Consider the equation $\mathrm{x}^2+4 \mathrm{x}-\mathrm{n}=0$, where $\mathrm{n} \in[20,100]$ is a natural number. Then the number of all distinct values of $n$, for which the given equation has integral roots, is equal to

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \mathrm{x}^2+4 \mathrm{x}+4=\mathrm{n}+4 \\ & (\mathrm{x}+2)^2=\mathrm{n}+4 \\ & \mathrm{x}=-2 \pm \sqrt{\mathrm{n}+4} \\ & \because 20 \leq \mathrm{n} \leq 100 \\ & \sqrt{24} \leq \sqrt{\mathrm{n}+4} \leq \sqrt{104} \\ & \Rightarrow \sqrt{\mathrm{n}+4} \in\{5,6,7,8,9,10\}
\end{aligned}$
$\therefore ' 6$ ' integral values of ' $n$ ' are possible /

Consider the equation $\mathrm{x}^2+4 \mathrm{x}-\mathrm{n}=0$, where $\mathrm{n} \in[20,100]$ is a natural…

Solution