Technique explanation
The Sopantyadvayamantyam sutra ("The ultimate and twice the penultimate") applies to a specific class of reciprocal equations where denominators are products of terms in an arithmetic progression (AP).
For equations of the form 1/(AB) + 1/(AC) = 1/(AD) + 1/(BC), where A, B, C, D are linear terms in an AP, the root can be found instantly using the relation:
D + 2C = 0
This rule collapses a complex rational equation into a simple linear one by leveraging the cancellation properties of AP terms.
Method Condition: Denominators must be products of terms in an arithmetic progression (A, B, C, D).
Standard fallback: Rationalizing the equation by finding a common denominator and solving the resulting polynomial.
Worked examples
Example 1: Standard AP Denominators
Solve: 1/[(x+1)(x+2)] + 1/[(x+1)(x+3)] = 1/[(x+1)(x+4)] + 1/[(x+2)(x+3)]
- Step 1: Verify AP: A=x+1, B=x+2, C=x+3, D=x+4. Common difference is 1.
- Step 2: Apply rule D + 2C = 0: (x+4) + 2(x+3) = 0 → 3x + 10 = 0
- Step 3: Solve for x: x = -10/3
Check: Substitute x = -10/3. LHS = 1/[(-7/3)(-4/3)] + 1/[(-7/3)(-1/3)] = 9/28 + 9/7 = 45/28; RHS = 1/[(-7/3)(2/3)] + 1/[(-4/3)(-1/3)] = -9/14 + 9/4 = 45/28. The identity holds.
Example 2: AP with Difference of 2
Solve: 1/[(x+1)(x+3)] + 1/[(x+1)(x+5)] = 1/[(x+1)(x+7)] + 1/[(x+3)(x+5)]
- Step 1: Verify AP: A=x+1, B=x+3, C=x+5, D=x+7. Common difference is 2.
- Step 2: Apply rule D + 2C = 0: (x+7) + 2(x+5) = 0 → 3x + 17 = 0
- Step 3: Solve for x: x = -17/3
Check: Substitute x = -17/3. LHS = 1/[(-14/3)(-8/3)] + 1/[(-14/3)(-2/3)] = 9/112 + 9/28 = 45/112; RHS = 1/[(-14/3)(4/3)] + 1/[(-8/3)(-2/3)] = -9/56 + 9/16 = 45/112. The identity holds.
Academy Course