Let `alpha and beta (a lt beta) " be the roots of the equation " x^(2) + bx + c = 0," where " b gt 0 and c lt 0 . `
Consider the following : 1. ` beta lt -alpha` 2. `beta lt |a|`
Which of the above is//are corect ?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To solve the problem, we need to analyze the roots of the quadratic equation \(x^2 + bx + c = 0\) where \(b > 0\) and \(c < 0\). The roots are denoted as \(\alpha\) and \(\beta\) with the condition that \(\alpha < \beta\).

### Step 1: Find the roots of the quadratic equation
The roots of the quadratic equation can be found using the quadratic formula:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = b\), and \(c = c\). Thus, the roots are:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4c}}{2} \]
Since \(c < 0\), the discriminant \(b^2 - 4c\) is positive, ensuring that the roots are real.

### Step 2: Determine the values of \(\alpha\) and \(\beta\)
Given that \(\alpha < \beta\), we can assign:
\[ \alpha = \frac{-b - \sqrt{b^2 - 4c}}{2}, \quad \beta = \frac{-b + \sqrt{b^2 - 4c}}{2} \]

### Step 3: Analyze the first statement: \(\beta < -\alpha\)
To check if \(\beta < -\alpha\), we first calculate \(-\alpha\):
\[ -\alpha = -\left(\frac{-b - \sqrt{b^2 - 4c}}{2}\right) = \frac{b + \sqrt{b^2 - 4c}}{2} \]
Now, we need to compare \(\beta\) and \(-\alpha\):
\[ \beta = \frac{-b + \sqrt{b^2 - 4c}}{2} < \frac{b + \sqrt{b^2 - 4c}}{2} \]
This simplifies to:
\[ -b + \sqrt{b^2 - 4c} < b + \sqrt{b^2 - 4c} \]
Subtracting \(\sqrt{b^2 - 4c}\) from both sides gives:
\[ -b < b \]
This is true since \(b > 0\). Thus, the first statement \(\beta < -\alpha\) is **correct**.

### Step 4: Analyze the second statement: \(\beta < |\alpha|\)
Next, we need to check if \(\beta < |\alpha|\):
Since \(\alpha\) is negative (as \(b > 0\) and \(c < 0\)), we have:
\[ |\alpha| = -\alpha = \frac{b + \sqrt{b^2 - 4c}}{2} \]
Now, we compare \(\beta\) and \(|\alpha|\):
\[ \beta = \frac{-b + \sqrt{b^2 - 4c}}{2} < \frac{b + \sqrt{b^2 - 4c}}{2} \]
This simplifies to:
\[ -b + \sqrt{b^2 - 4c} < b + \sqrt{b^2 - 4c} \]
Subtracting \(\sqrt{b^2 - 4c}\) from both sides gives:
\[ -b < b \]
This is again true since \(b > 0\). Thus, the second statement \(\beta < |\alpha|\) is also **correct**.

### Conclusion
Both statements are correct:
1. \(\beta < -\alpha\)
2. \(\beta < |\alpha|\)