## Online Course Discussion Forum

### II-A Algebra Lecture 9

Question 9.26:

Find all pairs (x,y)(x,y) of real numbers so that x+xy+y=4x+xy+y=4 and x2+xy+y2=8x2+xy+y2=8.

Solution: Let u=x+yu=x+y and v=xyv=xy, so we have u+v=4u+v=4 and u2v2=8u2−v2=8. Since u2v2=(uv)(u+v)u2−v2=(u−v)(u+v), we have (uv)=2(u−v)=2. Solving the system u+v=4u+v=4 and uv=2u−v=2 we get u=3,v=1u=3,v=1. Hence x,yx,y are roots of the equation t2ut+v=0t2−ut+v=0 or t23t+1t2−3t+1. Solving we have t=3±52t=3±52. Thus the pairs are (3+52,352),(352,3+52)(3+52,3−52),(3−52,3+52).

### My question is: why is u^2-v^2=8? Isn't it u^2-v=8? Also, where does the variable t come from? Thanks.

You are correct, it should be $u^2 - v = 8$. Thanks for catching that! Both lead to the same correct solutions, and that's probably why we didn't catch the mistake earlier. The solution is now updated in your hw.

Here's the updated solution:

Let $u = x + y$, and $v = xy$, so we have $u + v = 4$ and $u^2 - v = 8$. This leads to the quadratic equation $u^2 + u - 12 = 0$, which has solutions $u = -4, 3$, so $(u,v) = (-4, 8), (3, 1)$. $x$ and $y$ are roots of the equation $t^2 - ut + v = 0$. For $(u, v) = (-4, 8)$ this gives the equation $t^2 + 4t + 8 = 0$ which has no real solutions. For $(u,v) = (3, 1)$ this gives the equation $t^2 - 3t + 1$. Solving we have $\displaystyle t = \frac{3\pm \sqrt{5}}{2}$. Thus the pairs are $\displaystyle \left(\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2}\right), \left(\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right)$.

As to where does the $t$ come from: Since $u = x+y$ and $v = xy$, the equation $t^2 - ut + v = 0$ is the same as the equation $t^2 - (x+y)t + xy = 0$, that is, $(t - x)(t - y) = 0$, which clearly has solutions $x$ and $y$. If $u$ and $v$ had be chosen differently, this would not be true.