Question

If both the roots of $x^2+2(k+2)x+9k-1=0$ are negative, then $k$ 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 $(\frac{1}{9},\frac{5-\sqrt{5}}{2}]\cup [\frac{5+\sqrt{5}}{2},\infty )$

$x^2+2(k+2)x+9k-1=0$

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

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

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

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