Mathematics - Functions Question with Solution | TestHub

$\begin{array}{rl}& y=\frac{5-x}{x^2-3x+2}\\ & y{x}^2-3xy+2y+x-5=0\\ & y{z}^2+(-3y+1)x+(2y-5)=0\end{array}$
Case I : If $y=0$ (Accepted)
$\Rightarrow x=5$
Case II : If $y\ne 0$
$\begin{array}{rl}& \mathrm{D}\ge 0\\ & (-3y+1{)}^2-4(y)(2y-5)\ge 0\\ & 9y^2+1-6y-8y^2+20y\ge 0\\ & {\mathrm{y}}^2+14y+1\ge 0\\ & (y+7{)}^2-48\ge 0\\ & |y+7|\ge 4\sqrt{3}\\ & \Rightarrow y+7\ge 4\sqrt{3}\text{ or }\mathrm{y}+7\le -4\sqrt{3}\\ & \Rightarrow \mathrm{y}\ge 4\sqrt{3}-7\text{ or }\mathrm{y}\le -4\sqrt{3}-7\end{array}$
From Case I and Case II
$\mathrm{y}\in (-\infty ,-4\sqrt{3}-7]\cup [4\sqrt{3}-7,\infty )$
So $\alpha =-4\sqrt{3}-7$
$\beta =4\sqrt{3}-7$
$\begin{array}{rl}\Rightarrow a^2+b^2& =(-4\sqrt{3}-7{)}^2+(4\sqrt{3}-7{)}^2\\ & =2(48+49)\\ & =194\end{array}$