Symmetric Expressions of Zeroes α and β | Class 10 Maths

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 can be written using just these two, for example $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ and $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$. For $2x^2 - 5x + 3$, this gives $\alpha^2 + \beta^2 = \frac{13}{4}$ without finding the zeroes.

Sum and product of zeroes

For the quadratic $ax^2 + bx + c$ ($a \ne 0$) with zeroes $\alpha$ and $\beta$:

These are in NCERT Class 10, Polynomials (relationship between zeroes and coefficients). The same results hold for the roots of the equation $ax^2 + bx + c = 0$. In higher classes they are called Vieta's relations.

What is a symmetric expression?

An expression in $\alpha$ and $\beta$ is symmetric if swapping $\alpha$ and $\beta$ does not change it. For example, $\alpha^2 + \beta^2$ and $\frac{1}{\alpha} + \frac{1}{\beta}$ are symmetric, but $\alpha - \beta$ is not. Every symmetric expression can be written in terms of $\alpha + \beta$ and $\alpha\beta$.

Useful identities

Where the first one comes from:

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

So:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Problem 1: evaluate without finding the zeroes

If $\alpha$ and $\beta$ are the zeroes of $2x^2 - 5x + 3$, find (i) $\alpha^2 + \beta^2$, (ii) $\frac{1}{\alpha} + \frac{1}{\beta}$, (iii) $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.

Step 1: Sum and product. Here $a = 2$, $b = -5$, $c = 3$.

$\alpha + \beta = \frac{5}{2}, \qquad \alpha\beta = \frac{3}{2}$

Step 2: (i)

$\alpha^2 + \beta^2 = \left(\frac{5}{2}\right)^2 - 2 \times \frac{3}{2} = \frac{25}{4} - 3 = \frac{13}{4}$

Step 3: (ii)

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{5/2}{3/2} = \frac{5}{3}$

Step 4: (iii)

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{13/4}{3/2} = \frac{13}{6}$

Check: The zeroes of $2x^2 - 5x + 3 = (2x - 3)(x - 1)$ are $1$ and $\frac{3}{2}$. Then $1 + \frac{9}{4} = \frac{13}{4}$, $1 + \frac{2}{3} = \frac{5}{3}$ and $\frac{2}{3} + \frac{3}{2} = \frac{13}{6}$. ✓

Problem 2: form a quadratic with new zeroes

If $\alpha$ and $\beta$ are the zeroes of $x^2 - 3x + 2$, find a quadratic polynomial whose zeroes are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

Step 1: Sum and product of the old zeroes.

$\alpha + \beta = 3, \qquad \alpha\beta = 2$

Step 2: Sum of the new zeroes.

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3}{2}$

Step 3: Product of the new zeroes.

$\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{2}$

Step 4: Form the polynomial using $k[x^2 - (\text{sum})x + \text{product}]$:

$k\left(x^2 - \frac{3}{2}x + \frac{1}{2}\right)$

Taking $k = 2$:

$2x^2 - 3x + 1$

Check: The zeroes of $x^2 - 3x + 2$ are 1 and 2, so the new zeroes are 1 and $\frac{1}{2}$, and $2x^2 - 3x + 1 = (2x - 1)(x - 1)$. ✓

Summary table

Exam tips

• Write $\alpha + \beta$ and $\alpha\beta$ first, with the signs checked: the sum is $-\frac{b}{a}$, so a negative $b$ gives a positive sum.
• Do not find the zeroes unless the question asks for them; work with sum and product.
• Keep fractions as fractions until the end to avoid rounding errors.
• When forming a new polynomial, multiply by a suitable $k$ to clear fractions.

Common mistakes

• Writing $\alpha^2 + \beta^2 = (\alpha + \beta)^2$, forgetting to subtract $2\alpha\beta$.
• Taking $\alpha + \beta = \frac{b}{a}$ instead of $-\frac{b}{a}$.
• Mixing up $\frac{1}{\alpha} + \frac{1}{\beta}$ with $\frac{1}{\alpha + \beta}$.
• Using $+$ instead of $-$ for the middle term: the polynomial is $x^2 - (\text{sum})x + \text{product}$.

Frequently Asked Questions

What is the sum and product of zeroes of a quadratic polynomial?

For ax² + bx + c, the sum of zeroes is −b/a and the product is c/a.

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

Use α² + β² = (α + β)² − 2αβ and substitute the sum −b/a and product c/a.

What is 1/α + 1/β in terms of the coefficients?

1/α + 1/β = (α + β)/αβ = (−b/a)/(c/a) = −b/c.

How do you form a quadratic polynomial whose zeroes are 1/α and 1/β?

Find the new sum (α + β)/αβ and new product 1/αβ, then write k[x² − (sum)x + product].

What is a symmetric expression of zeroes?

It is an expression in α and β that stays the same when α and β are swapped, such as α² + β² or αβ. It can always be written using α + β and αβ.