Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~x^2+y^2=44^2-90xy+x^2y^2$
But from here I didn't know what to do. Could you help me in solving this equation?
 A: $$xy+x+y=44$$
$$\implies xy+x+y+1=45=3\times15$$
$$\implies (y+1)(x+1)=3\times15$$
therefore $y=2$ and $x=14$ , also it satisfies the second identity. Hence $x^2+y^2=196+4=200.$
A: Let $A = xy$ and $B = x + y$. Then we have
$$A + B = 44$$
$$AB = 448$$
By Viete's formula, $A$ and $B$ are the roots of
$$r^2 - 44r + 448 = 0$$
$$(r - 16)(r - 28) = 0$$
such that
$$(xy, x + y) = (16, 28),~(28, 16)$$ 
from which it is trivial do deduce $x^2 + y^2$:
$$\begin{align}x^2 + y^2 &= (x + y)^2 - 2xy
\\&= 28^2 - 2(16), 16^2 - 2(28)\\
&=752, 200\end{align}$$
A: Here is a solution using vieta's formula. 
Let $s=x+y$ and $p=xy$. Then $s+p=44$ and $sp=448$. Therefore $s$ and $p$ are the roots of the quadratic $z^2-44z+448=0$. Thus it follows that $s=16$ and $p=28$. Hence $$x^2+y^2=(x+y)^2-2xy=s^2-2p=256-56=\boxed{200}.$$
But note that if $s=28$ and $p=16$, then 
$$x^2+y^2=(x+y)^2-2xy=784-32=\boxed{752}.$$
A: Here is one suggestion. Let $a=x+y, b=xy$ then $a+b=44,  ab=448$
This gives $a=16, b=28, x^2+y^2=a^2-2b=200$ or $a=28, b=16, x^2+y^2= 752$
The intermediate step is to formulate and solve the quadratic satisfied by $a$ and $b$
A: $$(x+y)=44-xy..............1$$
$$xy(x+y)=448........................2$$
substitute 1 in 2
$$xy(44-xy)=448$$
$$x^2y^2-44xy+448=0$$
use the quadratic formula
$$xy=22\pm6$$
or
$$x^2y^2=(22\pm6)^2$$
now  square the equ.1
$$x^2+2xy+y^2=44^2-88xy+x^2y^2$$
$$x^2+y^2=44^2-90xy+x^2y^2$$
hence
$$x^2+y^2=44^2-90(22\pm6)+(22\pm6)^2$$
