First slide

Theory of equations

Question

$If b < 0, then the roots x_{1} and x_{2} of the equation 2 x^{2} + 6 x + b = 0 satisfy the condition (x_{1} / x_{2}) + (x_{2} / x_{1}) < k where k =$

Moderate

A

2
B

-2
C

0
D

4

Solution

$The discriminant of the quadratic equation 2 x^{2} + 6 x + b = 0 is$
$given by D = 36 - 8 b > 0 . Therefore, the given equation has real roots$
$We have \frac{x_{1}}{x_{2}} + \frac{x_{2}}{x_{1}} = \frac{x_{1}^{2} + x_{2}^{2}}{x_{1} x_{2}} = \frac{{(x_{1} + x_{2})}^{2} - 2 x_{1} x_{2}}{x_{1} x_{2}}$
$= \frac{(- 3)^{2} - 2 (\frac{b}{2})}{\frac{b}{2}} = \frac{2 (9 - b)}{b} = \frac{18}{b} - 2 < - 2 [∵ b < 0]$

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