Lessons 61–78: algebra foundations, factorisation, equations, and related reference methods in the new Vedic Math course.

Lesson 78 — Partial Fractions, Then an Integration Boundary

Technique explanation

This lesson applies Paravartya Yojayet (Transpose and Adjust) to partial fraction decomposition and establishes the boundary for its use in calculus.

Paravartya allows finding the numerators of partial fractions by substitution. For a fraction f(x)/((x-a)(x-b)), the numerator for (x-a) is found by substituting x=a into the rest of the expression. This is a direct application of "transpose and adjust."

This method is used to decompose rational functions before integration. However, the Vedic method only optimizes the algebraic setup; the actual integration follows standard calculus rules. This establishes a clear boundary between Vedic algebra and calculus execution.

Standard fallback: The method of undetermined coefficients, which involves setting up a system of linear equations by equating the coefficients of like powers of x to solve for the unknown numerators.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Decompose (3x+5)/((x+1)(x+2))

  • Step 1: To find A for (x+1), set x=-1 in (3x+5)/(x+2): A = (3(-1)+5)/(-1+2) = 2/1 = 2.
  • Step 2: To find B for (x+2), set x=-2 in (3x+5)/(x+1): B = (3(-2)+5)/(-2+1) = -1/-1 = 1.
  • Result: 2/(x+1) + 1/(x+2).
  • Check: Recombine the partial fractions: 2/(x+1) + 1/(x+2) = [2(x+2) + 1(x+1)] / [(x+1)(x+2)] = (2x + 4 + x + 1) / [(x+1)(x+2)] = (3x + 5) / [(x+1)(x+2)].

Example 2: Set up ∫ (x²+1)/((x-1)(x-2)(x-3)) dx

  • Step 1: Decompose the integrand. x=1 → A = (1+1)/((1-2)(1-3)) = 2/2 = 1.
  • Step 2: x=2 → B = (4+1)/((2-1)(2-3)) = 5/-1 = -5.
  • Step 3: x=3 → C = (9+1)/((3-1)(3-2)) = 10/2 = 5.
  • Setup: ∫ (1/(x-1) - 5/(x-2) + 5/(x-3)) dx. The integration is now a standard term-by-term step.
  • Check: Verify the numerator expansion: (x-2)(x-3) - 5(x-1)(x-3) + 5(x-1)(x-2) = (x²-5x+6) - 5(x²-4x+3) + 5(x²-3x+2) = x² - 5x + 6 - 5x² + 20x - 15 + 5x² - 15x + 10 = x² + 1.
Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 77 — Complex Merger Equations

Technique explanation

This lesson extends the merger principle to Complex Merger Equations, which have two terms on each side where the total numerator sums match (A + B = C + D).

In a complex merger, pair terms with equal numerators across the equals sign. For example, moving terms in 3/x + 5/(x+1) = 3/(x+2) + 5/(x+3) gives 3(1/x - 1/(x+2)) + 5(1/(x+1) - 1/(x+3)) = 0. This creates a structural cancellation that drops the degree from three to two.

This method simplifies the solving process by reducing the equation to a quadratic. If the numerator sums do not match, the cubic terms will not cancel.

Standard fallback: Combining all rational terms onto one side of the equation, finding a common denominator for the entire expression, and solving the resulting high-degree polynomial equation.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Solve 3/x + 5/(x+1) = 3/(x+2) + 5/(x+3)

  • Step 1: Pair identical numerators: 3(1/x - 1/(x+2)) + 5(1/(x+1) - 1/(x+3)) = 0.
  • Step 2: Simplify: 3(2)/(x(x+2)) + 5(2)/((x+1)(x+3)) = 0 → 6/(x² + 2x) + 10/(x² + 4x + 3) = 0.
  • Step 3: 6(x² + 4x + 3) + 10(x² + 2x) = 0 → 16x² + 44x + 18 = 0 → 8x² + 22x + 9 = 0.
  • Step 4: (2x + 1)(4x + 9) = 0 → x = -1/2, -9/4.
  • Check: Substitute x = -1/2 into the LHS: 3/(-0.5) + 5/(0.5) = -6 + 10 = 4. RHS: 3/(-0.5+2) + 5/(-0.5+3) = 3/1.5 + 5/2.5 = 2 + 2 = 4. Both sides are equal.

Example 2: Solve 2/(x+1) + 1/(x+2) = 2/(x+4) + 1/(x+5)

  • Step 1: Pair identical numerators: 2(1/(x+1) - 1/(x+4)) + 1(1/(x+2) - 1/(x+5)) = 0.
  • Step 2: Simplify: 2(3)/((x+1)(x+4)) + 1(3)/((x+2)(x+5)) = 0 → 6/(x² + 5x + 4) + 3/(x² + 7x + 10) = 0.
  • Step 3: 2(x² + 7x + 10) + (x² + 5x + 4) = 0 → 3x² + 19x + 24 = 0.
  • Step 4: x = (-19 ± √73)/6.
  • Check: Verify the quadratic 3x² + 19x + 24 = 0 derived from 2(x² + 7x + 10) + (x² + 5x + 4) = 0. Its roots are x = (-19 ± √(361 - 288))/6 = (-19 ± √73)/6. Substituting these into the paired form confirms the zero sum.
Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 76 — Merger Equations: Reduce the Degree

Technique explanation

This lesson covers Merger Equations, utilizing the principle that when numerators sum to a common total, terms can be merged to drop the degree of the equation by one.

A merger occurs when the sum of numerators on the left equals the numerator on the right (A + B = C). By splitting the right-hand term and regrouping, we can cancel a common denominator factor. This transforms a cubic cross-multiplication into a linear equation.

This method applies to rational equations of the form A/(x+a) + B/(x+b) = C/(x+c). If the condition A + B = C is not met, the highest-degree terms will not cancel, and the merger will not work.

Standard fallback: Standard cross-multiplication of all rational terms, which typically results in a cubic equation, followed by the use of the rational root theorem or synthetic division to find solutions.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Solve 3/(x+1) + 5/(x+2) = 8/(x+3)

  • Step 1: Check 3 + 5 = 8. Correct.
  • Step 2: Calculate merged coefficients: A' = 3(3 - 1) = 6, B' = 5(3 - 2) = 5.
  • Step 3: Solve 6/(x+1) + 5/(x+2) = 0 → 6(x + 2) + 5(x + 1) = 0 → 11x + 17 = 0.
  • Step 4: x = -17/11.
  • Check: Substitute x = -17/11 into the LHS: 3/(-17/11 + 1) + 5/(-17/11 + 2) = 3/(-6/11) + 5/(5/11) = -33/6 + 11 = 5.5. RHS: 8/(-17/11 + 3) = 8/(16/11) = 88/16 = 5.5.

Example 2: Solve 1/x + 1/(x+1) + 3/(x+2) = 5/(x+3)

  • Step 1: Check 1 + 1 + 3 = 5. Correct.
  • Step 2: Merge against c = 3: A' = 1(3 - 0) = 3, B' = 1(3 - 1) = 2, C' = 3(3 - 2) = 3.
  • Step 3: Solve 3/x + 2/(x+1) + 3/(x+2) = 0 → 3(x+1)(x+2) + 2x(x+2) + 3x(x+1) = 0.
  • Step 4: 8x² + 16x + 6 = 0 → 4x² + 8x + 3 = 0 → (2x+1)(2x+3)=0 → x = -1/2, -3/2.
  • Check: Substitute x = -1/2 into the LHS: 1/(-0.5) + 1/(0.5) + 3/(1.5) = -2 + 2 + 2 = 2. RHS: 5/(-0.5+3) = 5/2.5 = 2. Both sides are equal.
Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 75 — Elimination and Retention in Multivariable Factorisation

Technique explanation

This lesson introduces the sub-sutra Lopanasthapanabhyam (By Elimination and Retention). It provides a systematic method for factorising homogeneous quadratics in three or more variables by temporarily eliminating variables.

To factorise a complex expression like 2x² + 6y² + 3z² + 7xy + 11yz + 7zx, follow a three-step process: First, set z = 0 and factorise the remaining x, y terms. Second, set y = 0 and factorise the x, z terms. Third, splice the results by matching the common variable terms (usually x). Finally, verify the coefficient of the unused term (the yz term).

This method applies to homogeneous quadratics where every term has the same total degree. If the variables don't match perfectly during splicing, you must scale the factorisations accordingly.

Standard fallback: The method of undetermined coefficients or grouping terms by degree and variable, applying the AC method or quadratic formula to each variable subset independently to find factors.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Factorise 2x² + 6y² + 3z² + 7xy + 11yz + 7zx

  • Step 1 (z=0): 2x² + 7xy + 6y² factors into (2x + 3y)(x + 2y).
  • Step 2 (y=0): 2x² + 7zx + 3z² factors into (2x + z)(x + 3z).
  • Step 3 (Splice): Match the 2x and x terms from both sets to get (2x + 3y + z)(x + 2y + 3z).
  • Check: Expand the factors (2x + 3y + z)(x + 2y + 3z) = 2x(x + 2y + 3z) + 3y(x + 2y + 3z) + z(x + 2y + 3z) = (2x² + 4xy + 6xz) + (3xy + 6y² + 9yz) + (xz + 2yz + 3z²) = 2x² + 6y² + 3z² + 7xy + 11yz + 7zx.

Example 2: Factorise 3x² + 5xy + 2y² + 4xz + 3yz + z²

  • Step 1 (z=0): 3x² + 5xy + 2y² factors into (3x + 2y)(x + y).
  • Step 2 (y=0): 3x² + 4xz + z² factors into (3x + z)(x + z).
  • Step 3 (Splice): Match the 3x and x terms to get (3x + 2y + z)(x + y + z).
  • Check: Expand the factors (3x + 2y + z)(x + y + z) = 3x(x + y + z) + 2y(x + y + z) + z(x + y + z) = (3x² + 3xy + 3xz) + (2xy + 2y² + 2yz) + (xz + yz + z²) = 3x² + 5xy + 2y² + 4xz + 3yz + z².
Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 74 — Solve by Observation, Then Verify

Technique explanation

This lesson utilizes the sub-sutra Vilokanam (By Mere Observation). It encourages recognizing structural patterns in equations that reveal roots without exhaustive computation.

Vilokanam is the discipline of looking before computing. Many complex-looking equations are built from simple components. For example, if x + 1/x = 13/6, we observe that 13/6 = 3/2 + 2/3. Since the structure matches x + 1/x, we can immediately infer x = 3/2 or x = 2/3.

This method is best suited for equations with clear structural patterns, such as reciprocal or additive structures. If no pattern is obvious, expansion into standard polynomial form is the fallback.

Standard fallback: Full algebraic expansion of the terms, finding a common denominator for all fractions, and solving the resulting polynomial equation through standard root-finding techniques.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Solve x + 1/x = 17/4

  • Step 1: Observe the structure x + 1/x.
  • Step 2: Decompose 17/4 into 4 + 1/4.
  • Step 3: Compare x + 1/x = 4 + 1/4.
  • Step 4: Read off x = 4 or x = 1/4.
  • Check: Verify using the sum and product of roots. For x² - (17/4)x + 1 = 0, the sum is 4 + 1/4 = 17/4 and the product is 4 * 1/4 = 1. Both match the coefficients of the quadratic form of the equation.

Example 2: Solve (x+3)/(x+4) + (x+5)/(x+6) = 2

  • Step 1: Observe the sum equals 2.
  • Step 2: Split 2 into 1 + 1 and move to each term: ((x+3)/(x+4) - 1) + ((x+5)/(x+6) - 1) = 0.
  • Step 3: Simplify: -1/(x+4) + -1/(x+6) = 0.
  • Step 4: Solve x+4 + x+6 = 0 → 2x = -10 → x = -5.
  • Check: Substitute x = -5 into the original equation: (-5+3)/(-5+4) + (-5+5)/(-5+6) = -2/-1 + 0/1 = 2 + 0 = 2. The result matches the right-hand side exactly.
Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 73 — Only the Last Terms: Common-Expression Equations

Technique explanation

This lesson teaches the application of the sub-sutra Antyayoreva (Only the Last Terms). It addresses equations where the leading portions of the numerator and denominator are identical or proportional, allowing the solution to be determined by the constant terms alone.

When an equation takes the form (P + a)/(P + b) = c/d, cross-multiplication yields d(P + a) = c(P + b). Expanding this gives dP + ad = cP + cb. Rearranging to solve for P results in P(d - c) = cb - ad, or P = (ad - bc)/(c - d). If a, b, c, d are such that the constants cancel or simplify significantly, the "common expression" P is directly solvable.

This method applies only when the variable parts (leading terms) of the numerator and denominator are identical (P) or can be scaled to be identical. For cases where they differ, the standard cross-multiplication method should be used.

Standard fallback: Cross-multiplication of the rational terms to form a standard polynomial equation, followed by collecting terms and solving the resulting quadratic or linear equation using the quadratic formula or factoring.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Solve (x² + 5x + 6)/(x² + 5x + 4) = 3/2

  • Step 1: Identify P = x² + 5x.
  • Step 2: Constants are a=6, b=4, c=3, d=2.
  • Step 3: Apply P = (ad - bc)/(c - d) = (62 - 43)/(3 - 2) = (12 - 12)/1 = 0.
  • Step 4: Solve x² + 5x = 0 → x(x + 5) = 0 → x = 0, -5.
  • Check: Calculate the standard products: 2(x² + 5x + 6) = 2x² + 10x + 12 and 3(x² + 5x + 4) = 3x² + 15x + 12. Equating them gives 2x² + 10x + 12 = 3x² + 15x + 12, which simplifies to x² + 5x = 0, yielding x = 0 and x = -5.

Example 2: Solve (x² + 3x + 5)/(x² + 3x + 2) = 2

  • Step 1: Identify P = x² + 3x.
  • Step 2: Constants are a=5, b=2, c=2, d=1.
  • Step 3: Apply P = (ad - bc)/(c - d) = (51 - 22)/(2 - 1) = (5 - 4)/1 = 1.
  • Step 4: Solve x² + 3x = 1 → x² + 3x - 1 = 0. Using the quadratic formula, x = (-3 ± √13)/2.
  • Check: Substitute x = (-3 ± √13)/2 back into the common expression P = x² + 3x. Calculation: ((-3 ± √13)/2)² + 3((-3 ± √13)/2) = (9 ± 6√13 + 13)/4 + (-18 ± 6√13)/4 = (22 ± 6√13 - 18 ± 6√13)/4 = 4/4 = 1. Then (1+5)/(1+2) = 6/3 = 2.
Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 72 — Reciprocal Equation Patterns: Ultimate and Penultimate Terms

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.

Module
Vedic Math — Reference Algebra and Equations
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
Vedic Math: Foundations to Reference Applications

Lesson 71 — Part and Whole: Symmetry in Biquadratic Forms

Technique explanation

The Vyashtisamanshtih sutra ("Part and whole") suggests viewing an equation's components in relation to a common structure. This is particularly effective for solving biquadratic equations by identifying symmetric parts and substituting for the "whole."

By pairing factors strategically, we can create common terms that allow for a substitution, reducing a 4th-degree equation to a much simpler quadratic form.

Method Condition: Only factors whose constant terms sum to the same value will produce a common term for substitution.

Standard fallback: Full expansion to a quartic equation followed by numerical methods or the Rational Root Theorem.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Factor Pairing

Solve: (x+1)(x+2)(x+3)(x+4) = 120

  • Step 1: Pair factors: [(x+1)(x+4)] * [(x+2)(x+3)] = 120 → (x² + 5x + 4)(x² + 5x + 6) = 120
  • Step 2: Substitute u = x² + 5x: (u+4)(u+6) = 120 → u² + 10u - 96 = 0
  • Step 3: Solve for u: (u+16)(u-6) = 0 → u = 6 (ignoring complex roots from u=-16)
  • Step 4: Solve for x: x² + 5x = 6 → x² + 5x - 6 = 0 → (x+6)(x-1) = 0 → x = 1, -6

Check: Substitute x=-6: (-6+1)(-6+2)(-6+3)(-6+4) = (-5)(-4)(-3)(-2) = 120. The product is verified.

Example 2: Symmetry by Averages

Solve: (x+2)³ / (x+4)³ = (x+1) / (x+5)

  • Step 1: Identify center: Averages of {2, 4} and {1, 5} are both 3.
  • Step 2: Substitute y = x+3: (y-1)³ / (y+1)³ = (y-2) / (y+2)
  • Step 3: Observe symmetry: The equation is odd-symmetric; y=0 is a root.
  • Step 4: Solve for x: x+3 = 0 → x = -3

Check: Substitute x=-3 into both sides: LHS = (-3+2)³/(-3+4)³ = (-1)³/1³ = -1; RHS = (-3+1)/(-3+5) = -2/2 = -1. Both sides are equal.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 70 — Quadratic Roots from a Derivative–Discriminant Relation

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.

Module
Vedic Math — Reference Algebra and Equations
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.

Academy Course
Vedic Math: Foundations to Reference Applications

Lesson 69 — Complete a Cubic or a Biquadratic into Structure

Technique explanation

The Puranapuranabhyam sutra ("By the completion or non-completion") is used to solve higher-degree equations by completing them into perfect powers or symmetric structures. This method transforms complex cubics and biquadratics into solvable forms through strategic addition or subtraction of terms.

For example, an equation that is "near" a perfect cube can be rewritten in terms of that cube, allowing for a simple substitution that reduces the complexity of the problem.

Method Condition: Requires the equation to be "near" a perfect power or symmetric form.

Standard fallback: Application of the Rational Root Theorem or synthetic division.

Module
Vedic Math — Reference Algebra and Equations
Worked examples

Example 1: Cubic Completion

Solve: x³ + 6x² + 11x + 6 = 0

  • Step 1: Identify nearest perfect cube: (x+2)³ = x³ + 6x² + 12x + 8
  • Step 2: Rewrite the equation: (x+2)³ - (x+2) = 0
  • Step 3: Substitute y = x+2: y³ - y = 0 → y(y-1)(y+1) = 0 → y = 0, 1, -1
  • Step 4: Solve for x: x+2=0 → x=-2; x+2=1 → x=-1; x+2=-1 → x=-3

Check: Expand (x+1)(x+2)(x+3) = (x²+3x+2)(x+3) = x³+3x²+3x²+9x+2x+6 = x³+6x²+11x+6. The expansion matches the original polynomial.

Example 2: Biquadratic Symmetry

Solve: (x+3)⁴ + (x+5)⁴ = 706

  • Step 1: Substitute average y = x+4: (y-1)⁴ + (y+1)⁴ = 706
  • Step 2: Expand and simplify: 2y⁴ + 12y² + 2 = 706 → y⁴ + 6y² - 352 = 0
  • Step 3: Solve as quadratic in y²: (y²+22)(y²-16) = 0 → y² = 16 → y = ±4
  • Step 4: Solve for x: x+4=4 → x=0; x+4=-4 → x=-8

Check: Substitute x=-8: (-8+3)⁴ + (-8+5)⁴ = (-5)⁴ + (-3)⁴ = 625 + 81 = 706. The solution is verified.

Academy Course
Vedic Math: Foundations to Reference Applications
Vedic Math — Reference Algebra and Equations