So on my math homework I had this problem:
$\left( \frac{x^2}{2}-5x+50 \right) \left( \frac{y^2}{2}-5y+50 \right) = 674^2$
$x+y=10$
I was thinking about substituting in $x$ for $y$ because $y=10-x$ and then just expanding everything, but that seems really algebraically bashy, so is there a better way to solve this? Or if the solution is to substitute, how would I solve it?
One solution without bash is by solving $xy$. Notice that the first equation is symmetric w.r.t. $x,y$. So, we have
$$674^2=\left( \frac{x^2}{2}-5x+50 \right) \left( \frac{y^2}{2}-5y+50 \right) \\=\frac{x^2y^2}4-\frac{5}2xy(x+y)+25(x^2+y^2)+25xy-250(x+y)+2500$$
Notice that $(x^2+y^2)=(x+y)^2-2xy$, we have $$674^2=\left( \frac{x^2}{2}-5x+50 \right) \left( \frac{y^2}{2}-5y+50 \right) \\=\frac{x^2y^2}4-\frac{5}2xy(x+y)+25(x+y)^2-25xy-250(x+y)+2500$$
Substitute in $x+y=10$, we have
$$674^2=\frac{x^2y^2}4-25xy+2500-25xy-2500+2500$$
Or, $$x^2y^2-200xy+100^2-1348^2=0$$ So, we have $xy=100\pm 1348$, or, $xy=-1248$ or $xy=1448$
So, $x,y$ must satisfy one of these by Vieta:
$$t^2-10t-1248=0 \text{ or }t^2-10t+1448=0$$
The first equation yields $x,y=5\pm\sqrt{1273}$ and the second equation yields $x,y=5\pm\sqrt{1423}i$.
P.S. I have checked my solution twice, and put it into number empire, but there are no integer solutions... where are they?
This is what the problem looks like now
And this is the result