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Sum and Product of Zeroes

Class 10Polynomials

Polynomials are fundamental building blocks in algebra, and understanding their zeroes is crucial for solving equations and analysing functions. The zeroes (or roots) of a polynomial are the values of the variable for which the polynomial equals zero. In Class 10, we study a powerful 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 directly determined from the coefficients a, b, and c without actually finding the zeroes. This relationship, derived from the factor theorem, provides a shortcut for verifying roots, forming polynomials, and understanding the structure of quadratic expressions. This concept extends to cubic polynomials as well, making it one of the most versatile tools in the Polynomials chapter.

What is Sum and Product of Zeroes of a Polynomial?

Let us consider a polynomial and define its zeroes and the relationships between them.

Zeroes of a Polynomial: A zero (or root) of a polynomial p(x) is a value of x, say x = k, such that p(k) = 0. Geometrically, the zeroes are the x-coordinates of the points where the graph of the polynomial intersects the x-axis.

For a Quadratic Polynomial ax^2 + bx + c (where a is not equal to 0):

If alpha and beta are the two zeroes of the polynomial, then:

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

Product of zeroes: alpha * beta = c/a

These relationships mean that you can find the sum and product of the roots of any quadratic polynomial just by looking at the coefficients, without solving the equation.

For a Cubic Polynomial ax^3 + bx^2 + cx + d (where a is not equal to 0):

If alpha, beta, and gamma are the three zeroes of the polynomial, then:

Sum of zeroes: alpha + beta + gamma = -b/a

Sum of products taken two at a time: alpha*beta + beta*gamma + gamma*alpha = c/a

Product of zeroes: alpha * beta * gamma = -d/a

These are known as Vieta's formulas (named after the French mathematician Francois Viete). They establish a direct connection between the roots and coefficients of any polynomial. Note the alternating signs: -b/a, +c/a, -d/a.

For a Linear Polynomial ax + b: The single zero is x = -b/a. There is no sum or product relationship since there is only one zero.

Sum and Product of Zeroes Formula

Quadratic Polynomial: ax^2 + bx + c

If alpha and beta are the zeroes, then:

alpha + beta = -b/a (Sum of zeroes = negative of coefficient of x divided by coefficient of x^2)

alpha * beta = c/a (Product of zeroes = constant term divided by coefficient of x^2)

Cubic Polynomial: ax^3 + bx^2 + cx + d

If alpha, beta, and gamma are the zeroes, then:

alpha + beta + gamma = -b/a

alpha*beta + beta*gamma + gamma*alpha = c/a

alpha * beta * gamma = -d/a

Memory Aid: For a quadratic ax^2 + bx + c with zeroes alpha and beta:

Sum = -(coefficient of x) / (coefficient of x^2) = -b/a

Product = (constant term) / (coefficient of x^2) = c/a

Derivation and Proof

The relationship between zeroes and coefficients can be derived by comparing two forms of the same polynomial.

Derivation for Quadratic Polynomial:

Let p(x) = ax^2 + bx + c be a quadratic polynomial with zeroes alpha and beta.

Since alpha and beta are zeroes, (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 needed because the leading coefficient of p(x) is a, not 1.

Now, let us expand the right side:

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)

Comparing coefficients with ax^2 + bx + c:

Coefficient of x^2: a = a (consistent)

Coefficient of x: b = -a(alpha + beta), which gives alpha + beta = -b/a

Constant term: c = a(alpha*beta), which gives alpha*beta = c/a

This completes the derivation for quadratic polynomials.

Derivation for Cubic Polynomial:

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

Since alpha, beta, and gamma are zeroes:

p(x) = a(x - alpha)(x - beta)(x - gamma)

Expanding step by step:

First, (x - alpha)(x - beta) = x^2 - (alpha + beta)x + alpha*beta

Then, multiply by (x - gamma):

= x^3 - gamma*x^2 - (alpha + beta)x^2 + (alpha + beta)*gamma*x + alpha*beta*x - alpha*beta*gamma

= x^3 - (alpha + beta + gamma)x^2 + (alpha*beta + beta*gamma + gamma*alpha)x - alpha*beta*gamma

Including the leading coefficient a:

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

Comparing with ax^3 + bx^2 + cx + d:

b = -a(alpha + beta + gamma), so alpha + beta + gamma = -b/a

c = a(alpha*beta + beta*gamma + gamma*alpha), so alpha*beta + beta*gamma + gamma*alpha = c/a

d = -a(alpha*beta*gamma), so alpha*beta*gamma = -d/a

These derivations show that the formulas arise naturally from expanding the factored form of the polynomial.

Types and Properties

The sum and product relationships apply to different types of polynomials based on their degree:

1. Linear Polynomial (Degree 1): ax + b has exactly one zero: x = -b/a. There is no sum or product relationship as there is only a single zero.

2. Quadratic Polynomial (Degree 2): ax^2 + bx + c can have at most 2 zeroes. The sum and product of these two zeroes are related to the coefficients by the formulas alpha + beta = -b/a and alpha * beta = c/a. The zeroes may be real and distinct, real and equal, or complex (not studied in Class 10).

3. Cubic Polynomial (Degree 3): ax^3 + bx^2 + cx + d can have at most 3 zeroes. The three symmetric functions of the zeroes (sum, sum of pairwise products, and product) are related to the coefficients. At least one zero is always real for a cubic polynomial with real coefficients.

Special Cases:

  • If one zero of a quadratic is 0, then the product of zeroes is 0, which means c = 0. The polynomial becomes ax^2 + bx.
  • If both zeroes are equal (say alpha = beta), then the sum = 2*alpha = -b/a and the product = alpha^2 = c/a.
  • If the zeroes are reciprocals of each other (alpha = 1/beta), then their product = 1, which means c/a = 1 or c = a.
  • If the zeroes are negatives of each other (alpha = -beta), then their sum = 0, which means -b/a = 0, so b = 0.

Methods

Method 1: Finding Sum and Product from a Given Polynomial

Given a polynomial, identify the coefficients and apply the formulas directly.

For ax^2 + bx + c: Sum = -b/a, Product = c/a.

This method does not require finding the actual zeroes.

Method 2: Verification of Zeroes

If you are given specific zeroes and a polynomial, verify by checking:

(a) Each zero satisfies the polynomial (substitution gives 0).

(b) The sum of given zeroes equals -b/a.

(c) The product of given zeroes equals c/a.

Method 3: Finding Remaining Zeroes When Some Are Known

If one zero of a quadratic is known (say alpha), the other zero beta can be found using:

beta = -b/a - alpha (from the sum formula), or beta = c/(a * alpha) (from the product formula).

Method 4: Forming a Polynomial from Its Zeroes

If the zeroes alpha and beta are given, the quadratic polynomial is:

p(x) = k[x^2 - (alpha + beta)x + alpha*beta] where k is any non-zero constant.

This method is covered in detail in the separate topic 'Forming Quadratic Polynomial from Zeroes'.

Solved Examples

Example 1: Example 1: Find the sum and product of zeroes of 6x^2 - 7x - 3

Problem: Find the sum and product of zeroes of the polynomial p(x) = 6x^2 - 7x - 3.

Solution:

Comparing with ax^2 + bx + c: a = 6, b = -7, c = -3.

Sum of zeroes = -b/a = -(-7)/6 = 7/6.

Product of zeroes = c/a = -3/6 = -1/2.

Verification: Factoring, 6x^2 - 7x - 3 = (2x - 3)(3x + 1). Zeroes are x = 3/2 and x = -1/3.

Sum = 3/2 + (-1/3) = 9/6 - 2/6 = 7/6. Matches -b/a.

Product = (3/2)(-1/3) = -3/6 = -1/2. Matches c/a.

Answer: Sum = 7/6, Product = -1/2.

Example 2: Example 2: Find sum and product of zeroes of 4x^2 + 12x + 9

Problem: Find the sum and product of zeroes of 4x^2 + 12x + 9. Also find the zeroes.

Solution:

Here a = 4, b = 12, c = 9.

Sum of zeroes = -b/a = -12/4 = -3.

Product of zeroes = c/a = 9/4.

To find the zeroes, note that 4x^2 + 12x + 9 = (2x + 3)^2.

Setting (2x + 3)^2 = 0, we get 2x + 3 = 0, so x = -3/2.

Both zeroes are equal: alpha = beta = -3/2.

Verification: Sum = -3/2 + (-3/2) = -3. Product = (-3/2)(-3/2) = 9/4. Both match.

Answer: Sum = -3, Product = 9/4. Both zeroes are -3/2 (a repeated root).

Example 3: Example 3: Find a polynomial whose sum and product of zeroes are given

Problem: Find a quadratic polynomial whose sum of zeroes is 1/4 and product of zeroes is -1.

Solution:

Using the relation p(x) = k[x^2 - (sum of zeroes)x + (product of zeroes)]:

p(x) = k[x^2 - (1/4)x + (-1)] = k[x^2 - x/4 - 1]

To eliminate the fraction, let k = 4:

p(x) = 4x^2 - x - 4.

Verification: For 4x^2 - x - 4, a = 4, b = -1, c = -4.

Sum = -b/a = -(-1)/4 = 1/4. Correct.

Product = c/a = -4/4 = -1. Correct.

Answer: One such polynomial is 4x^2 - x - 4. (Any non-zero scalar multiple is also valid.)

Example 4: Example 4: Verify zeroes and coefficient relationships for a cubic polynomial

Problem: Verify the relationship between zeroes and coefficients for the polynomial p(x) = 2x^3 + x^2 - 5x + 2, given that its zeroes are 1, -2, and 1/2.

Solution:

Here a = 2, b = 1, c = -5, d = 2. Zeroes: alpha = 1, beta = -2, gamma = 1/2.

Verification 1: Check that these are actually zeroes.

p(1) = 2(1) + 1 - 5 + 2 = 0. Correct.

p(-2) = 2(-8) + 4 + 10 + 2 = -16 + 16 = 0. Correct.

p(1/2) = 2(1/8) + 1/4 - 5/2 + 2 = 1/4 + 1/4 - 5/2 + 2 = 1/2 - 5/2 + 2 = -2 + 2 = 0. Correct.

Verification 2: Sum of zeroes = 1 + (-2) + 1/2 = -1/2. And -b/a = -1/2. Matches.

Verification 3: Sum of products taken two at a time = (1)(-2) + (-2)(1/2) + (1/2)(1) = -2 + (-1) + 1/2 = -5/2. And c/a = -5/2. Matches.

Verification 4: Product of zeroes = (1)(-2)(1/2) = -1. And -d/a = -2/2 = -1. Matches.

Answer: All three relationships are verified.

Example 5: Example 5: Find the zeroes given the sum, product, and one zero

Problem: If one zero of 3x^2 - 8x + k is 2, find the value of k and the other zero.

Solution:

Since x = 2 is a zero of 3x^2 - 8x + k:

p(2) = 3(4) - 8(2) + k = 0

12 - 16 + k = 0

k = 4.

The polynomial is 3x^2 - 8x + 4 with a = 3, b = -8, c = 4.

Sum of zeroes = -b/a = 8/3.

Since one zero is 2, the other zero = 8/3 - 2 = 8/3 - 6/3 = 2/3.

Verification: Product of zeroes = 2 * (2/3) = 4/3. And c/a = 4/3. Matches.

Answer: k = 4, and the other zero is 2/3.

Example 6: Example 6: Find a quadratic polynomial with zeroes that are reciprocals

Problem: If the zeroes of a quadratic polynomial are 3 and 1/3, find the polynomial.

Solution:

Sum of zeroes = 3 + 1/3 = 10/3.

Product of zeroes = 3 * (1/3) = 1.

The polynomial is p(x) = k[x^2 - (10/3)x + 1].

Taking k = 3 to clear the fraction: p(x) = 3x^2 - 10x + 3.

Verification: 3x^2 - 10x + 3 = (3x - 1)(x - 3). Zeroes: x = 1/3 and x = 3. Correct.

Note: Since the product of zeroes is 1, the zeroes are reciprocals of each other. This means c = a in the polynomial (both are 3 here).

Answer: 3x^2 - 10x + 3.

Example 7: Example 7: Relationship between zeroes for a special condition

Problem: If the zeroes of the polynomial x^2 - px + q are such that one zero is double the other, express p^2 in terms of q.

Solution:

Let the zeroes be alpha and 2*alpha.

From the polynomial x^2 - px + q (here a = 1, b = -p, c = q):

Sum of zeroes: alpha + 2*alpha = -(-p)/1 = p. So 3*alpha = p, giving alpha = p/3.

Product of zeroes: alpha * 2*alpha = q/1 = q. So 2*alpha^2 = q.

Substituting alpha = p/3 into 2*alpha^2 = q:

2(p/3)^2 = q

2 * p^2/9 = q

2p^2 = 9q

p^2 = 9q/2.

Answer: p^2 = 9q/2.

Example 8: Example 8: Sum and product for a polynomial with irrational zeroes

Problem: Find the sum and product of zeroes of x^2 - 3. Are the zeroes rational or irrational?

Solution:

Writing x^2 - 3 as x^2 + 0*x + (-3): a = 1, b = 0, c = -3.

Sum of zeroes = -b/a = 0/1 = 0.

Product of zeroes = c/a = -3/1 = -3.

Finding the zeroes: x^2 - 3 = 0, so x^2 = 3, giving x = +sqrt(3) and x = -sqrt(3).

Verification: Sum = sqrt(3) + (-sqrt(3)) = 0. Correct.

Product = sqrt(3) * (-sqrt(3)) = -3. Correct.

Both zeroes are irrational. Note that irrational zeroes of a quadratic with rational coefficients always occur in conjugate pairs (one positive, one negative of the same surd).

Answer: Sum = 0, Product = -3. The zeroes are sqrt(3) and -sqrt(3), both irrational.

Example 9: Example 9: Find k if one zero of kx^2 - 5x + 1 is reciprocal of the other

Problem: For what value of k will the zeroes of the polynomial kx^2 - 5x + 1 be reciprocals of each other?

Solution:

Let the zeroes be alpha and 1/alpha.

Product of zeroes = alpha * (1/alpha) = 1.

From the polynomial: Product of zeroes = c/a = 1/k.

Setting these equal: 1/k = 1, so k = 1.

Verification: With k = 1, the polynomial is x^2 - 5x + 1. Sum of zeroes = 5, Product = 1.

Using the quadratic formula: x = (5 + sqrt(21))/2 and x = (5 - sqrt(21))/2.

Product = [(5 + sqrt(21))/2] * [(5 - sqrt(21))/2] = (25 - 21)/4 = 4/4 = 1. Confirmed, they are reciprocals.

Answer: k = 1.

Example 10: Example 10: Find the zeroes and verify for a cubic polynomial

Problem: Find all zeroes of x^3 - 4x^2 + x + 6 and verify the sum-product relationships.

Solution:

Try x = -1: (-1)^3 - 4(-1)^2 + (-1) + 6 = -1 - 4 - 1 + 6 = 0. So x = -1 is a zero.

Divide x^3 - 4x^2 + x + 6 by (x + 1) using synthetic division or long division:

x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6) = (x + 1)(x - 2)(x - 3).

The zeroes are -1, 2, and 3.

Here a = 1, b = -4, c = 1, d = 6.

Verification:

Sum = -1 + 2 + 3 = 4. And -b/a = -(-4)/1 = 4. Matches.

Sum of products two at a time = (-1)(2) + (2)(3) + (3)(-1) = -2 + 6 - 3 = 1. And c/a = 1/1 = 1. Matches.

Product = (-1)(2)(3) = -6. And -d/a = -6/1 = -6. Matches.

Answer: Zeroes are -1, 2, 3. All relationships verified.

Real-World Applications

The sum and product of zeroes relationships have widespread applications both within mathematics and in real-world problem solving.

Forming Polynomials: When the roots of an equation are known (from experimental data, for instance), the polynomial can be reconstructed using the sum and product relationships. This is used in curve fitting and interpolation in data science.

Solving Systems of Equations: Many algebraic problems involve relationships between roots (e.g., one root is double the other, or roots are reciprocals). The sum and product formulas convert these into equations in the coefficients, making the problems solvable without finding the actual roots.

Physics and Engineering: In circuit analysis, the roots of the characteristic equation of a circuit determine its behaviour (oscillating, decaying, etc.). The sum and product of roots relate to physical parameters like resistance, capacitance, and inductance.

Competitive Examinations: JEE, NTSE, and Olympiad problems frequently test the ability to use Vieta's formulas for higher-degree polynomials and to derive relationships between roots and coefficients.

Error Checking: When solving a quadratic equation, you can verify your answers by checking that their sum equals -b/a and their product equals c/a. This is a quick and reliable verification method.

Key Points to Remember

  • For a quadratic polynomial ax^2 + bx + c with zeroes alpha and beta: Sum = -b/a, Product = c/a.
  • For a cubic polynomial ax^3 + bx^2 + cx + d with zeroes alpha, beta, gamma: Sum = -b/a, Sum of pairwise products = c/a, Product = -d/a.
  • These relationships are called Vieta's formulas.
  • The relationships hold regardless of whether the zeroes are rational, irrational, or complex.
  • If zeroes are equal (repeated root), the polynomial is a perfect square: a(x - alpha)^2.
  • If zeroes are reciprocals, then c = a in the quadratic polynomial.
  • If zeroes are negatives of each other, then b = 0 (the coefficient of x is zero).
  • The formulas can be derived by expanding the factored form a(x - alpha)(x - beta) and comparing coefficients.
  • A quadratic polynomial can be written as k[x^2 - (sum)x + (product)] for any non-zero constant k.
  • Always verify your answers by checking both the sum and product relationships.

Practice Problems

  1. Find the sum and product of zeroes of 5x^2 + 13x - 6 and verify by finding the actual zeroes.
  2. If the zeroes of x^2 + px + 12 are in the ratio 1:3, find the value of p.
  3. Find a quadratic polynomial whose zeroes are (3 + sqrt(5)) and (3 - sqrt(5)).
  4. If alpha and beta are zeroes of 2x^2 - 7x + 3, find the value of alpha^2 + beta^2.
  5. One zero of 3x^2 + kx - 2 is 1/3. Find k and the other zero.
  6. If the sum of zeroes of kx^2 + 2x + 3k is equal to their product, find the value of k.
  7. Find the zeroes of 2x^3 - x^2 - 13x - 6 and verify all three coefficient relationships.
  8. For what values of k does kx^2 - 6x + 1 have zeroes that are equal? Find the equal zeroes.

Frequently Asked Questions

Q1. What are zeroes of a polynomial?

Zeroes (or roots) of a polynomial p(x) are the values of x for which p(x) = 0. For example, the zeroes of x^2 - 5x + 6 are x = 2 and x = 3, because p(2) = 4 - 10 + 6 = 0 and p(3) = 9 - 15 + 6 = 0. Geometrically, zeroes are the x-coordinates where the graph of the polynomial crosses the x-axis.

Q2. 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 (alpha + beta) equals -b/a, and the product of zeroes (alpha * beta) equals c/a. These relationships allow you to determine the sum and product without actually finding the individual zeroes.

Q3. Can a quadratic polynomial have only one zero?

A quadratic polynomial always has exactly two zeroes (counting multiplicity). However, both zeroes can be equal (a repeated root). For example, x^2 - 4x + 4 = (x - 2)^2 has two equal zeroes, both equal to 2. In some cases, the zeroes may be complex numbers (not real), which are not studied in Class 10.

Q4. What happens if the coefficient of x is zero in a quadratic?

If b = 0 in ax^2 + bx + c, then the sum of zeroes is -b/a = 0. This means the zeroes are negatives of each other. For example, x^2 - 9 = (x - 3)(x + 3) has zeroes 3 and -3, which sum to 0.

Q5. How do you verify if given values are the correct zeroes of a polynomial?

Two methods: (1) Substitute each value into the polynomial and check that the result is zero. (2) Check that the sum and product of the given values match -b/a and c/a respectively. If both conditions are satisfied, the values are correct zeroes.

Q6. What are Vieta's formulas?

Vieta's formulas (named after Francois Viete) express the relationships between the roots of a polynomial and its coefficients. For a quadratic ax^2 + bx + c: sum = -b/a, product = c/a. For a cubic ax^3 + bx^2 + cx + d: sum = -b/a, sum of pairwise products = c/a, product = -d/a. These extend to polynomials of any degree.

Q7. Can the sum of zeroes be zero?

Yes. The sum of zeroes is -b/a, so if b = 0, the sum is zero. This occurs when the zeroes are negatives of each other. For example, x^2 - 4 has zeroes 2 and -2 with sum = 0. Similarly, x^2 - 7 has zeroes sqrt(7) and -sqrt(7) with sum = 0.

Q8. Can the product of zeroes be negative?

Yes. The product of zeroes is c/a. If a and c have opposite signs, the product is negative. A negative product means the zeroes have opposite signs (one positive, one negative). For example, x^2 - x - 6 = (x - 3)(x + 2) has zeroes 3 and -2, with product = -6.

Q9. How is this topic different from solving quadratic equations?

Solving a quadratic equation means finding the actual numerical values of the zeroes. The sum and product relationships give you information about the zeroes (their sum and product) without finding the individual values. Sometimes the sum and product are sufficient to answer a question without actually solving the equation.

Q10. Do the sum and product formulas work for polynomials with irrational zeroes?

Yes. The formulas alpha + beta = -b/a and alpha * beta = c/a hold regardless of whether the zeroes are rational, irrational, or even complex. For example, x^2 - 2 has zeroes sqrt(2) and -sqrt(2). Sum = 0 = -0/1, and Product = -2 = -2/1. Both formulas work perfectly.

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