Question

If $$ S = \left\{ m \in \mathbb{R} : x^2 - 2(1 + 3m)x + 7(3 + 2m) = 0 \text{ has distinct roots} \right\} $$ then the number of elements in $ S $ is:

• A. 2
• B. 3
• C. 4
• D. Infinite

Hint:

For distinct roots of a quadratic, ensure \( D>0 \). Use inequality solving techniques to determine valid parameter ranges.

Answer 1

The Correct Option is D

We are given a quadratic in \( x \) with parameter \( m \), and are asked to find how many real values of \( m \) make the equation have distinct roots. Let us write the general quadratic: \[ x^2 - 2(1 + 3m)x + 7(3 + 2m) = 0 \] Compute the discriminant: \[ D = b^2 - 4ac \] Where:
- \( a = 1 \) 
- \( b = -2(1 + 3m) = -2 - 6m \) 
- \( c = 7(3 + 2m) = 21 + 14m \) So: \[ D = (-2 - 6m)^2 - 4(1)(21 + 14m) = (36m^2 + 24m + 4) - (84 + 56m) = 36m^2 - 32m - 80 \] We want: \[ D > 0 \Rightarrow 36m^2 - 32m - 80 > 0 \] This is a quadratic inequality. Since the parabola opens upward (coefficient of \( m^2 \) is positive), the expression is positive outside the roots. 
Solve: \[ 36m^2 - 32m - 80 = 0 \Rightarrow m = \frac{32 \pm \sqrt{(-32)^2 + 4 \cdot 36 \cdot 80}}{2 \cdot 36} \]
$= \frac{32 \pm \sqrt{1024 + 11520}}{72} $
$= \frac{32 \pm \sqrt{12544}}{72} $
$= \frac{32 \pm 112}{72} $ 
So, \[ m = \frac{144}{72} = 2,\quad m = \frac{-80}{72} = -\frac{10}{9} \] Therefore, the quadratic has distinct roots for: \[ m < -\frac{10}{9} \quad \text{or} \quad m > 2 \] This is an infinite number of real values of \( m \), hence: \[ \boxed{\text{Infinite}} \]