Alpha-Beta Identities for Zeroes of Polynomials | Class 10

Quick answer: If $\alpha$ and $\beta$ are the zeroes of $ax^2 + bx + c$, then $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. Any symmetric expression in $\alpha$ and $\beta$ can be written using just these two values, for example $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ and $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$. So you can find their values without finding the zeroes.

For a quadratic polynomial $p(x) = ax^2 + bx + c$, the sum and product of its zeroes are easy to find: $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.

Board questions often go one step further. They ask for values such as $\alpha^2 + \beta^2$, $\frac{1}{\alpha} + \frac{1}{\beta}$ or $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$, or ask you to form a new polynomial whose zeroes are related to $\alpha$ and $\beta$. The trick is to rewrite each expression using only the sum and the product of the zeroes.

In CBSE Class 10 Mathematics, Chapter 2 (Polynomials), these questions on the relationship between zeroes and coefficients are common in board exams.

What You Will Learn

• The relationship between the zeroes and coefficients of a quadratic polynomial
• What a symmetric expression is
• Key identities for $\alpha^2 + \beta^2$, $(\alpha - \beta)^2$, $\frac{1}{\alpha} + \frac{1}{\beta}$, $\alpha^3 + \beta^3$ and more
• A solved problem evaluating three expressions
• How to form a new quadratic polynomial whose zeroes are $2\alpha + 1$ and $2\beta + 1$

1. Sum and Product of Zeroes

For a quadratic polynomial $p(x) = ax^2 + bx + c$ (with $a \ne 0$) whose zeroes are $\alpha$ and $\beta$:

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

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

Key idea: An expression is symmetric if it stays the same when $\alpha$ and $\beta$ are swapped. Every symmetric expression can be written using only $(\alpha + \beta)$ and $\alpha\beta$, so you never need to find $\alpha$ and $\beta$ separately.

2. Key Identities

3. Solved Problem 1: Evaluating Expressions

Question

If $\alpha$ and $\beta$ are the zeroes of $p(x) = 2x^2 - 5x + 2$, find:

(i) $\frac{1}{\alpha} + \frac{1}{\beta}$

(ii) $\alpha^2 + \beta^2$

(iii) $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$

Solution

Step 1: Find the sum and product. Here $a = 2$, $b = -5$, $c = 2$.

$\alpha + \beta = -\frac{-5}{2} = \frac{5}{2} \qquad \alpha\beta = \frac{2}{2} = 1$

Step 2: Part (i).

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/2}{1} = \frac{5}{2}$

Step 3: Part (ii).

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{5}{2}\right)^2 - 2(1) = \frac{25}{4} - 2 = \frac{17}{4}$

Step 4: Part (iii).

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{17/4}{1} = \frac{17}{4}$

Answers: (i) $\frac{5}{2}$ (ii) $\frac{17}{4}$ (iii) $\frac{17}{4}$

Check: $2x^2 - 5x + 2 = (2x - 1)(x - 2)$, so the zeroes are $\frac{1}{2}$ and $2$. Then $\alpha^2 + \beta^2 = \frac{1}{4} + 4 = \frac{17}{4}$. ✓

4. Solved Problem 2: Forming a New Polynomial

Question

If $\alpha$ and $\beta$ are the zeroes of $f(x) = x^2 - 5x + 6$, find a quadratic polynomial whose zeroes are $2\alpha + 1$ and $2\beta + 1$.

Solution

Step 1: Find the sum and product of the original zeroes. Here $a = 1$, $b = -5$, $c = 6$.

$\alpha + \beta = 5 \qquad \alpha\beta = 6$

Step 2: Sum of the new zeroes.

$(2\alpha + 1) + (2\beta + 1) = 2(\alpha + \beta) + 2 = 2(5) + 2 = 12$

Step 3: Product of the new zeroes.

$(2\alpha + 1)(2\beta + 1) = 4\alpha\beta + 2(\alpha + \beta) + 1 = 4(6) + 2(5) + 1 = 35$

Step 4: Form the polynomial. A quadratic polynomial with a given sum $S$ and product $P$ of zeroes is $k\left[x^2 - Sx + P\right]$, where $k$ is any non-zero constant. Taking $k = 1$:

$p(x) = x^2 - 12x + 35$

Answer: The required polynomial is $x^2 - 12x + 35$ (or any non-zero multiple of it).

Check: The zeroes of $x^2 - 5x + 6$ are $2$ and $3$, so the new zeroes are $5$ and $7$. Indeed, $x^2 - 12x + 35 = (x - 5)(x - 7)$. ✓

5. Summary Table

6. Exam Tips and Common Mistakes

Exam Tip: When forming a new polynomial, write the general form $k\left[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})\right]$ before substituting.

Exam Tip: If you can factorise the polynomial easily, use the actual zeroes to check your answer.

Common Mistake: Sign errors in $\alpha + \beta = -\frac{b}{a}$. If $b$ is negative (say $b = -5$), then $-\frac{b}{a} = +\frac{5}{a}$.

Common Mistake: Writing $\alpha^2 + \beta^2 = (\alpha + \beta)^2$. You must subtract $2\alpha\beta$.

Frequently Asked Questions

What is the sum and product of the zeroes of ax² + bx + c?

The sum is $-\frac{b}{a}$ and the product is $\frac{c}{a}$.

How do you find α² + β² without finding the zeroes?

Use $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$, then substitute the sum and product of the zeroes.

How do you find 1/α + 1/β?

Use $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$, which is the sum of the zeroes divided by their product.

How do you form a quadratic polynomial from its zeroes?

If the sum of the zeroes is $S$ and the product is $P$, the polynomial is $k\left[x^2 - Sx + P\right]$, where $k$ is any non-zero constant.

What is a symmetric expression?

An expression in α and β that stays the same when α and β are swapped, such as α² + β² or αβ.