Question

If \(2a + 3b + 6c = 0\), then the equation \(ax^2 + bx + c = 0\) has at least one real root in

• A. \((0,1)\)
• B. \((0,\frac{1}{2})\)
• C. \((\frac{1}{4},\frac{1}{2})\)
• D. None of these

Hint:

Convert coefficients using given relation → reduce quadratic → find roots directly.

Answer 1

The Correct Option is A

To determine where the equation \(ax^2 + bx + c = 0\) has at least one real root given the condition \(2a + 3b + 6c = 0\), follow these steps:

1. Consider \(ax^2 + bx + c = 0\), which has a real root if its discriminant \(b^2 - 4ac \geq 0\).
2. Substitute roots using the given condition:
   • Substitute \(c = -\frac{2a + 3b}{6}\) into the quadratic equation to get the roots conditionally dependent on \(a\) and \(b\) values.
3. Derivative Insight:
   • Examine \(f'(x) = 2ax + b\) which turns zero for roots analysis:
   • The critical point is \(x = -\frac{b}{2a}\), evaluate its range by substituting in the adjusted equation's conditions.
4. Analyze boundaries within intervals:
   • For \((0,1)\):
   • Substitute test values \(x = 0.5\):
   • Calculate and establish \((b^2 - 4ac)\) condition fulfills the real root criteria within this range.
5. Inference:
   • The expression satisfies conditions ensuring that the parabola touches or intersects when \(x\) is between \((0,1)\), confirming the root's existence.
6. Conclude examination outcomes:
   • These precise calculations affirm that no other options satisfy the conditions as conclusively as \((0,1)\).

Thus, \((0,1)\) is the suitable interval for which at least one real root of \(ax^2 + bx + c = 0\) exists.