Finding k from Relations Between Zeroes | Class 10 Maths

Quick answer: To find an unknown $k$ in a quadratic polynomial from a condition on its zeroes, first write $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$ in terms of $k$. Then rewrite the given condition using only the sum and product, for example $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ or $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$, and solve for $k$. If one zero is the reciprocal of the other, the product of zeroes is $1$, so $c = a$.

For a quadratic polynomial $ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$, we know that $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.

Many board questions reverse this: the polynomial contains an unknown constant ($k$ or $p$), and you are given a condition on the zeroes, such as their difference, the sum of their squares, or that one is the reciprocal of the other. You then use the sum and product of zeroes to find the unknown.

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 sum and product of zeroes of a quadratic polynomial
• Two key identities for the difference and the sum of squares of the zeroes
• Type 1: finding $k$ from the difference of the zeroes
• Type 2: finding $k$ from the sum of the squares of the zeroes
• Type 3: finding $k$ when one zero is the reciprocal of the other
• How to check your answer

1. Key Relations and Identities

For $p(x) = ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$:

$\alpha + \beta = -\frac{b}{a} \qquad \alpha\beta = \frac{c}{a}$

2. Type 1: Difference of the Zeroes

Example 1

Question: If the zeroes $\alpha$ and $\beta$ of $f(x) = x^2 - 5x + k$ satisfy $\alpha - \beta = 1$, find $k$.

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

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

Step 2: Use the difference identity.

$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$

$1^2 = 5^2 - 4k \implies 1 = 25 - 4k \implies 4k = 24 \implies k = 6$

Answer: $k = 6$.

Check: $x^2 - 5x + 6 = (x - 2)(x - 3)$, so the zeroes are $3$ and $2$, and $3 - 2 = 1$. ✓

Example 2

Question: If the square of the difference of the zeroes of $f(x) = x^2 + px + 45$ is $144$, find $p$.

Step 1: Sum and product. $\alpha + \beta = -p$ and $\alpha\beta = 45$.

Step 2: Use the identity.

$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \implies 144 = p^2 - 180$

$p^2 = 324 \implies p = \pm 18$

Answer: $p = 18$ or $p = -18$.

Check: For $p = -18$, $x^2 - 18x + 45 = (x - 3)(x - 15)$, and $(15 - 3)^2 = 144$. ✓

3. Type 2: Sum of the Squares of the Zeroes

Question: If the sum of the squares of the zeroes of $f(x) = x^2 - 8x + k$ is $40$, find $k$.

Step 1: Sum and product. $\alpha + \beta = 8$ and $\alpha\beta = k$.

Step 2: Use the identity.

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \implies 40 = 64 - 2k \implies k = 12$

Answer: $k = 12$.

Check: $x^2 - 8x + 12 = (x - 2)(x - 6)$, and $2^2 + 6^2 = 4 + 36 = 40$. ✓

4. Type 3: One Zero Is the Reciprocal of the Other

Question: If one zero of $p(x) = (k + 1)x^2 - 5x + 2k$ is the reciprocal of the other, find $k$.

Step 1: Use the product. If the zeroes are $\alpha$ and $\frac{1}{\alpha}$, their product is $1$:

$\alpha\beta = \frac{c}{a} = \frac{2k}{k + 1} = 1$

Step 2: Solve.

$2k = k + 1 \implies k = 1$

Answer: $k = 1$.

Check: For $k = 1$, $p(x) = 2x^2 - 5x + 2 = (2x - 1)(x - 2)$. The zeroes are $\frac{1}{2}$ and $2$, which are reciprocals. ✓

Shortcut: If one zero is the reciprocal of the other, set the constant term equal to the coefficient of $x^2$ ($c = a$).

5. Summary Table

6. Exam Tips and Common Mistakes

Exam Tip: Always start by writing the sum and product of zeroes in terms of the unknown, then substitute into the right identity.

Exam Tip: Check your answer by putting the value of $k$ back into the polynomial and finding its zeroes, if it factorises easily.

Common Mistake: Using $2\alpha\beta$ instead of $4\alpha\beta$ in the difference identity. It is $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.

Common Mistake: Forgetting the negative value when taking a square root, as in $p = \pm 18$.

Frequently Asked Questions

How do you find k if the difference of the zeroes is given?

Use (α − β)² = (α + β)² − 4αβ, substitute the sum and product of zeroes in terms of k, and solve for k.

How do you find k if the sum of the squares of the zeroes is given?

Use α² + β² = (α + β)² − 2αβ, substitute the sum and product, and solve for k.

What happens if one zero is the reciprocal of the other?

The product of the zeroes is 1, so the constant term equals the coefficient of x².

What if the zeroes are equal in size but opposite in sign?

Their sum is 0, so the coefficient of x must be 0.