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How to Use the Sridharacharya Formula to Solve Quadratic Equations

Updated: June 2, 2026

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Category: Advanced Math

“Mathematics is the language in which God has written the universe.” — Galileo

What Is the Sridharacharya Formula?

If you’ve ever stared at a quadratic equation and thought, “How on earth do I solve this?” — you’re not alone. The good news: a brilliant Indian mathematician solved that problem over 1,100 years ago.

The Sridharacharya formula (also written as Shreedhara Acharya’s formula or the quadratic formula) is a universal method to find the roots of any quadratic equation of the form:

ax² + bx + c = 0 → x = (−b ± √(b² − 4ac)) / 2a

Where a, b, and c are real number coefficients and a ≠ 0. The two values of x you get are called the roots (or solutions) of the equation.

Who Was Sridharacharya? (A Quick History Lesson)

Sridharacharya Information Table

c. 870–930 AD	Sridhara (Sridharacharya) was born in India. He was a mathematician, Sanskrit scholar, and philosopher.
Key Work	He introduced the method of "completing the square" to solve quadratic equations — the very foundation of what we call the Sridharacharya formula today.
Legacy	His formula is now a standard tool in algebra taught worldwide, referenced in textbooks from Class 10 all the way to competitive entrance exams like JEE.

Breaking Down the Formula-Every Part Explained

Symbol	Name	What It Does
a	Leading Coefficient	Determines the parabola's width. Must be ≠ 0.
b	Linear Coefficient	Influences where roots are on the number line.
c	Constant Term	The y-intercept of the parabola.
±	Plus or Minus	Gives TWO solutions — one with + and one with −.
b² − 4ac	The Discriminant (D)	The key to predicting the nature of roots (see table below).
√(b² − 4ac)	Square Root of D	The spread of roots from the centre −b/2a.
2a	Denominator	Normalises the result relative to the leading coefficient.

The Discriminant-Your Root Predictor

Before you even solve the equation, the discriminant D = b² − 4ac tells you exactly what kind of roots to expect:

Discriminant (D)	Condition	Nature of Roots
D > 0	b² − 4ac > 0	Two distinct real roots
D = 0	b² − 4ac = 0	Two equal real roots
D < 0	b² − 4ac < 0	No real roots (complex)

Sridharacharya Formula Proof (Step-by-Step Derivation)

The Sridharacharya formula proof is derived using the “completing the square” technique. Here’s every algebraic step, visualised clearly:

1

Start with the standard form

ax² + bx + c = 0 (where a ≠ 0)

2

Divide every term by a

x² + (b/a)x + c/a = 0

3

Move the constant to the right side

x² + (b/a)x = −c/a

4

Complete the square — add (b/2a)² to both sides

x² + (b/a)x + b²/4a² = b²/4a² − c/a

5

Write the left side as a perfect square

(x + b/2a)² = (b² − 4ac) / 4a²

6

Take the square root of both sides

x + b/2a = ± √(b² − 4ac) / 2a

7

Isolate x — we arrive at the formula!

x = (−b ± √(b² − 4ac)) / 2a ✓

The Sridharacharya Formula: x = (−b ± √(b² − 4ac)) / 2a

How to Use the Sridharacharya Formula-5 Simple Steps

ST
EP
1

Write in standard form

Rearrange your equation so it looks like ax² + bx + c = 0

ST
EP
2

Identify a, b, c

Read off the coefficients. Watch for negatives — they trip most students up!

ST
EP
3

Calculate the discriminant D

Work out D = b² − 4ac. This tells you how many real roots there are.

ST
EP
4

Plug into the formula

Substitute a, b, c and D into x = (−b ± √D) / 2a

ST
EP
5

Simplify both roots

Calculate x₁ (using +) and x₂ (using −) separately and simplify.

Sridharacharya Formula Examples-Fully Solved

Below are four sridharacharya formula with example problems, ranging from easy to challenging:

Example 1 (Easy): Solve x² + 5x + 6 = 0

Step 1 — Identify: a = 1, b = 5, c = 6

Step 2 — Discriminant: D = 5² − 4(1)(6) = 25 − 24 = 1

Step 3 — D > 0, so two distinct real roots exist.

Step 4 — Apply formula: x = (−5 ± √1) / 2(1) = (−5 ± 1) / 2

x₁ = (−5 + 1)/2 = −4/2 = −2

x₂ = (−5 − 1)/2 = −6/2 = −3

Roots: x = −2 and x = −3 ✓

Example 2 (Equal Roots): Solve x² − 4x + 4 = 0

Step 1 — Identify: a = 1, b = −4, c = 4

Step 2 — Discriminant: D = (−4)² − 4(1)(4) = 16 − 16 = 0

Step 3 — D = 0, so two equal real roots exist.

Step 4 — x = (−(−4) ± √0) / 2(1) = 4/2 = 2

Both roots: x = 2 (repeated root) ✓

Example 3 (Decimal Answer): Solve 3x² − x − 7 = 0

Step 1 — Identify: a = 3, b = −1, c = −7

Step 2 — Discriminant: D = (−1)² − 4(3)(−7) = 1 + 84 = 85

Step 3 — D > 0, two distinct real roots exist.

Step 4 — x = (1 ± √85) / 6 = (1 ± 9.22) / 6

x₁ = (1 + 9.22)/6 ≈ 1.70

x₂ = (1 − 9.22)/6 ≈ −1.37

Roots: x ≈ 1.70 and x ≈ −1.37 ✓

Example 4 (No Real Roots): Solve x² + x + 1 = 0

Step 1 — Identify: a = 1, b = 1, c = 1

Step 2 — Discriminant: D = 1² − 4(1)(1) = 1 − 4 = −3

Step 3 — D < 0, so there are no real roots.

The equation has two complex (imaginary) roots. This is important for higher-level maths.

x = (−1 ± i√3) / 2 → complex numbers ✓

5 Common Mistakes to Avoid

Common Mistakes and Fixes Table

#	Mistake	Fix
1	Forgetting that −b means negate b (sign error)	Always write out the sign of b explicitly before substituting.
2	Squaring b incorrectly when b is negative	(−3)² = 9, not −9. Bracket negative values before squaring.
3	Only computing one root (forgetting the ± )	Always solve for both x₁ (+) and x₂ (−) separately.
4	Not rearranging to standard form first	Move all terms to one side: ax² + bx + c = 0 before starting.
5	Dividing only the numerator by 2a, not the full expression	The entire (−b ± √D) is divided by 2a — use parentheses!

Real-World Applications of the Sridharacharya Formula

The Sridharacharya formula isn’t just an exam tool — it’s actively used across industries:

Sridharacharya Formula Applications Table

Field	How Sridharacharya Formula Helps
Physics — Projectile Motion	Find when a ball hits the ground: h = ut - ½gt²
Engineering — Bridge Design	Calculate stress curves in parabolic arches
Finance — Break-even Analysis	Solve for units when revenue = cost² + cost
Computer Graphics	Compute curve intersections in 2D/3D rendering
Agriculture	Optimise rectangular field area given a fixed perimeter

Quick Reference Card

SRIDHARACHARYA FORMULA

x = (−b ± √(b² − 4ac)) / 2a

For equation: ax² + bx + c = 0 (a ≠ 0)

D > 0 → 2 real roots D = 0 → 1 repeated root D < 0 → no real roots

Conclusion

The Sridharacharya formula is one of the most powerful and elegant tools in mathematics. Whether you’re solving simple classroom problems or applying it to real-world engineering and physics challenges, mastering this formula opens up a wide world of problem-solving.

Remember the three key steps: identify your coefficients, check the discriminant, and apply the formula carefully. With the Sridharacharya formula with example walkthroughs above and the proof in your toolkit, you’re fully equipped for exams and beyond.

You can opt for our Advanced Math or Vedic Math+Mental Math courses. Our Math Quiz for grades 3rd, 4th, 5th, and 6th helps in further exciting and engaging in mathematics with hands-on lessons.

Frequently Asked Questions (FAQs)

Sridharacharya Formula FAQ

Q1: What is the Sridharacharya formula used for?

A: It is used to find the roots (solutions) of any quadratic equation of the form ax² + bx + c = 0. It works even when factorisation is difficult or impossible.

Q2: Is the Sridharacharya formula the same as the quadratic formula?

A: Yes — they are the same formula. 'Sridharacharya formula' is the name used in the Indian curriculum, honouring the mathematician Sridhara, who first derived it around the 9th century.

Q3: What happens when the discriminant is negative?

A: When b² – 4ac < 0, the square root is of a negative number, meaning there are no real roots. The equation has two complex (imaginary) solutions.

Q4: Can the Sridharacharya formula solve all quadratic equations?

A: Yes — unlike factorisation (which only works cleanly for certain equations), the Sridharacharya formula works for all quadratic equations, including those with irrational or complex roots.

Q5: How do I remember the Sridharacharya formula easily?

A: A popular mnemonic: 'Negative b, plus or minus the square root of b squared minus 4ac, all over 2a.' Writing it out a few times while doing practice problems is the fastest way to memorise it.

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