Question

If $b<0$, then the roots $x_1$ and $x_2$ of the equation $2x^2+6x+b=0$, satisfy the condition $\left(\frac{x_1}{x_2}\right)+\left(\frac{x_2}{x_1}\right)<K$, where $K$ is equal to

• A. $2$
• B. $-2$
• C. $0$
• D. $4$

Answer 1

The correct option is B $-2$

The discriminant of the quadratic equation $2x^2+6x+b=0$ is given by $D=36-8b>0$.
Therefore, the given equation has real roots.
we have,
$\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{x_1^2+x_2^2}{x_1\cdot x_2}=\frac{{(x_1+x_2)}^2-2x_1{x}_2}{x_1\cdot x_2}$
$=\frac{{(-3)}^2-2\left(b/2\right)}{\left(b/2\right)}=\frac{18}{b}-2<-2$ $[\because b<0]$