Show that the largest root of the following equation is $Show that the largest root of the following equation is $<n;n>2$.
$x^2-(n-2)x-2(n-2)$.
On solving I got the largest root $a=\frac{(n-2)+\sqrt {(n-2)^2+8(n-2)}}{2}$ but unable to show how is it $<n$.
I tried for various values of $n$ where the result is holding .But to prove it.
Please help
 A: Hint: notice that
$$
\boldsymbol{(n-2)^2 + 8(n-2)} < (n-2)^2 + 8(n-2) + 16 = ((n-2) + 4)^2 = n+2.
$$
Now apply this to
$$
a = \frac{n-2 + \sqrt{\boldsymbol{(n-2)^2 + 8(n-2)}}}{2}
$$
A: I believe that $n$ is a natural number. 
For $n=1$ our equation has no roots.
For $n=2$ it's obvious. 
Let $n>2$ and $f(x)=x^2-(n-2)x-2(n-2)$.
Hence, it's obvious that $a>\frac{n-2}{2}$ and $f$ an increasing function on $\left[\frac{n-2}{2},+\infty\right)$.
Let $a\geq n$.
Thus, $$0=a^2-(n-2)a-2(n-2)\geq n^2-(n-2)n-2(n-2)$$
or $0\geq4$, which is contradiction.
Id est, $a<n$ and we are done!
A: Assuming $n \ge 2$ . . .

One can just simplify the inequality $a < n$ until it admits to being true:
\begin{align*}
&a < n\\[5pt]
\iff\;&\frac{(n-2)+\sqrt {(n-2)^2+8(n-2)}}{2} < n\\[6pt]
\iff\;&(n-2) + \sqrt {(n-2)^2+8(n-2)} < 2n\\[6pt]
\iff\;&\sqrt {(n-2)^2+8(n-2)} < n+2\\[6pt]
\iff\;&(n-2)^2+8(n-2) < (n+2)^2\qquad\text{[since $n \ge 2$]}\\[5pt]
\iff\;&(n^2 - 4n + 4) + (8n-16) < n^2 + 4n + 4\\[5pt]
\iff\;&n^2 + 4n -12< n^2 + 4n + 4\\[5pt]
\iff\;&-12 < 4\\[5pt]
\end{align*}
which is clearly true.
A: Let $f(x)=x^2-(n-2)x-2(n-2).$
Two observations:


*

*Discriminant of the equation $f(x)=0$, $(n-2)^2+8(n-2)>0$, since $n>2$. Then the equation must have two real roots. Then $\min f(x)<0$, and minima is $\frac{n-2}{2}<n$.





*$f(n)=n^2-(n-2)n-2(n-2)=4>0$.



Then due to continuity of $f(x)$ largest root must lie in $(\frac{n-2}{2},n)$.
A: Hint:   expressing in terms of $x-n\,$:
$$x^2-(n-2)x-2(n-2) \;=\; (x-n)^2 + (n + 2) (x-n) + 4$$ 
The RHS is clearly positive for $\,x \ge n\,$.
