Consider the equation x^2+4 x-n=0, where n [20,100] is a natural number.

Question

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

Accepted Answer

$$\begin{aligned}
& x^2+4 x-n=0 \text { has integer roots } \
& \Rightarrow x=\frac{-4 \pm \sqrt{16+4 n}}{2}=-2 \pm \sqrt{4+n}
\end{aligned}$$

For $x$ to be integer $4+n$ must be perfect squares

$$\begin{aligned}
& n \in[20,100] \
& n+4 \in[24,104]=S
\end{aligned}$$

$\left\{25,36, \ldots 10^2\right\} \in S \Rightarrow 5^2, 6^2, \ldots 10^2 \Rightarrow 6$ values of $n$

Solutions · 1