Quadratic Polynomials with Conjugate Zeroes | Class 10

Quick answer: If a quadratic polynomial has rational coefficients and one zero is $p + \sqrt{q}$ (with $\sqrt{q}$ irrational), the other zero is its conjugate $p - \sqrt{q}$. Their sum $2p$ and product $p^2 - q$ are rational, so the polynomial is $k[x^2 - 2px + (p^2 - q)]$. For example, a polynomial with zero $3 + \sqrt{5}$ is $x^2 - 6x + 4$.

Syllabus note: The rationalised NCERT Class 10 Polynomials chapter covers the relationship between zeroes and coefficients of a quadratic. The "conjugate roots" result is extra (it is proved in higher classes), and the division algorithm used in Problem 2 was removed from the rationalised Class 10 book. Treat both as enrichment and check your syllabus.

Zeroes and coefficients: the basic formula

If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial, then the polynomial is:

$p(x) = k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]$

where $k$ is any non-zero constant. So you only need the sum and product of the zeroes.

Why irrational zeroes come in conjugate pairs

Take $\alpha = p + \sqrt{q}$ and $\beta = p - \sqrt{q}$, where $p$ and $q$ are rational and $\sqrt{q}$ is irrational.

Sum:

$\alpha + \beta = 2p$

Product:

$\alpha\beta = p^2 - (\sqrt{q})^2 = p^2 - q$

Both are rational, so the polynomial $k[x^2 - 2px + (p^2 - q)]$ has rational coefficients. If the second zero were anything else, the sum or product would usually contain $\sqrt{q}$, and the coefficients would not all be rational.

Problem 1: polynomial with zero 3 + √5

Find a quadratic polynomial with rational coefficients, one of whose zeroes is $3 + \sqrt{5}$.

Step 1: Find the other zero. With rational coefficients, the other zero is $3 - \sqrt{5}$.

Step 2: Sum of zeroes.

$(3 + \sqrt{5}) + (3 - \sqrt{5}) = 6$

Step 3: Product of zeroes.

$(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4$

Step 4: Write the polynomial.

$p(x) = k(x^2 - 6x + 4)$

Answer: Taking $k = 1$, the polynomial is $x^2 - 6x + 4$.

Problem 2: all zeroes of a degree-4 polynomial (enrichment)

Find all the zeroes of $2x^4 - 9x^3 + 5x^2 + 3x - 1$, given that two of its zeroes are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.

Step 1: Quadratic factor from the known zeroes.

Sum $= 4$, product $= 4 - 3 = 1$. So $x^2 - 4x + 1$ is a factor.

Step 2: Divide. Dividing $2x^4 - 9x^3 + 5x^2 + 3x - 1$ by $x^2 - 4x + 1$:

So the quotient is $2x^2 - x - 1$ and the remainder is 0.

Step 3: Factorise the quotient.

$2x^2 - x - 1 = (2x + 1)(x - 1)$

So the remaining zeroes are $x = -\frac{1}{2}$ and $x = 1$.

Answer: The four zeroes are $2 + \sqrt{3}$, $2 - \sqrt{3}$, $1$ and $-\frac{1}{2}$.

Check by multiplying back: $(x^2 - 4x + 1)(2x^2 - x - 1) = 2x^4 - 9x^3 + 5x^2 + 3x - 1$. ✓

Summary table

Exam tips

• Always state "since the coefficients are rational, the other zero is the conjugate" before using it.
• Use $(a + b)(a - b) = a^2 - b^2$ for the product; it removes the surd at once.
• Write the polynomial with $k$ first, then give the simplest case $k = 1$.
• If a division gives a non-zero remainder when it should be zero, recheck the signs in each subtraction.

Common mistakes

• Changing the sign of the rational part instead of the surd: the conjugate of $3 + \sqrt{5}$ is $3 - \sqrt{5}$, not $-3 + \sqrt{5}$.
• Using the conjugate rule when the coefficients are not rational. The rule needs rational coefficients.
• Writing the polynomial as $x^2 + (\alpha + \beta)x + \alpha\beta$; the middle sign must be minus.
• Forgetting that $k$ can be any non-zero constant, so many polynomials are correct.

Frequently Asked Questions

What is the conjugate of p + √q?

The conjugate of p + √q is p − √q. Only the sign of the surd part changes.

Why do irrational zeroes come in conjugate pairs?

For a polynomial with rational coefficients, the sum and product of the zeroes must be rational. Pairing p + √q with p − √q gives a rational sum 2p and a rational product p² − q.

What is a quadratic polynomial whose one zero is 3 + √5?

With rational coefficients, the other zero is 3 − √5. The sum is 6 and the product is 4, so the polynomial is x² − 6x + 4 (or any non-zero multiple of it).

How do you find all zeroes of a degree-4 polynomial when two are given?

Form the quadratic factor from the two given zeroes, divide the polynomial by it, and factorise the quotient to get the other two zeroes.

Is the division algorithm for polynomials in the Class 10 syllabus?

The division algorithm for polynomials was removed from the rationalised NCERT Class 10 book, so questions like Problem 2 are now enrichment.