Relationship Between Zeroes and Coefficients for CBSE Class 10

In algebra, the zeroes of a polynomial (the values of $x$ that make $P(x) = 0$) and its coefficients (the numerical multipliers of the powers of $x$) are intimately connected. The relationship between zeroes and coefficients allows us to find the sum and product of zeroes directly without solving the equation, verify our factorisation, and reconstruct a polynomial when only its roots or root properties are known.

In CBSE Class 10 Mathematics, questions testing this relationship are among the most frequently asked in board examinations, carrying anywhere from 2 to 4 marks.

What You Will Learn

• Algebraic derivation of the relationship for quadratic polynomials
• The Sum of Zeroes ($\alpha + \beta = -b/a$) and Product of Zeroes ($\alpha\beta = c/a$)
• Reconstructing a quadratic polynomial from the sum and product of its zeroes
• Relationships for cubic polynomials (sum, pairwise product sum, product)
• Solving symmetric expressions involving zeroes ($\alpha^2 + \beta^2, \frac{1}{\alpha} + \frac{1}{\beta}$, etc.)
• Step-by-step solved CBSE board exam questions and verification procedures

1. Quadratic Polynomial: Derivation of Relationships

Let $P(x) = ax^2 + bx + c$ be a quadratic polynomial where $a, b, c \in \mathbb{R}$ and $a \ne 0$. Let $\alpha$ (alpha) and $\beta$ (beta) be the zeroes of $P(x)$.

By the Factor Theorem, if $\alpha$ and $\beta$ are zeroes, then $(x - \alpha)$ and $(x - \beta)$ are factors of $P(x)$. Therefore: $ax^2 + bx + c = k(x - \alpha)(x - \beta)$ where $k$ is a non-zero constant.

Expanding the right-hand side: $ax^2 + bx + c = k[x^2 - (\alpha + \beta)x + \alpha\beta]$ $ax^2 + bx + c = kx^2 - k(\alpha + \beta)x + k\alpha\beta$

Equating the coefficients of like powers of $x$ on both sides:

1. Coefficient of $x^2$: $a = k$
2. Coefficient of $x$: $b = -k(\alpha + \beta) = -a(\alpha + \beta) \implies \alpha + \beta = -\frac{b}{a}$
3. Constant term: $c = k\alpha\beta = a\alpha\beta \implies \alpha\beta = \frac{c}{a}$

2. The Fundamental Quadratic Formulas

1. Sum of Zeroes

$\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$

2. Product of Zeroes

$\alpha\beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$

Important: <u>Notice the negative sign in the sum formula ($-\frac{b}{a}$). Forgetting this negative sign is the single most common student error in board exams!</u>

3. Forming a Quadratic Polynomial

If the sum of zeroes $S = \alpha + \beta$ and the product of zeroes $P = \alpha\beta$ are known, the quadratic polynomial is given by:

$P(x) = k[x^2 - (\alpha + \beta)x + \alpha\beta] = k[x^2 - Sx + P]$

where $k$ is any non-zero real constant.

Example:

If the sum of zeroes is $S = -3$ and product of zeroes is $P = 2$: $P(x) = x^2 - (-3)x + 2 = x^2 + 3x + 2$

4. Cubic Polynomial Relationships

For a cubic polynomial $P(x) = ax^3 + bx^2 + cx + d$ ($a \ne 0$) with zeroes $\alpha, \beta, \gamma$:

1. Sum of Zeroes: $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{\text{Coefficient of } x^2}{\text{Coefficient of } x^3}$
2. Sum of Products Taken Two at a Time: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{Coefficient of } x}{\text{Coefficient of } x^3}$
3. Product of All Zeroes: $\alpha\beta\gamma = -\frac{d}{a} = -\frac{\text{Constant term}}{\text{Coefficient of } x^3}$

Remember: Notice the alternating sign pattern for coefficients:

• Sum ($x^2$ term) $\implies$ Negative ($-b/a$)
• Pairwise product ($x$ term) $\implies$ Positive ($+c/a$)
• Product of all three (constant) $\implies$ Negative ($-d/a$)

5. Solved CBSE Board Exam Questions

Solved Example 1: Finding Zeroes and Verifying Relationships

Problem: Find the zeroes of the quadratic polynomial $P(x) = x^2 - 2x - 8$, and verify the relationship between the zeroes and the coefficients.

Solution:

1. Find Zeroes by Splitting the Middle Term: $x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2)$ Set $P(x) = 0 \implies (x - 4)(x + 2) = 0$. Thus, zeroes are $\alpha = 4$ and $\beta = -2$.

2. Identify Coefficients from $ax^2 + bx + c$: $a = 1, \quad b = -2, \quad c = -8$

3. Verify Sum of Zeroes:

   • From zeroes: $\alpha + \beta = 4 + (-2) = 2$
   • From coefficients: $-\frac{b}{a} = -\frac{-2}{1} = 2$
   • Sum is verified: $\alpha + \beta = -\frac{b}{a} = 2$.

4. Verify Product of Zeroes:

   • From zeroes: $\alpha\beta = 4 \times (-2) = -8$
   • From coefficients: $\frac{c}{a} = \frac{-8}{1} = -8$
   • Product is verified: $\alpha\beta = \frac{c}{a} = -8$.

Solved Example 2: Symmetric Functions of Zeroes

Problem: If $\alpha$ and $\beta$ are the zeroes of the polynomial $P(x) = 2x^2 - 5x + 7$, find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ and $\alpha^2 + \beta^2$.

Solution: From $P(x) = 2x^2 - 5x + 7$, we have $a = 2, b = -5, c = 7$: $\alpha + \beta = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$ $\alpha\beta = \frac{c}{a} = \frac{7}{2}$

1. Evaluate $\frac{1}{\alpha} + \frac{1}{\beta}$: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{\frac{5}{2}}{\frac{7}{2}} = \frac{5}{7}$

2. Evaluate $\alpha^2 + \beta^2$: Use the algebraic identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ $\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2\left(\frac{7}{2}\right) = \frac{25}{4} - 7 = \frac{25 - 28}{4} = -\frac{3}{4}$

6. Summary and Revision Cheat Sheet

Exam Tip: When a question gives roots like $\alpha$ and $\frac{1}{\alpha}$ (reciprocal roots), immediately use the product formula: $\alpha \cdot \frac{1}{\alpha} = 1 = \frac{c}{a} \implies a = c$!

Common Mistake: In questions asking to form a quadratic polynomial, students often write $x^2 + Sx + P$ instead of $x^2 - Sx + P$. Always remember the negative sign in front of the sum $S$!