Question

Let $\left[a\right]$ denotes the greatest integer less than or equal to $a$. Given that the quadratic equation $x^2+[a^2-5a+b+4]x+b=0$ has roots $-5$ and $1$, which of the following is true

• A. $a\in (-1,\frac{5-3\sqrt{5}}{2})\cup (\frac{5+3\sqrt{5}}{2},6)$
• B. $a\in (-1,\frac{5-3\sqrt{5}}{2})$
• C. $a\in (\frac{5+3\sqrt{5}}{2},6)$
• D. $a\in (-1,\frac{5-3\sqrt{5}}{2})\cup (\frac{5-3\sqrt{5}}{2},6)$

Answer 1

Answer: (A), (B), (C)

$a\in (-1,\frac{5-3\sqrt{5}}{2})$
$a\in (\frac{5+3\sqrt{5}}{2},6)$
$a\in (-1,\frac{5-3\sqrt{5}}{2})\cup (\frac{5+3\sqrt{5}}{2},6)$

Since $-5$ and $1$ are the roots of the equation

$x^2+[a^2-5a+b+4]x+b=0$

$\therefore -5+1=-[a^2-5a+b+4]and-5\times 1=b$

$\Rightarrow [a^2-5a+b+4]=4andb=-5$

$\Rightarrow [a^2-5a-5+4]=4$

$\Rightarrow [a^2-5a-1]=4\Rightarrow 4\le a^2-5a-1<5$

$\Rightarrow a^2-5a-1\ge 4and{a}^2-5a-1<5$

$\Rightarrow a^2-5a-5\ge 0and{a}^2-5a-6<0$

$\Rightarrow a\le \frac{5-3\sqrt{5}}{2}ora\ge \frac{5+3\sqrt{5}}{2}and-1,6$

$\therefore aϵ(-1,\frac{5-3\sqrt{5}}{2})\cup (\frac{5+3\sqrt{5}}{2},6)$