Relationship Between Zeroes and Coefficients of a Polynomial – Class 10 Maths with Solved Questions

Short Notes

Relationship Between Zeroes and Coefficients – Class 10

Introduction:
In Class 10, understanding the relationship between the zeroes (roots) of a polynomial and its coefficients is fundamental. This relationship helps in connecting the roots to the polynomial constants.

What is a Polynomial?

A polynomial in one variable \( x \) of degree \( n \) has the form:

where \( a_n, a_{n-1}, \dots, a_0 \) are constants and \( a_n \neq 0 \).

Zeroes (Roots) of a Polynomial:
Zeroes or roots of a polynomial are the values of \( x \) such that \( p(x) = 0 \).

Relationship for Quadratic Polynomials:

Consider quadratic polynomial:

If the roots are \( \alpha \) and \( \beta \), then:

• Sum of zeroes: \[ \alpha + \beta = -\frac{b}{a} \]
• Product of zeroes: \[ \alpha \beta = \frac{c}{a} \]

Derivation:
From factorized form:

Comparing with \( ax^2 + bx + c = 0 \) gives the relations above.

Relationship for Cubic Polynomials:

For cubic polynomial:

With roots \( \alpha, \beta, \gamma \):

• \( \alpha + \beta + \gamma = -\frac{b}{a} \)
• \( \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \)
• \( \alpha \beta \gamma = -\frac{d}{a} \)

Why is this Useful?

• Find sum/product of roots without directly solving the polynomial.
• Form polynomials when roots are known.
• Analyze and understand polynomials better.

Example 1:
Find sum and product of zeroes of \( 2x^2 – 5x + 3 = 0 \).

• Sum of zeroes: \[ \alpha + \beta = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2} \]
• Product of zeroes: \[ \alpha \beta = \frac{c}{a} = \frac{3}{2} \]

Example 2:
If one zero of the quadratic polynomial \( x^2 – 3x + k = 0 \) is 2, find the value of \( k \).

Let the roots be \( \alpha = 2 \), and \( \beta \) unknown:

• Using sum of zeroes: \[ \alpha + \beta = -\frac{-3}{1} = 3 \implies 2 + \beta = 3 \implies \beta = 1 \]
• Using product of zeroes: \[ \alpha \beta = k \implies 2 \times 1 = k \implies k = 2 \]

Example 3:
Form a quadratic polynomial whose zeroes are 4 and -3.

• Sum of zeroes: \( 4 + (-3) = 1 \)
• Product of zeroes: \( 4 \times (-3) = -12 \)

Required polynomial is:

Summary Table:

Practice Questions

Questions:

1. Find the sum and product of zeroes of the polynomial \( x^2 + 7x + 12 = 0 \).
2. If one zero of the polynomial \( x^2 – 5x + k = 0 \) is 3, find the value of \( k \).
3. Form a quadratic polynomial whose zeroes are 2 and -5.
4. For the polynomial \( 3x^2 + 2x – 1 = 0 \), find the sum and product of its zeroes.
5. One zero of the polynomial \( 4x^2 – x + 1 = 0 \) is \( \frac{1}{4} \). Find the other zero.
6. Find the polynomial equation whose roots are 4 and 5.
7. The sum of zeroes of a quadratic polynomial is 6 and the product is 8. Write the polynomial.
8. Given that the zeroes of \( x^2 + px + 20 = 0 \) are equal, find the value of \( p \).
9. Find the sum of zeroes of the polynomial \( 5x^2 – 15x + 20 = 0 \).
10. If the zeroes of \( x^2 – 3x + m = 0 \) are reciprocal of each other, find the value of \( m \).
11. A quadratic polynomial has zeroes 1 and -3. Form the polynomial.
12. For \( 2x^2 + bx + 5 = 0 \), if the zeroes are equal, find \( b \).
13. Find the product of the zeroes of the polynomial \( x^2 – 7x + 10 = 0 \).
14. Write a quadratic polynomial whose zeroes are 0 and 4.
15. If the sum of zeroes of the polynomial \( ax^2 + bx + c = 0 \) is 5 and product is 6, write the polynomial.

Answers:

1. Sum = -7, Product = 12
2. Sum of zeroes = 5, so other zero = 5 – 3 = 2; Product = \( k = 3 \times 2 = 6 \)
3. \( x^2 + 3x – 10 = 0 \) (since sum = 2 + (-5) = -3 and product = 2 \times -5 = -10)
4. Sum = \(-\frac{2}{3}\), Product = \(-\frac{1}{3}\)
5. Let the other zero be \( \alpha \). Sum = \( \frac{1}{4} + \alpha = -\frac{-1}{4} = \frac{1}{4} \Rightarrow \alpha = 0 \). So other zero is 0.
6. \( x^2 – 9x + 20 = 0 \)
7. \( x^2 – 6x + 8 = 0 \)
8. \( \text{Discriminant} = p^2 – 4 \times 1 \times 20 = 0 \Rightarrow p^2 = 80 \Rightarrow p = \pm 4 \sqrt{5} \)
9. Sum = \(\frac{15}{5} = 3\)
10. Sum = 3; Product = m; For reciprocal roots, product = 1, so \( m = 1 \)
11. \( x^2 + 2x – 3 = 0 \)
12. Discriminant = 0, so \( b^2 – 4 \times 2 \times 5 = 0 \implies b^2 = 40 \implies b = \pm 2 \sqrt{10} \)
13. Product = 10
14. \( x^2 – 4x = 0 \) or \( x(x – 4) = 0 \)
15. \( x^2 – 5x + 6 = 0 \)