Question

Let $x_1$ and $x_2$ be the roots of the equation $2x^2+6x+b=0.$ If $\frac{x_1}{x_2}+\frac{x_2}{x_1}=k$ and $b<0,$ then the minimum integral value of $|k|$ is

Answer 1

$2x^2+6x+b=0$
Discriminant of the equation
$\Delta =36-8b>0(\because b<0)$
So, the given equation has two real and distinct roots.
Sum and product of the roots,
$x_1+x_2=-3$ and $x_1{x}_2=\frac{b}{2}$

Now, $k=\frac{x_1}{x_2}+\frac{x_2}{x_1}$
$=\frac{x_1^2+x_2^2}{x_1{x}_2}=\frac{(x_1+x_2{)}^2-2x_1{x}_2}{x_1{x}_2}=\frac{(x_1+x_2{)}^2}{x_1{x}_2}-2$
$\Rightarrow k=\frac{18}{b}-2$

We know that,
$k<-2\Rightarrow |k|>2$
Hence, the minimum integral value of $|k|$ is $3.$