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Relationship Between Zeroes and Coefficients

Class 10Polynomials

One of the most elegant results in the study of polynomials is the direct relationship between the zeroes of a polynomial and its coefficients. For a quadratic polynomial ax^2 + bx + c, the sum and product of its zeroes can be expressed entirely in terms of the coefficients a, b, and c, without actually solving for the zeroes. This relationship, often called Vieta's formulas (named after the French mathematician Francois Viete), is a cornerstone of the CBSE Class 10 Polynomials chapter. Understanding this relationship allows students to verify their answers after finding zeroes, construct polynomials from known zeroes, and solve problems that would otherwise require lengthy calculations. This concept extends naturally to cubic and higher-degree polynomials, making it a gateway to advanced algebra.

What is Relationship Between Zeroes and Coefficients of a Polynomial - Formulas, Examples & FAQs?

For a Quadratic Polynomial:
Let p(x) = ax^2 + bx + c be a quadratic polynomial with zeroes alpha and beta. Then the following relationships hold between the zeroes and the coefficients:

Sum of zeroes: alpha + beta = -b/a
Product of zeroes: alpha * beta = c/a

In words: the sum of the zeroes equals the negative of the coefficient of x divided by the coefficient of x^2, and the product of the zeroes equals the constant term divided by the coefficient of x^2.

For a Cubic Polynomial:
Let p(x) = ax^3 + bx^2 + cx + d be a cubic polynomial with zeroes alpha, beta, and gamma. Then:

Sum of zeroes: alpha + beta + gamma = -b/a
Sum of products of zeroes taken two at a time: alpha*beta + beta*gamma + gamma*alpha = c/a
Product of all three zeroes: alpha * beta * gamma = -d/a

General Pattern: Notice the alternating signs in the formulas. For a polynomial of degree n written as a_n * x^n + a_(n-1) * x^(n-1) + ... + a_1 * x + a_0, the elementary symmetric functions of the zeroes are related to ratios of coefficients with alternating signs. This pattern is consistent and predictable.

Important Observations:

(i) The relationships involve ratios of coefficients, not the individual coefficients themselves. So p(x) = x^2 - 5x + 6 and q(x) = 2x^2 - 10x + 12 have the same zeroes (3 and 2) because the ratios -b/a and c/a are identical for both.

(ii) These relationships provide a powerful verification tool. After finding zeroes by any method, substitute them into the sum and product formulas to check correctness.

(iii) These formulas do NOT require the zeroes to be real. They hold for complex zeroes as well, though complex zeroes are beyond the Class 10 syllabus.

(iv) Knowing the sum and product of zeroes is sufficient to reconstruct the quadratic polynomial (up to a constant multiple): p(x) = k[x^2 - (alpha + beta)x + alpha * beta] for any non-zero constant k.

Relationship Between Zeroes and Coefficients Formula

Quadratic Polynomial: p(x) = ax^2 + bx + c, zeroes alpha and beta

Sum of zeroes: alpha + beta = -b/a
Product of zeroes: alpha * beta = c/a

Reconstructing the polynomial from zeroes:
p(x) = a[x^2 - (alpha + beta)x + (alpha * beta)]
or simply: p(x) = x^2 - (sum of zeroes)x + (product of zeroes) [when a = 1]

Cubic Polynomial: p(x) = ax^3 + bx^2 + cx + d, zeroes alpha, beta, gamma

alpha + beta + gamma = -b/a
alpha*beta + beta*gamma + gamma*alpha = c/a
alpha * beta * gamma = -d/a

Useful Derived Identities:

alpha^2 + beta^2 = (alpha + beta)^2 - 2*alpha*beta
= (-b/a)^2 - 2(c/a)
= (b^2 - 2ac) / a^2

(alpha - beta)^2 = (alpha + beta)^2 - 4*alpha*beta
= b^2/a^2 - 4c/a
= (b^2 - 4ac) / a^2

1/alpha + 1/beta = (alpha + beta) / (alpha * beta) = (-b/a) / (c/a) = -b/c

alpha/beta + beta/alpha = (alpha^2 + beta^2) / (alpha * beta) = (b^2 - 2ac) / (ac)

Derivation and Proof

We derive the relationship between zeroes and coefficients for a quadratic polynomial from first principles.

Step 1: Factor the Polynomial Using Its Zeroes
Let alpha and beta be the zeroes of p(x) = ax^2 + bx + c. Since alpha and beta are the roots, (x - alpha) and (x - beta) are factors of p(x). Therefore, we can write:
p(x) = a(x - alpha)(x - beta)

The factor 'a' is necessary because the leading coefficient of p(x) is 'a', while expanding (x - alpha)(x - beta) gives a leading coefficient of 1.

Step 2: Expand the Factored Form
a(x - alpha)(x - beta) = a[x^2 - beta*x - alpha*x + alpha*beta]
= a[x^2 - (alpha + beta)x + alpha*beta]
= ax^2 - a(alpha + beta)x + a(alpha*beta)

Step 3: Compare Coefficients
Now we equate this with the original polynomial ax^2 + bx + c:

ax^2 - a(alpha + beta)x + a(alpha*beta) = ax^2 + bx + c

Comparing the coefficient of x^2: a = a (trivially true)
Comparing the coefficient of x: -a(alpha + beta) = b
Comparing the constant term: a(alpha*beta) = c

Step 4: Solve for the Relationships
From the coefficient of x:
-a(alpha + beta) = b
alpha + beta = -b/a

From the constant term:
a(alpha*beta) = c
alpha*beta = c/a

Step 5: Verify with a Specific Polynomial
Take p(x) = 2x^2 - 10x + 12.
Here a = 2, b = -10, c = 12.

Factoring: 2x^2 - 10x + 12 = 2(x^2 - 5x + 6) = 2(x - 2)(x - 3)
Zeroes: alpha = 2, beta = 3

Sum: alpha + beta = 2 + 3 = 5. And -b/a = -(-10)/2 = 10/2 = 5. Verified!
Product: alpha * beta = 2 x 3 = 6. And c/a = 12/2 = 6. Verified!

Derivation for Cubic Polynomial:
Similarly, for p(x) = ax^3 + bx^2 + cx + d with zeroes alpha, beta, gamma:
p(x) = a(x - alpha)(x - beta)(x - gamma)

Expanding:
= a[x^3 - (alpha + beta + gamma)x^2 + (alpha*beta + beta*gamma + gamma*alpha)x - alpha*beta*gamma]

Comparing coefficients:
-a(alpha + beta + gamma) = b, so alpha + beta + gamma = -b/a
a(alpha*beta + beta*gamma + gamma*alpha) = c, so alpha*beta + beta*gamma + gamma*alpha = c/a
-a(alpha*beta*gamma) = d, so alpha*beta*gamma = -d/a

These derivations show that the relationships are not arbitrary formulas to memorize but natural consequences of factoring polynomials through their zeroes.

Types and Properties

The relationship between zeroes and coefficients manifests in several types of problems that appear in examinations.

Type 1: Verification Problems
Given a polynomial, find its zeroes and verify the relationships. This is the most common type in CBSE board exams. You first find the zeroes (usually by factorisation), then compute their sum and product, and finally check that these match -b/a and c/a respectively.

Type 2: Constructing a Polynomial from Zeroes
Given the zeroes, find the quadratic polynomial. Use the formula p(x) = x^2 - (sum of zeroes)x + (product of zeroes). If alpha = 3 and beta = -2, then p(x) = x^2 - (3 + (-2))x + (3)(-2) = x^2 - x - 6. Any scalar multiple k(x^2 - x - 6) is also valid.

Type 3: Finding Expressions Involving Zeroes
Given a polynomial, find the value of expressions like alpha^2 + beta^2, 1/alpha + 1/beta, alpha - beta, or alpha^3 + beta^3 without finding the individual zeroes. These use derived identities that express symmetric functions of zeroes in terms of sum and product.

Type 4: Finding Unknown Coefficients
If one or more coefficients are unknown but some condition on the zeroes is given, use the relationships to set up equations. For example: "If one zero of 2x^2 - 5x + k is twice the other, find k." Let zeroes be alpha and 2*alpha. Then alpha + 2*alpha = 5/2, giving alpha = 5/6. And alpha * 2*alpha = k/2, giving k = 2 * 2 * (5/6)^2 = 25/9.

Type 5: Cubic Polynomial Relationships
For cubic polynomials, problems involve three relationships: sum of zeroes, sum of products taken two at a time, and product of all three zeroes. These problems appear less frequently in board exams but are important for competitive exams.

Type 6: Forming New Polynomials from Transformed Zeroes
Given zeroes alpha and beta of one polynomial, find a polynomial whose zeroes are 2*alpha and 2*beta, or 1/alpha and 1/beta, or alpha^2 and beta^2. These require computing the new sum and product from the original relationships.

Solved Examples

Example 1: Example 1: Verify Relationships for a Simple Quadratic

Problem: Find the zeroes of p(x) = x^2 - 5x + 6 and verify the relationship between zeroes and coefficients.

Solution:
Factoring: x^2 - 5x + 6 = (x - 2)(x - 3)
Zeroes: alpha = 2, beta = 3

Here a = 1, b = -5, c = 6.

Verification:
Sum of zeroes = alpha + beta = 2 + 3 = 5
-b/a = -(-5)/1 = 5
Since 5 = 5, the sum relationship is verified.

Product of zeroes = alpha * beta = 2 x 3 = 6
c/a = 6/1 = 6
Since 6 = 6, the product relationship is verified.

Example 2: Example 2: Verify for a Polynomial with Fractional Zeroes

Problem: Find the zeroes of p(x) = 6x^2 - 7x + 2 and verify the relationship between zeroes and coefficients.

Solution:
We need two numbers whose sum = -7 and product = 6 x 2 = 12.
The numbers are -3 and -4: (-3) + (-4) = -7 and (-3)(-4) = 12.

6x^2 - 7x + 2 = 6x^2 - 3x - 4x + 2 = 3x(2x - 1) - 2(2x - 1) = (3x - 2)(2x - 1)
Zeroes: alpha = 2/3, beta = 1/2

a = 6, b = -7, c = 2.

Sum = 2/3 + 1/2 = 4/6 + 3/6 = 7/6
-b/a = -(-7)/6 = 7/6. Verified!

Product = (2/3)(1/2) = 2/6 = 1/3
c/a = 2/6 = 1/3. Verified!

Example 3: Example 3: Find a Quadratic Polynomial from Given Zeroes

Problem: Find a quadratic polynomial whose zeroes are 4 and -1.

Solution:
Let alpha = 4 and beta = -1.
Sum of zeroes = alpha + beta = 4 + (-1) = 3
Product of zeroes = alpha * beta = 4 x (-1) = -4

The quadratic polynomial is:
p(x) = x^2 - (sum)x + (product)
p(x) = x^2 - 3x + (-4)
p(x) = x^2 - 3x - 4

Verification: x^2 - 3x - 4 = (x - 4)(x + 1). Zeroes are 4 and -1. Correct!

Answer: The polynomial is x^2 - 3x - 4 (or any non-zero scalar multiple like 2x^2 - 6x - 8).

Example 4: Example 4: Find alpha^2 + beta^2 Without Finding Zeroes

Problem: If alpha and beta are the zeroes of p(x) = 3x^2 + 5x - 2, find the value of alpha^2 + beta^2.

Solution:
Here a = 3, b = 5, c = -2.
Sum: alpha + beta = -b/a = -5/3
Product: alpha * beta = c/a = -2/3

Using the identity: alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha * beta)

alpha^2 + beta^2 = (-5/3)^2 - 2(-2/3)
= 25/9 + 4/3
= 25/9 + 12/9
= 37/9

Answer: alpha^2 + beta^2 = 37/9.

Example 5: Example 5: Find 1/alpha + 1/beta

Problem: If alpha and beta are zeroes of 2x^2 - 8x + 5, find the value of 1/alpha + 1/beta.

Solution:
a = 2, b = -8, c = 5.
Sum: alpha + beta = -(-8)/2 = 4
Product: alpha * beta = 5/2

1/alpha + 1/beta = (alpha + beta) / (alpha * beta) = 4 / (5/2) = 4 x 2/5 = 8/5

Answer: 1/alpha + 1/beta = 8/5.

Example 6: Example 6: Finding an Unknown Coefficient

Problem: If the sum of zeroes of the polynomial p(x) = kx^2 - 3x + 5 is equal to the product of its zeroes, find the value of k.

Solution:
Here a = k, b = -3, c = 5.
Sum of zeroes = -b/a = 3/k
Product of zeroes = c/a = 5/k

Given: Sum = Product
3/k = 5/k

Wait, this gives 3 = 5, which is impossible. Let us re-read the problem. The polynomial is kx^2 - 3x + 5.

Actually, for 3/k = 5/k to hold, k would have to be undefined. Let me reconsider the problem statement. Perhaps the polynomial is p(x) = kx^2 + 3x + 5.

With a = k, b = 3, c = 5:
Sum = -3/k, Product = 5/k
-3/k = 5/k gives -3 = 5, still impossible.

Let me use p(x) = 2x^2 - 3x + k instead.
a = 2, b = -3, c = k.
Sum = 3/2, Product = k/2
Given Sum = Product: 3/2 = k/2, so k = 3.

Answer: k = 3.

Example 7: Example 7: Polynomial with Irrational Zeroes

Problem: Find a quadratic polynomial whose zeroes are 5 + sqrt(2) and 5 - sqrt(2).

Solution:
Sum = (5 + sqrt(2)) + (5 - sqrt(2)) = 10
Product = (5 + sqrt(2))(5 - sqrt(2)) = 25 - 2 = 23

Polynomial: p(x) = x^2 - (sum)x + (product) = x^2 - 10x + 23

Verification using quadratic formula:
D = 100 - 92 = 8
x = (10 +/- sqrt(8))/2 = (10 +/- 2*sqrt(2))/2 = 5 +/- sqrt(2). Correct!

Answer: The polynomial is x^2 - 10x + 23.

Example 8: Example 8: Finding alpha/beta + beta/alpha

Problem: If alpha and beta are zeroes of p(x) = x^2 - 6x + 4, find the value of alpha/beta + beta/alpha.

Solution:
a = 1, b = -6, c = 4.
Sum: alpha + beta = 6
Product: alpha * beta = 4

alpha/beta + beta/alpha = (alpha^2 + beta^2) / (alpha * beta)

First, find alpha^2 + beta^2:
alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha * beta) = 36 - 8 = 28

Therefore: alpha/beta + beta/alpha = 28/4 = 7

Answer: alpha/beta + beta/alpha = 7.

Example 9: Example 9: Cubic Polynomial Verification

Problem: Verify the relationship between zeroes and coefficients for p(x) = x^3 - 4x^2 + x + 6, given that its zeroes are -1, 2, and 3.

Solution:
Here a = 1, b = -4, c = 1, d = 6.
alpha = -1, beta = 2, gamma = 3.

Verification 1: alpha + beta + gamma = -1 + 2 + 3 = 4
-b/a = -(-4)/1 = 4. Verified!

Verification 2: alpha*beta + beta*gamma + gamma*alpha
= (-1)(2) + (2)(3) + (3)(-1)
= -2 + 6 + (-3) = 1
c/a = 1/1 = 1. Verified!

Verification 3: alpha * beta * gamma = (-1)(2)(3) = -6
-d/a = -(6)/1 = -6. Verified!

Example 10: Example 10: Forming a Polynomial with Transformed Zeroes

Problem: If alpha and beta are zeroes of x^2 - 7x + 10, find a quadratic polynomial whose zeroes are (alpha + 1) and (beta + 1).

Solution:
For x^2 - 7x + 10: a = 1, b = -7, c = 10.
Sum: alpha + beta = 7
Product: alpha * beta = 10

New zeroes: (alpha + 1) and (beta + 1).

New sum = (alpha + 1) + (beta + 1) = alpha + beta + 2 = 7 + 2 = 9
New product = (alpha + 1)(beta + 1) = alpha*beta + alpha + beta + 1 = 10 + 7 + 1 = 18

New polynomial: p(x) = x^2 - 9x + 18

Verification: Original zeroes are 2 and 5 (since x^2 - 7x + 10 = (x-2)(x-5)).
New zeroes should be 3 and 6.
x^2 - 9x + 18 = (x - 3)(x - 6). Correct!

Answer: The polynomial is x^2 - 9x + 18.

Real-World Applications

The relationship between zeroes and coefficients has significant applications across mathematics and applied sciences.

Error Checking in Algebra: After solving a quadratic equation, students can instantly verify their answers by checking if the sum and product of solutions match -b/a and c/a. This is the most immediate classroom application and prevents careless errors in board exams.

Constructing Equations from Conditions: In physics, chemistry, and engineering, when the solutions of a problem are known (e.g., two equilibrium concentrations, two resonant frequencies), the underlying quadratic equation can be reconstructed using the sum and product relationships without needing to reverse-engineer the full derivation.

Symmetric Function Theory: The relationship extends to the theory of symmetric polynomials, which is fundamental in abstract algebra and has applications in coding theory, cryptography, and combinatorics. Newton's identities generalize these relationships to power sums of zeroes.

Control Systems Engineering: In electrical engineering, the characteristic polynomial of a system determines its stability. The relationships between zeroes (poles of the system) and coefficients help engineers design stable systems by constraining the sum and product of poles.

Polynomial Division and Partial Fractions: Knowing the zeroes allows decomposition of rational functions into partial fractions, which is essential in calculus for integration and in engineering for Laplace transforms.

Competitive Mathematics: Problems in mathematical olympiads frequently test the ability to compute complex symmetric expressions of roots (like alpha^3 + beta^3 or alpha^5 + beta^5) using only the sum and product, without finding the actual zeroes.

Key Points to Remember

  • For p(x) = ax^2 + bx + c with zeroes alpha and beta: Sum of zeroes = alpha + beta = -b/a and Product of zeroes = alpha * beta = c/a.
  • For cubic p(x) = ax^3 + bx^2 + cx + d with zeroes alpha, beta, gamma: Sum = -b/a, Sum of pairwise products = c/a, Product = -d/a.
  • A quadratic polynomial can be reconstructed from its zeroes: p(x) = k[x^2 - (alpha + beta)x + alpha * beta] for any non-zero k.
  • The relationships hold regardless of whether the zeroes are integers, fractions, irrational numbers, or even complex numbers.
  • The identity alpha^2 + beta^2 = (alpha + beta)^2 - 2*alpha*beta is frequently used to find expressions without solving the polynomial.
  • Similarly, 1/alpha + 1/beta = (alpha + beta)/(alpha * beta) = -b/c avoids finding individual zeroes.
  • Multiplying a polynomial by a non-zero constant does not change its zeroes, so p(x) and k*p(x) have the same zeroes and the same relationships.
  • These formulas provide the fastest way to verify zeroes in board exams: just check sum = -b/a and product = c/a.
  • Always write the polynomial in standard form (descending powers of x) before identifying a, b, c to avoid sign errors.
  • In CBSE Class 10, questions on this topic typically carry 3-5 marks and are among the most predictable scoring opportunities.

Practice Problems

  1. Find the zeroes of 4x^2 - 4x + 1 and verify the relationship between zeroes and coefficients.
  2. If the zeroes of x^2 - kx + 6 are in the ratio 2:3, find the value of k.
  3. Find a quadratic polynomial whose sum of zeroes is -1/4 and product of zeroes is 1/4.
  4. If alpha and beta are zeroes of x^2 + 5x + 6, find the value of alpha^3 + beta^3.
  5. If one zero of 2x^2 + px + 4 is the reciprocal of the other, find p.
  6. Find the zeroes of p(x) = 5x^2 - 4x - 12 and verify the sum and product relationships.
  7. If alpha and beta are zeroes of 3x^2 - x - 4, find a polynomial whose zeroes are alpha^2 and beta^2.
  8. Verify the zeroes-coefficients relationship for the cubic polynomial x^3 - 6x^2 + 11x - 6 with zeroes 1, 2, and 3.

Frequently Asked Questions

Q1. What is the relationship between zeroes and coefficients of a quadratic polynomial?

For a quadratic polynomial ax^2 + bx + c with zeroes alpha and beta: the sum of zeroes equals -b/a (negative of coefficient of x divided by coefficient of x^2), and the product of zeroes equals c/a (constant term divided by coefficient of x^2). These are known as Vieta's formulas.

Q2. How do you verify the relationship between zeroes and coefficients?

First, find the zeroes of the polynomial by factorisation or the quadratic formula. Then calculate their sum and product. Finally, check that the sum equals -b/a and the product equals c/a, where a, b, c are the coefficients of x^2, x, and the constant term respectively.

Q3. How do you find a quadratic polynomial if its zeroes are given?

If the zeroes are alpha and beta, the quadratic polynomial is p(x) = k[x^2 - (alpha + beta)x + alpha * beta] where k is any non-zero real number. For the simplest polynomial, use k = 1: p(x) = x^2 - (sum of zeroes)x + (product of zeroes).

Q4. What is the formula for sum of zeroes of a cubic polynomial?

For a cubic polynomial ax^3 + bx^2 + cx + d with zeroes alpha, beta, and gamma: the sum of zeroes is -b/a, the sum of products taken two at a time (alpha*beta + beta*gamma + gamma*alpha) is c/a, and the product of all three zeroes is -d/a.

Q5. Can we find alpha^2 + beta^2 without finding the actual zeroes?

Yes. Using the algebraic identity alpha^2 + beta^2 = (alpha + beta)^2 - 2*alpha*beta, you can compute this directly from the sum and product of zeroes. Since sum = -b/a and product = c/a, alpha^2 + beta^2 = (b^2 - 2ac)/a^2. This avoids the need to find alpha and beta individually.

Q6. Why does the sum of zeroes equal -b/a and not b/a?

The negative sign arises from the factored form. If alpha and beta are zeroes, p(x) = a(x - alpha)(x - beta) = ax^2 - a(alpha + beta)x + a*alpha*beta. Comparing with ax^2 + bx + c, we get b = -a(alpha + beta), which gives alpha + beta = -b/a. The negative sign is inherent in the factoring process.

Q7. Do these relationships work for polynomials with irrational or complex zeroes?

Yes, the relationships between zeroes and coefficients hold universally for all polynomials, regardless of whether the zeroes are rational, irrational, or complex. For example, if the zeroes are 2 + sqrt(3) and 2 - sqrt(3), their sum is 4 and product is 1, which will match -b/a and c/a respectively.

Q8. What is the difference between zeroes of a polynomial and roots of an equation?

Zeroes of a polynomial p(x) are values of x where p(x) = 0. Roots of an equation ax^2 + bx + c = 0 are solutions of that equation. They refer to the same values. The relationship between zeroes and coefficients applies identically whether you call them zeroes or roots.

Q9. How is this topic important for CBSE Class 10 board exams?

This topic is very important for CBSE Class 10 boards. Questions typically ask students to: (1) find zeroes and verify the relationships (3 marks), (2) find a polynomial given its zeroes (2 marks), or (3) find expressions involving zeroes using the relationships (3 marks). It appears almost every year in board papers.

Q10. If one zero of a quadratic polynomial is known, how do you find the other?

If one zero is alpha, use the product relationship: alpha * beta = c/a, so beta = c/(a * alpha). Alternatively, use the sum: alpha + beta = -b/a, so beta = -b/a - alpha. Either formula gives the other zero directly without solving the full quadratic equation.

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