Question

If the roots of the equation \(x^2 + 2ax + b = 0\) are real, distinct and differ utmost by \(2m\), then b lies in the interval

• A. \((a^2, a^2 + m^2]\)
• B. \((a^2 + m^2, a^2)\)
• C. \([a^2, a^2 + 2m^2]\)
• D. \([a^2 - m^2, a^2)\)

Hint:

For a quadratic equation \(Ax^2 + Bx + C = 0\): Roots are real and distinct if the discriminant \( \Delta = B^2 - 4AC>0 \). The difference between roots is given by \( |\alpha - \beta| = \frac{\sqrt{\Delta}}{|A|} \). Alternatively, \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \).

Answer 1

The Correct Option is D

Step 1: Use the conditions for real and distinct roots.
For a quadratic equation \(Ax^2 + Bx + C = 0\), the roots are real and distinct if the discriminant \(\Delta = B^2 - 4AC > 0\).
Given the equation \(x^2 + 2ax + b = 0\), here \(A = 1\), \(B = 2a\), and \(C = b\).
So, the discriminant is:
\[ \Delta = (2a)^2 - 4(1)(b) = 4a^2 - 4b. \] Since the roots are real and distinct, we must have: \[ 4a^2 - 4b > 0 \] Divide by 4: \[ a^2 - b > 0 \] This implies: \[ b < a^2. \quad \text{(Condition 1)} \] Step 2: Use the condition that the roots differ utmost by \(2m\).
Let the roots of the equation be \(\alpha\) and \(\beta\).
From Vieta's formulas for \(x^2 + 2ax + b = 0\):
Sum of roots: \(\alpha + \beta = -2a\)
Product of roots: \(\alpha \beta = b\)
The difference between the roots is \(|\alpha - \beta|\). We know the identity: \[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta. \] Substitute the sum and product of roots: \[ (\alpha - \beta)^2 = (-2a)^2 - 4(b) = 4a^2 - 4b. \] So, the absolute difference between the roots is: \[ |\alpha - \beta| = \sqrt{4a^2 - 4b} = 2\sqrt{a^2 - b}. \] We are given that the roots differ utmost by \(2m\), which means \(|\alpha - \beta| \le 2m\).
\[ 2\sqrt{a^2 - b} \le 2m. \] Divide by 2: \[ \sqrt{a^2 - b} \le m. \] Since both sides are non-negative (because \(a^2 - b > 0\) from Condition 1 and \(m\) is a positive quantity), we can square both sides without changing the inequality direction: \[ (\sqrt{a^2 - b})^2 \le m^2 \] \[ a^2 - b \le m^2. \] Rearrange the inequality to solve for \(b\): \[ -b \le m^2 - a^2 \] Multiply by -1 and reverse the inequality sign: \[ b \ge a^2 - m^2. \quad \text{(Condition 2)} \] Step 3: Combine the conditions for \(b\).
From Condition 1, we have \(b < a^2\).
From Condition 2, we have \(b \ge a^2 - m^2\).
Combining these two inequalities, we find the interval for \(b\):
\[ a^2 - m^2 \le b < a^2. \] Thus, \(b\) lies in the interval \( [a^2 - m^2, a^2) \).