Question

If b < 0, then the roots x₁ and x₂ of the equation

2x² + 6x + b = 0, satisfy the condition (x₁/x₂) + (x₂/x₁) < – k where k is equal to

• A. – 3
• B. – 5
• C. – 6
• D. –

Answer 1

Answer: (D)

–

Answer 2

The discriminant of the quadratic equation 2x² + 6x + b = 0 is

given by D = 36 – 8b > 0. Therefore, the given equation has real roots.

We have

$\frac{x_{1}}{x_{2}}$ + $\frac{x_{2}}{x_{1}}$ = $\frac{x_{1}^{2} + x_{2}^{2}}{x_{1}x_{2}}$ = $\frac{(x_{1} + x_{2})^{2} - 2x_{1}x_{2}}{x_{1}x_{2}}$

= $\frac{( - 3)^{2} - 2(b/2)}{b/2}$= $\frac{18}{b}$ – 2 < – 2