Question

If the quadratic equation $x^2+[a^2-5a+b+4]x+b=0$ has roots $-5$ and $1$, then maximum value of $\left[a\right]$ (where $\left[a\right]$ denote the greatet integer function) is

Answer 1

​​​​​​Since, $-5$ and $1$ are the roots of the equation.
$x^2+[a^2-5a+b+4]x+b=0$
$\therefore -5+1=-[a^2-5a+b+4]$ and $-5\times 1=b$
$\Rightarrow [a^2-5a+b+4]=4$ and b = -5
$\Rightarrow [a^2-5a-5+4]=4$
$\Rightarrow [a^2-5a-1]=4\Rightarrow 4\le a^2-5a-1<5$
$\Rightarrow a^2-5a-1\ge 4$ and $a^2-5a-1<5$
$\Rightarrow a^2-5a-5\ge 0$ and $a^2-5a-6<0$
$\Rightarrow a\le \frac{5-3\sqrt{5}}{2}\text{or}a\ge \frac{5+3\sqrt{5}}{2}\text{and}-1<a<6$
$\therefore a\in (-1,\frac{5-3\sqrt{5}}{2}]\cup [\frac{5+3\sqrt{5}}{2},6)$
$\therefore [a{]}_{max}=5$