Could someone help with an explanation on how to treat the equation below, please? A solution I have read uses the discriminant to find the range i.e. $9-20y^2-16y>or = 0$. To me this means treating the variable Y as a constant , is that "acceptable" ? does anyone have a reference for these equations ?
$yx^2+5y-3x+4=0$
rewrite as $yx^2-3x+(5y+4)=0$
thanks ralph
Yes, I take from $$yx^2-3x+(5y+4)\ge 0~~~(1)$$ A quadratic $f(x)=Ax^2+Bx+C$ is positive definite for all real values of $x$ if $B^2\le 4AC$ and $A>0$ So (1) will hold for all real values of $x$ if $y>0$ and $B^2 \le 4AC$. So we get $$9 \le 4y(5y+4) \implies 20y^2+6y-9 \ge 0 $$ $$ \implies \left(x-\frac{4+\sqrt{61}}{10}\right) \left(x-\frac{4-\sqrt{61}}{10}\right) \ge 0 $$ So $$\implies y \ge \frac{4+\sqrt{61}}{10} ~or~ y \le \frac{4-\sqrt{61}}{10}$$ But due to the condition $y>0$, we finally get $$ y \ge \frac{4+\sqrt{61}}{10}.$$
