Question

If both roots of the equation $x^2+ax+2=0$ lie in the interval $(0,3),$ then the range of values of $a$ is

• A. $(-6,0)$
• B. $(-\frac{11}{3},\infty )$
• C. $(-\frac{11}{3},-2\sqrt{2}]$
• D. $(-\infty ,-2\sqrt{2}]\cup [2\sqrt{2},\infty )$

Answer 1

The correct option is C $(-\frac{11}{3},-2\sqrt{2}]$

Given the quadratic equation: $x^2+ax+2=0$
Let $\alpha ,\beta$ be the roots of the equation.
Now $0<\alpha ,\beta <3$ which can be represented as:


Let $f\left(x\right)=x^2+ax+2$

Now, for this condition to happen, we have $3$ conditions that need to be followed, that are:

$i)D\ge 0\Rightarrow a^2-8\ge 0\Rightarrow (a-2\sqrt{2}\left)\right(a+2\sqrt{2})\ge 0\Rightarrow a\in (-\infty ,-2\sqrt{2}]\cup [2\sqrt{2},\infty )\cdots (1)$

$ii)0<\frac{\text{Sum of roots}}{2}<3\Rightarrow 0<-\frac{a}{2}<3\Rightarrow a\in (-6,0)\cdots (2)$

$iii\left)f\right(0)>0\&f(3)>0$
$\Rightarrow 2>0\&9+3a+2>0\Rightarrow 3a+11>0\Rightarrow a\in (-\frac{11}{3},\infty )\cdots \left(3\right)$

Thus, from $\left(1\right),\left(2\right)\&\left(3\right),$ we get:
$a\in \{(-\infty ,-2\sqrt{2}]\cup [2\sqrt{2},\infty )\}\cap (-6,0)\cap (-\frac{11}{3},\infty )$

Thus, we get the solution as: $a\in (-\frac{11}{3},-2\sqrt{2}]$