Question

If both the roots of $x^2+2(a+2)x+9a-1=0$ are negative, then ${}^{\prime }{a}^{\prime }$ lies in

• A. $(\frac{1}{9},1]\cup [4,\infty )$
• B. $[\frac{1}{9},1]\cup [4,\infty )$
• C. $(\frac{1}{9},\frac{5-\sqrt{5}}{2}]\cup [\frac{5+\sqrt{5}}{2},\infty )$
• D. $(-2,\infty )$

Answer 1

The correct option is C $(\frac{1}{9},\frac{5-\sqrt{5}}{2}]\cup [\frac{5+\sqrt{5}}{2},\infty )$

$x^2+2(a+2)x+9a-1=0$
The required conditions are,
$\left(i\right)D\ge 0\Rightarrow 4(a+2{)}^2-4(9a-1)\ge 0\Rightarrow a^2-5a+5\ge 0$
Now,
$a^2-5a+5=0\Rightarrow a=\frac{5\pm \sqrt{25-20}}{2}\Rightarrow a=\frac{5\pm \sqrt{5}}{2}$
$a\in (-\infty ,\frac{5-\sqrt{5}}{2}]\cup [\frac{5+\sqrt{5}}{2},\infty )\cdots \left(1\right)$

$\left(ii\right)-\frac{b}{2a}<0\Rightarrow -\frac{2(a+2)}{2}<0\Rightarrow a+2>0\Rightarrow a\in (-2,\infty )\cdots \left(2\right)$

$\left(iii\right)f\left(0\right)>0\Rightarrow 9a-1>0\Rightarrow a\in (\frac{1}{9},\infty )\cdots \left(3\right)$

From equations $\left(1\right),\left(2\right)\text{and}\left(3\right)$,
$a\in (\frac{1}{9},\frac{5-\sqrt{5}}{2}]\cup [\frac{5+\sqrt{5}}{2},\infty )$