Quadratic Equations
Location of Roots
Grade 11
Question:
<p>If the equation <span class="math">\(ax^2 + bx + c = 0\)</span>, where <span class="math">\(a, b, c \in \mathbb{R}\)</span> and <span class="math">\(a > 0\)</span>, has two real roots <span class="math">\(\alpha\)</span> and <span class="math">\(\beta\)</span> such that <span class="math">\(\alpha < -2\)</span> and <span class="math">\(\beta > 2\)</span>, then</p>
<p>(A) <span class="math">\(c < 0\)</span></p>
<p>(B) <span class="math">\(a - |b| + c < 0\)</span></p>
<p>(C) <span class="math">\(4a + 2|b| + c < 0\)</span></p>
<p>(D) <span class="math">\(6a + |b| + c < 0\)</span></p>
Step-by-Step Solution
Key Concept: Since both roots lie on opposite sides of 2 (α < 2 < β), the parabola f(x) = ax² + bx + c opens upward (a > 0) and must be negative at x = 2. This gives f(2) < 0, leading directly to 4a + 2b + c < 0.
**Step 1:** Define the function and analyze the root condition.
Let $f(x) = ax^2 + bx + c$. The given condition states that the equation $f(x) = 0$ has two real roots, $\alpha$ and $\beta$, such that $\alpha < 2 < \beta$. This means that the value $x=2$ lies strictly between the two roots of the quadratic equation.
**Step 2:** Determine the opening direction of the parabola.
Since $a > 0$, the parabola represented by $f(x) = ax^2 + bx + c$ opens upwards. For an upward-opening parabola, the function values are negative for $x$ values between its roots and positive for $x$ values outside its roots.
**Step 3:** Evaluate the function at $x=2$.
As $x=2$ lies between the roots $\alpha$ and $\beta$, the value of the function at $x=2$ must be negative.
$$f(2) < 0$$
Substitute $x=2$ into the function definition:
$$a(2)^2 + b(2) + c < 0$$
$$4a + 2b + c < 0$$
**Step 4:** Conclusion.
The condition that must hold is $4a + 2b + c < 0$.
Correct Answer: A