Mathematics - Quadratic Equation Question with Solution | TestHub

$y{x}^2-y=ax+1\Rightarrow y{x}^2-ax-y-1=0$

$$\begin{array}{rlrlrlrl}& \Rightarrow {\mathrm{a}}^2+4(\mathrm{y})(\mathrm{y}+1)\ge 0& \Rightarrow 4{\mathrm{y}}^2+4\mathrm{y}+{\mathrm{a}}^2\ge 0& \Rightarrow 16-4\times 4{\mathrm{a}}^2\le 0& \Rightarrow 1-{\mathrm{a}}^2\le 0& \Rightarrow {\mathrm{a}}^2-1\ge 0& \mathrm{a}\in (-\infty ,-1]\cup [1,\infty )\text{ but }-1\ge \mathrm{a}\ge -5& \text{ so }\mathrm{a}=-5,-4,-3,-2,-1\end{array}$$

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