Question

If $b < 0$, then the roots $x_{1}$ and $x_{2}$ of the equation $2 x^{2}+6 x+b=0$, 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

Answer: (B)

The discriminant of the quadratic equation $2 x^{2}+6 x+b=0$ is given by $D=36-8 b>0$.
Therefore, the given equation has real roots.
we have, $\frac{x_{1}}{x_{2}}=\frac{x_{2}}{x_{1}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1} \cdot x_{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1} \cdot x_{2}}$
$=\frac{(-3)^{2}-2(b / 2)}{(b / 2)}=\frac{18}{b}-2<-2$
$[\because b<0]$