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

Hint:

Use Vieta's formulas: \(x_1+x_2 = -b/a\), \(x_1x_2 = c/a\).

Answer 1

The Correct Option is B

Step 1: Formula / Definition}
\[ \frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{x_1^2 + x_2^2}{x_1x_2} = \frac{(x_1+x_2)^2 - 2x_1x_2}{x_1x_2} \]
Step 2: Calculation / Simplification}
\(x_1 + x_2 = -\frac{6}{2} = -3\)
\(x_1x_2 = \frac{b}{2}\)
\[ \frac{x_1}{x_2} + \frac{x_2}{x_1} = \frac{(-3)^2 - 2(b/2)}{b/2} = \frac{9 - b}{b/2} = \frac{18}{b} - 2 \]
Since \(b<0\), \(\frac{18}{b}<0 \Rightarrow \frac{18}{b} - 2<-2\)
\(\therefore K = -2\)
Step 3: Final Answer
\[ K = -2 \]