Algebra Craze 2

Let $m > \sqrt{2}$ .

$\begin{aligned} x+y &=& mX - 1 \\ xy &=& \dfrac{1}{2}((m^2 - 1)X^2 - mX - 2) \end{aligned}$

$\implies$

$\begin{aligned} (x+y)^2 &=& m^2X^2 - 2mX + 1\\ 2xy &=& (m^2 - 1)X^2 - mX - 2 \end{aligned}$

$\implies (x + y)^2 - 2xy = x^2 + y^2 = X^2 - mX + 3$ .

$x = \dfrac{(m^2 - 1)X^2 - mX - 2}{2y} \implies 2y^2 - 2(mX - 1)y + ((m^2 - 1)^2 - mX - 2) = 0$ which has real value solutions whenever the discriminant $D = 20 - (4m^2 - 8)X^2 \geq 0 \implies |X|^2 \leq \dfrac{5}{m^2 - 2} \implies |X| \leq \sqrt{\dfrac{5}{m^2 - 2}} \implies$

$x^2 + y^2 = \dfrac{5}{m^2 - 2} - \sqrt{\dfrac{5}{m^2 - 2}}m + 3 = 3 - \sqrt{\dfrac{5}{m^2 - 1}}m + 1 \implies \dfrac{5}{m^2 - 2} = 1 \implies m = \sqrt{7}$

$\implies x^2 + y^2 = (4 - \sqrt{7})$ and $|X| \leq 1$ .

Using $m = \sqrt{7}$ and $X = 1 \implies$

$\begin{aligned} x+y &=& \sqrt{7} - 1 \\ y &=& \dfrac{4 - \sqrt{7}}{2x} \end{aligned}$

$\implies 2x^2 - 2(\sqrt{7} - 1)x + 4 - \sqrt{7} = 0 \implies x_{0} = \dfrac{\sqrt{7} - 1}{2} \implies y_{0} = \dfrac{4 - \sqrt{7}}{\sqrt{7} - 1} = \dfrac{\sqrt{7} - 1}{2} \implies 3x_{0} + 7y_{0} = 5(\sqrt{7} - 1)$ $\approx \boxed{8.22875656}$ .

To show $x + y = \sqrt{7} - 1$ is tangent to $y = \dfrac{4 - \sqrt{7}}{2x}$ at $(\dfrac{\sqrt{7} - 1}{2},\dfrac{\sqrt{7} - 1}{2})$ .

$y = \dfrac{4 - \sqrt{7}}{2x} \implies \dfrac{dy}{dx}|_{x = \dfrac{\sqrt{7} - 1}{2}} =$ $-\dfrac{4 - \sqrt{7}}{2x^2}|_{x = \dfrac{\sqrt{7} - 1}{2}} = \dfrac{-2(4 - \sqrt{7})}{(\sqrt{7} - 1)^2} =$

$-\dfrac{(4 - \sqrt{7})(\sqrt{7} + 1)}{3(\sqrt{7} - 1)} = \dfrac{-3(\sqrt{7} - 1)}{3(\sqrt{7} - 1)} = -1$

$\implies y - (\dfrac{\sqrt{7} -1}{2}) = -1(x - (\dfrac{\sqrt{7} - 1}{2}) \implies \boxed{x + y = \sqrt{7} - 1}$ .