Question

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

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. Infinite

Hint:

For a quadratic equation to have distinct real roots, always use: \[ D=b^2-4ac\gt 0. \] If the resulting inequality gives intervals, then the set contains infinitely many values.

Answer 1

The Correct Option is D

Step 1: Write down the quadratic and the condition for distinct real roots.
The quadratic is $x^2 - 2(1+3m)x + 7(3+2m) = 0$. For distinct real roots, the discriminant must be strictly positive: $D > 0$.
Step 2: Compute the discriminant.
$D = [2(1+3m)]^2 - 4 \cdot 1 \cdot 7(3+2m) = 4(1+3m)^2 - 28(3+2m)$.
Step 3: Expand and simplify $D > 0$.
$4(1+6m+9m^2) - 28(3+2m) > 0$
$4 + 24m + 36m^2 - 84 - 56m > 0$
$36m^2 - 32m - 80 > 0$
$9m^2 - 8m - 20 > 0$.
Step 4: Find the roots of $9m^2 - 8m - 20 = 0$.
Using the quadratic formula: $m = \dfrac{8 \pm \sqrt{64 + 720}}{18} = \dfrac{8 \pm \sqrt{784}}{18} = \dfrac{8 \pm 28}{18}$. So $m = \dfrac{36}{18} = 2$ or $m = \dfrac{-20}{18} = -\dfrac{10}{9}$.
Step 5: Determine where $9m^2-8m-20 > 0$.
The parabola opens upward, so $9m^2-8m-20>0$ when $m < -\dfrac{10}{9}$ or $m > 2$. This means $S = \left(-\infty, -\dfrac{10}{9}\right) \cup (2, +\infty)$.
Step 6: Count elements in $S$.
The set $S$ is an infinite union of real intervals, so it contains infinitely many real numbers $m$. Therefore the number of elements in $S$ is infinite. \[ \boxed{\text{Infinite}} \]