $$(2u-1)^2(4u+7)=0,$$ which gives $$u=\frac{1}{2},$$
$$x+y=1,$$ $$xy=-2$$ and we got the answer:
$$\{(-1,2),(2,-1)\}.$$
$$u^2\geq v^2$$ it's
$$\left(\frac{x+y}{2}\right)^2\geq xy$$ or
$$(x-y)^2\geq0.$$
