Sum and Product of Roots of a Quadratic Equation
The relationship between the roots and coefficients of a quadratic equation is one of the most elegant results in algebra. Given a quadratic equation ax2 + bx + c = 0, there exist simple formulae that relate the sum and product of its roots directly to the coefficients a, b, and c — without requiring us to actually find the roots themselves. These relationships, often called Vieta's formulae (named after the French mathematician Francois Viete), are incredibly powerful tools. They allow us to verify solutions, construct new equations from given roots, find expressions involving the roots without solving the equation, and solve a wide variety of examination problems efficiently. For CBSE Class 10 students, understanding the sum and product of roots deepens the comprehension of quadratic equations beyond mere mechanical solving. This topic regularly appears in board examinations, both as standalone questions and as parts of multi-step problems. In this comprehensive study, we will derive the formulae, explore their applications, and work through an extensive collection of examples to develop mastery over this important concept.
What is Sum and Product of Roots?
Let alpha (α) and beta (β) be the two roots of the quadratic equation ax2 + bx + c = 0 (where a ≠ 0). Then:
Sum of roots: α + β = -b/a
Product of roots: α × β = c/a
These formulae connect the roots of the equation to its coefficients without requiring the roots to be computed explicitly. They hold true regardless of whether the roots are rational, irrational, equal, or even complex.
The formulae arise from the fact that if α and β are roots of ax2 + bx + c = 0, then the quadratic can be factored as:
a(x - α)(x - β) = 0
Expanding: a[x2 - (α + β)x + αβ] = 0, which gives ax2 - a(α + β)x + aαβ = 0.
Comparing with ax2 + bx + c = 0:
- Coefficient of x: b = -a(α + β), so α + β = -b/a
- Constant term: c = aαβ, so αβ = c/a
These relationships are symmetric in α and β — the sum and product do not depend on which root is called α and which is called β.
Sum and Product of Roots of a Quadratic Equation Formula
Primary Formulae:
| Relationship | Formula |
|---|---|
| Sum of roots (α + β) | -b/a |
| Product of roots (αβ) | c/a |
Derived Expressions (frequently used):
| Expression | In terms of sum (S) and product (P) |
|---|---|
| α2 + β2 | (α + β)2 - 2αβ = S2 - 2P |
| (α - β)2 | (α + β)2 - 4αβ = S2 - 4P |
| |α - β| | √(S2 - 4P) = √D / |a| |
| α2 - β2 | (α + β)(α - β) |
| 1/α + 1/β | (α + β) / αβ = S/P |
| α/β + β/α | (α2 + β2) / αβ = (S2 - 2P)/P |
Forming a Quadratic from Given Roots:
If the roots are α and β, the quadratic equation is:
x2 - (α + β)x + αβ = 0
i.e., x2 - (sum)x + (product) = 0
Derivation and Proof
We derive the sum and product relationships from the quadratic formula.
Step 1: The roots of ax2 + bx + c = 0 are given by:
α = (-b + √(b2 - 4ac)) / (2a)
β = (-b - √(b2 - 4ac)) / (2a)
Step 2 — Sum of Roots:
α + β = [(-b + √D) / (2a)] + [(-b - √D) / (2a)]
= (-b + √D - b - √D) / (2a)
= -2b / (2a)
= -b/a
Notice that the √D terms cancel out. The sum of roots depends only on a and b, not on c directly through the discriminant.
Step 3 — Product of Roots:
αβ = [(-b + √D) / (2a)] × [(-b - √D) / (2a)]
Using the identity (m + n)(m - n) = m2 - n2 with m = -b and n = √D:
= [(-b)2 - (√D)2] / (2a)2
= [b2 - D] / (4a2)
= [b2 - (b2 - 4ac)] / (4a2)
= 4ac / (4a2)
= c/a
Both formulae are derived purely from the quadratic formula and hold for all quadratic equations, regardless of the nature of roots.
Methods
How to Use Sum and Product of Roots:
Type 1: Finding Sum and Product Given an Equation
Write the equation in standard form ax2 + bx + c = 0, identify a, b, c, and compute -b/a and c/a.
Type 2: Verifying Roots
If you have found roots α and β (by factorisation or formula), verify them by checking that α + β = -b/a and αβ = c/a.
Type 3: Finding Expressions Involving Roots Without Solving
To find expressions like α2 + β2, 1/α + 1/β, α3 + β3, etc., express them in terms of (α + β) and αβ using algebraic identities, then substitute the known sum and product.
Type 4: Forming a New Equation from Given Roots
If roots are given (say p and q), the equation is x2 - (p + q)x + pq = 0.
Type 5: Finding Unknown Coefficients
If one root is known, use the sum or product formula to find the other root or an unknown coefficient.
Type 6: Finding an Equation Whose Roots Are Related to Given Roots
If α, β are roots of a given equation, find a new equation whose roots are (say) 2α, 2β or α2, β2 or 1/α, 1/β. Compute the new sum and product, then form the equation.
Key Identities to Remember:
- (α + β)2 = α2 + 2αβ + β2
- α2 + β2 = (α + β)2 - 2αβ
- α3 + β3 = (α + β)3 - 3αβ(α + β)
- (α - β)2 = (α + β)2 - 4αβ
Solved Examples
Example 1: Finding Sum and Product from a Given Equation
Problem: Find the sum and product of the roots of 3x2 - 7x + 4 = 0.
Solution:
Here a = 3, b = -7, c = 4.
Sum of roots = -b/a = -(-7)/3 = 7/3
Product of roots = c/a = 4/3
Verification: Solving by factorisation: 3x2 - 3x - 4x + 4 = 0 → 3x(x-1) - 4(x-1) = 0 → (x-1)(3x-4) = 0. Roots are 1 and 4/3.
Sum = 1 + 4/3 = 7/3 ✓. Product = 1 × 4/3 = 4/3 ✓
Answer: Sum = 7/3, Product = 4/3.
Example 2: Finding a Specific Expression Without Solving
Problem: If α and β are roots of 2x2 + 3x - 5 = 0, find α2 + β2.
Solution:
Sum: α + β = -3/2. Product: αβ = -5/2.
Using the identity: α2 + β2 = (α + β)2 - 2αβ
= (-3/2)2 - 2(-5/2)
= 9/4 + 5
= 9/4 + 20/4
= 29/4
Answer: α2 + β2 = 29/4.
Example 3: Finding the Reciprocal Sum
Problem: If α and β are roots of x2 - 6x + 8 = 0, find 1/α + 1/β.
Solution:
Sum: α + β = 6. Product: αβ = 8.
1/α + 1/β = (α + β) / (αβ) = 6/8 = 3/4
Verification: Roots are 2 and 4. 1/2 + 1/4 = 3/4 ✓
Answer: 1/α + 1/β = 3/4.
Example 4: Finding the Unknown Coefficient Using One Root
Problem: If one root of x2 - 9x + k = 0 is 5, find k and the other root.
Solution:
Let the roots be 5 and β. Sum: 5 + β = 9 → β = 4.
Product: 5 × 4 = k → k = 20.
Verification: x2 - 9x + 20 = 0 → (x - 4)(x - 5) = 0 → x = 4, 5 ✓
Answer: k = 20 and the other root is 4.
Example 5: Forming a Quadratic Equation from Given Roots
Problem: Form the quadratic equation whose roots are 3 + √2 and 3 - √2.
Solution:
Sum = (3 + √2) + (3 - √2) = 6
Product = (3 + √2)(3 - √2) = 9 - 2 = 7
The equation is: x2 - (sum)x + (product) = 0
x2 - 6x + 7 = 0
Verification: D = 36 - 28 = 8 = (2√2)2. Roots = (6 ± 2√2)/2 = 3 ± √2 ✓
Answer: x2 - 6x + 7 = 0.
Example 6: Forming an Equation with Modified Roots
Problem: If α and β are roots of x2 - 5x + 6 = 0, form the equation whose roots are 2α and 2β.
Solution:
For the original equation: α + β = 5, αβ = 6.
New roots are 2α and 2β:
New sum = 2α + 2β = 2(α + β) = 2(5) = 10
New product = (2α)(2β) = 4αβ = 4(6) = 24
New equation: x2 - 10x + 24 = 0
Verification: Original roots are 2 and 3. New roots should be 4 and 6.
x2 - 10x + 24 = (x - 4)(x - 6) = 0 → x = 4, 6 ✓
Answer: x2 - 10x + 24 = 0.
Example 7: Finding the Difference of Roots
Problem: Without solving, find the absolute difference |α - β| for 2x2 - 7x + 3 = 0.
Solution:
Sum: α + β = 7/2. Product: αβ = 3/2.
(α - β)2 = (α + β)2 - 4αβ = (7/2)2 - 4(3/2) = 49/4 - 6 = 49/4 - 24/4 = 25/4
|α - β| = √(25/4) = 5/2
Verification: Roots are 3 and 1/2. |3 - 1/2| = 5/2 ✓
Answer: |α - β| = 5/2.
Example 8: Finding α/β + β/α
Problem: If α and β are roots of 3x2 + 2x - 1 = 0, find α/β + β/α.
Solution:
Sum: α + β = -2/3. Product: αβ = -1/3.
α/β + β/α = (α2 + β2) / αβ
First: α2 + β2 = (α + β)2 - 2αβ = 4/9 - 2(-1/3) = 4/9 + 2/3 = 4/9 + 6/9 = 10/9
Then: α/β + β/α = (10/9) / (-1/3) = (10/9) × (-3/1) = -30/9 = -10/3
Answer: α/β + β/α = -10/3.
Example 9: Equation Whose Roots Are Reciprocals of Given Roots
Problem: If α and β are roots of 4x2 - 3x + 7 = 0, form the equation whose roots are 1/α and 1/β.
Solution:
Original: α + β = 3/4, αβ = 7/4.
New sum: 1/α + 1/β = (α + β)/αβ = (3/4)/(7/4) = 3/7
New product: (1/α)(1/β) = 1/αβ = 4/7
New equation: x2 - (3/7)x + 4/7 = 0
Multiply by 7: 7x2 - 3x + 4 = 0
Note: This is the original equation with coefficients reversed: the original is 4x2 - 3x + 7, and the reciprocal-root equation is 7x2 - 3x + 4 (swap a and c).
Answer: 7x2 - 3x + 4 = 0.
Example 10: Finding α³ + β³ Without Solving
Problem: If α and β are roots of x2 - 4x + 2 = 0, find α3 + β3.
Solution:
Sum: α + β = 4. Product: αβ = 2.
Using the identity: α3 + β3 = (α + β)3 - 3αβ(α + β)
= (4)3 - 3(2)(4)
= 64 - 24
= 40
Answer: α3 + β3 = 40.
Real-World Applications
The sum and product of roots formulae have numerous applications in mathematics:
Quick Verification: After solving a quadratic equation, the fastest way to check correctness is to verify that the sum and product of your roots match -b/a and c/a respectively.
Constructing Equations: In many problems, the roots are described by conditions (e.g., 'roots differ by 3', 'one root is twice the other') rather than given explicitly. Sum and product formulae convert these conditions into solvable equations.
Higher Algebra: These relationships generalise to polynomials of any degree (Vieta's formulae). For a cubic ax3 + bx2 + cx + d = 0, the sum of roots is -b/a, the sum of products taken two at a time is c/a, and the product of all roots is -d/a.
Symmetric Functions: Any symmetric expression in the roots (one that does not change when α and β are swapped) can be expressed in terms of the sum and product. This is a powerful principle in algebra.
Coordinate Geometry: When finding intersection points of a line and a conic section, the sum and product of roots gives information about the midpoint and the product of distances, without finding the actual intersection points.
Physics: In oscillatory systems, the sum and product of roots of the characteristic equation relate to damping and natural frequency of the system.
Key Points to Remember
- For ax2 + bx + c = 0, the sum of roots α + β = -b/a and the product αβ = c/a.
- These formulae hold regardless of whether roots are rational, irrational, equal, or complex.
- A quadratic with roots α and β can be written as x2 - (α + β)x + αβ = 0, i.e., x2 - (sum)x + (product) = 0.
- Use algebraic identities to express α2 + β2, α3 + β3, 1/α + 1/β, etc., in terms of (α + β) and αβ. The key identities are: α2 + β2 = (α+β)2 - 2αβ and α3 + β3 = (α+β)3 - 3αβ(α+β).
- To verify roots, check that their sum = -b/a and product = c/a. This is faster than re-solving the equation.
- If one root is known, the other can be found using sum or product: other root = (-b/a) - known root, or other root = (c/a) / known root.
- For an equation whose roots are reciprocals of the original, swap a and c: if original is ax2 + bx + c = 0, the reciprocal-root equation is cx2 + bx + a = 0.
- The signs of the sum and product reveal the signs of the roots: if product > 0 and sum > 0, both roots are positive. If product > 0 and sum < 0, both roots are negative. If product < 0, roots have opposite signs.
- These relationships are a special case of Vieta's formulae for polynomials of any degree.
- The sum and product approach often avoids unnecessary computation — you can answer questions about the roots without finding the roots themselves.
- The sum of roots equals the negation of the ratio of the linear coefficient to the leading coefficient. This is a common source of sign errors — always remember the negative sign.
- When forming equations with modified roots (like 2α, 2β or α+1, β+1), compute the new sum and product from the original sum and product, then form the new equation.
Practice Problems
- Find the sum and product of roots of 5x² - 8x + 3 = 0. Verify by solving.
- If α and β are roots of x² + 7x + 12 = 0, find α² + β² and α³ + β³.
- One root of 2x² - kx + 6 = 0 is 3. Find k and the other root.
- Form the quadratic equation whose roots are 2 + √5 and 2 - √5.
- If α and β are roots of x² - 3x + 1 = 0, find α/β + β/α.
- If the roots of x² - px + q = 0 differ by 4, express p in terms of q.
Frequently Asked Questions
Q1. What are the sum and product of roots formulae?
For the quadratic equation ax² + bx + c = 0, the sum of roots is -b/a and the product of roots is c/a. These relate the roots to the coefficients directly.
Q2. Why is the sum of roots -b/a and not b/a?
The negative sign comes from the derivation. When we expand a(x - α)(x - β) = 0, the coefficient of x becomes -a(α + β). Setting this equal to b gives α + β = -b/a. The negative sign is essential and commonly a source of errors.
Q3. Do these formulae work when roots are irrational or complex?
Yes, these formulae work for all types of roots — rational, irrational, equal, or complex. They are derived from the general quadratic formula and hold universally.
Q4. How do I form a quadratic equation from two given roots?
If the roots are α and β, compute their sum S = α + β and product P = αβ. The equation is x² - Sx + P = 0. Multiply through to clear fractions if needed.
Q5. How do I find α² + β² without solving the equation?
Use the identity α² + β² = (α + β)² - 2αβ. Compute the sum and product from -b/a and c/a, then substitute.
Q6. Can I use these formulae to check my solutions?
Yes! After solving a quadratic equation, add your roots and check against -b/a. Multiply your roots and check against c/a. If either doesn't match, you have an error.
Q7. What if the product of roots is negative?
If αβ < 0, the roots have opposite signs (one positive, one negative). If αβ > 0, both roots have the same sign (both positive or both negative, determined by the sign of the sum).
Q8. What are Vieta's formulae?
Vieta's formulae generalize the sum and product of roots to polynomials of any degree. For a quadratic, they give sum = -b/a and product = c/a. For a cubic, they also give the sum of products of roots taken two at a time. The Class 10 formulae are the simplest case.
Q9. Is this topic important for CBSE exams?
Yes, questions on sum and product of roots appear regularly in CBSE Class 10 exams. They may ask you to find the sum/product, verify roots, form new equations, or find expressions involving roots.
Q10. How do I form an equation whose roots are double the roots of a given equation?
If α and β are roots of the original equation with sum S and product P, the new roots 2α and 2β have sum 2S and product 4P. The new equation is x² - 2Sx + 4P = 0.
Related Topics
- Quadratic Formula
- Nature of Roots of Quadratic Equation
- Relationship Between Zeroes and Coefficients
- Forming Quadratic Polynomial from Zeroes
- Quadratic Equations
- Standard Form of Quadratic Equation
- Solving Quadratic Equations by Factorisation
- Factorisation of Quadratic Expressions
- Completing the Square Method
- Discriminant of Quadratic Equation
- Word Problems on Quadratic Equations
- Applications of Quadratic Equations
- Roots and Graphs of Quadratic Equations
