Technique explanation
The Chalana-Kalanabhyam sutra ("By calculus" / "Sequential motion") reveals a profound link between calculus and algebra. For any quadratic equation ax² + bx + c = 0, the roots can be found using the first derivative and the discriminant (D).
The relationship states that at the roots of the equation, the derivative is equal to the positive or negative square root of the discriminant:
2ax + b = ±√D
This method provides a geometric insight into the behavior of the derivative at the x-intercepts and is useful for identifying repeated roots.
Method Condition: Specifically for quadratic equations; explains the link between calculus and algebra.
Standard fallback: The standard quadratic formula x = [-b ± √(b² - 4ac)] / 2a.
Worked examples
Example 1: Rational Roots
Solve: 2x² + 5x + 3 = 0
- Step 1: Calculate discriminant (D): 5² - 4(2)(3) = 25 - 24 = 1
- Step 2: Find the derivative: d/dx(2x² + 5x + 3) = 4x + 5
- Step 3: Set up relation: 4x + 5 = ±√1 = ±1
- Step 4: Solve for x: 4x+5=1 → x=-1; 4x+5=-1 → x=-1.5
Check: Substitute x=-1.5 into the original equation: 2(-1.5)² + 5(-1.5) + 3 = 2(2.25) - 7.5 + 3 = 4.5 - 7.5 + 3 = 0. The root is correct.
Example 2: Repeated Roots
Find if x² - 6x + 9 = 0 has a repeated root.
- Step 1: Discriminant (D): (-6)² - 4(1)(9) = 36 - 36 = 0
- Step 2: Derivative: 2x - 6 = ±√0 = 0
- Step 3: Solve: 2x = 6 → x = 3
Since D=0, the derivative vanishes at the root, confirming x=3 is a repeated root.
Check: Expand (x-3)² = x² - 2(3)x + 3² = x² - 6x + 9. The square matches the original equation, confirming the repeated root.
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