Question

Quadratic equation $x^2+[a^2-5a+b+4]x+b=0$ has roots $-5$ and $1$, then number of integral values of $a$ are
Note: $[.]$ denotes the greatest integer function.

• A. $0$
• B. $1$
• C. $3$
• D. None of these

Answer 1

The correct option is A $0$

Since $-5$ and $1$ are the roots of 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 $b=-5$
$\Rightarrow [a^2-5a-1]=4$
$4\le a^2-5a-1<5$
$\Rightarrow a^2-5a-5\ge 0$ and $a^2-5a-6<0$
$\Rightarrow a\in (-1,\frac{5-3\sqrt{5}}{2}]\cup \left[\frac{5+3\sqrt{5}}{2},6\right)$
Number of integral values of $a$ is $0$ as $\frac{5-3\sqrt{5}}{2}<0$ and $\frac{5+3\sqrt{5}}{2}>5$