Question

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

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

Answer 1

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

Given: $x^2+2(a+2)x+9a-1=0$
Let $\alpha$ and $\beta$ be the roots of the quadratic equation. Then, according to the given conditions the graph of the quadratic expression $y=x^2+2(a+2)x+9a-1$ will be:


Thus, 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, using the quadratic formulae for $a^2-5a+5=0,$
We get the roots as: $a=\frac{5\pm \sqrt{25-20}}{2}=\frac{5\pm \sqrt{5}}{2}$
$\Rightarrow a^2-5a+5\ge 0\Rightarrow (a-\frac{5+\sqrt{5}}{2})(a-\frac{5-\sqrt{5}}{2})\ge 0a\in (-\infty ,\frac{5-\sqrt{5}}{2}]\cup [\frac{5+\sqrt{5}}{2},\infty )\cdots \left(1\right)$

$\left(ii\right)-\frac{b}{2a}<0\text{Here,}a=1,b=2(a+2)\Rightarrow -\frac{2(a+2)}{2}<0\Rightarrow a+2>0\Rightarrow a\in (-2,\infty )\cdots \left(2\right)$

$\left(iii\right)f\left(0\right)>0\Rightarrow 0^2+2(a+x)0+9a-1>0\Rightarrow 9a-1>0\Rightarrow a\in (\frac{1}{9},\infty )\cdots \left(3\right)$

From equations $\left(1\right),\left(2\right)\&\left(3\right),$ we get:


$a\in (\frac{1}{9},\frac{5-\sqrt{5}}{2}]\cup [\frac{5+\sqrt{5}}{2},\infty )$