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Qudratic Polynomials in Class 10: Relationship Between Zeroes and Coefficients

September 15, 2026 10th Mathematics(English) No Comments
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1 1.Qudratic Polynomials in Class 10: Relationship Between Zeroes and Coefficients
1.1 2.Relationship Between the Zeros and Co-efficients of a Quadratic Polynomial
1.2 3.Problems Based on Qudratic Polynomials in Class 10
1.2.1 4. Practice questions of quadratic polynomials in class 10 for students
1.2.2 5.Key Points Practice Questions of Qudratic Polynomials in Class 10
1.2.3 6.Frequently Asked Questions Related to Qudratic Polynomials in Class 10: Relationship Between Zeroes and Coefficients
1.2.4 Q:1.What is meant by the zeros of a polynomial?
1.2.5 Q:2.What is the relationship between the zeros of a Polynomial and its coefficients?
1.2.6 Q:3.Write the relationship between the zeros and their Coefficents of a cubic Polynomial
1.3 7.छात्र-छात्राओं से आज प्रश्न (Today’s Question to students):

1.Qudratic Polynomials in Class 10: Relationship Between Zeroes and Coefficients

In this article “Quddratic Polynomials in Class 10” we will read to fiind out zeroes of Polynomials and relationship between zeroes and Co-efficients of Variables.

Also Read This Article:- Distance Between Two Points Class 9

2.Relationship Between the Zeros and Co-efficients of a Quadratic Polynomial

In previous class,We know how to factorise quadratic polynomials by splitting the middle term, so that the product of two terms become equal to the product of first and third terms. The zeroes of a quadratic polynomial may be real or imaginary.
General Form: Let \alpha and \beta be the zeroes of a quadratic polynomial f(x) = ax^2 + bx + c
Then, (x - \alpha) and (x - \beta) be factors of f(x),
\therefore f(x) = k(x - \alpha)(x - \beta), where k is constant.
\Rightarrow ax^2 + bx + c = k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]
\Rightarrow ax^2 + bx + c = kx^2 - k(\alpha + \beta)x + k\alpha\beta
Comparing the coefficients of x^2, x and constant terms on both sides, we get
a = k, b = -k(\alpha + \beta) and c = k\alpha\beta
\Rightarrow \alpha + \beta = -\frac{b}{a} and \alpha\beta = \frac{c}{a}
Hence, sum of zeroes
(\alpha + \beta) = -\frac{b}{a} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}
Product of zeroes= \frac{c}{a} = \frac{\text{constant term}}{\text{coefficient of } x^2}

Also Read This Article:- Linear Polynomials Class 9

3.Problems Based on Qudratic Polynomials in Class 10

Example: 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Example: 1 (i) x^2 - 2x - 8
Solution: x^2 - 2x - 8
\Rightarrow x^2 - 4x + 2x - 8
\Rightarrow x(x - 4) + 2(x - 4)
\Rightarrow (x + 2)(x - 4)
Zeroes: x + 2 = 0 \Rightarrow x = -2
x - 4 = 0 \Rightarrow x = 4
\therefore x = -2, 4
Sum of zeroes = -2 + 4 = 2
Product of zeroes = (-2)(4) = -8
By Formula when ax^2 + bx + c
Sum of zeroes = -\frac{b}{a}
= -\frac{(-2)}{1} = 2
Product of zeroes = \frac{c}{a} = \frac{-8}{1} = -8
Example: 1 (ii) 4s^2 - 4s + 1
Solution: 4s^2 - 4s + 1
\Rightarrow 4s^2 - 2s - 2s + 1
\Rightarrow 2s(2s - 1) - 1(2s - 1)
\Rightarrow (2s - 1)^2
Zeroes are: (2s - 1)^2 = 0
\Rightarrow 2s - 1 = 0 \Rightarrow s = 1/2
2s = 1 \Rightarrow s = \frac{1}{2}
s = \frac{1}{2}, \frac{1}{2}
Sum of zeroes = \frac{1}{2} + \frac{1}{2} = \frac{1+1}{2} = 1
Product of zeroes = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{4}
By Formula when polynomial is ax^2 + bx + c
Sum of zeroes = -\frac{b}{a} = -\frac{(-4)}{4} = 1
Product of zeroes = \frac{c}{a} = \frac{1}{4}
Example: 1 (iii) 6x^2 - 3 - 7x
Solution: 6x^2 - 3 - 7x
\Rightarrow 6x^2 - 7x - 3
\Rightarrow 6x^2 - 9x + 2x - 3
\Rightarrow 3x(2x - 3) + 1(2x - 3)
\Rightarrow (3x + 1)(2x - 3)
Zeroes are: 3x + 1 = 0 \Rightarrow x = -\frac{1}{3}
2x - 3 = 0 \Rightarrow x = \frac{3}{2}
Sum of zeroes = -\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6}
Product of zeroes = \left(-\frac{1}{3}\right)\left(\frac{3}{2}\right) = -\frac{1}{2}
By Formula compare to polynomial ax^2 + bx + c
Sum of zeroes = -\frac{b}{a} = -\frac{(-7)}{6} = \frac{7}{6}
Product of zeroes = \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}
Example: 1 (iv) 4u^2 + 8u
Solution: 4u^2 + 8u
\Rightarrow 4u(u + 2)
Zeroes are: 4u(u + 2) = 0
\Rightarrow u = 0, \quad u + 2 = 0 \Rightarrow u = -2
Relationship between zeroes and coefficients
Sum of zeroes = 0 - 2 = -2
Product of zeroes = (0)(-2) = 0
By Formula: Compare to standard form of polynomial ax^2 + bx + c
Sum of zeroes = -\frac{b}{a} = -\frac{8}{4} = -2
Product of zeroes = \frac{c}{a} = \frac{0}{4} = 0
Example: 1 (v) t^2 - 15
Solution: t^2 - 15
\Rightarrow t^2 - (\sqrt{15})^2
= (t - \sqrt{15})(t + \sqrt{15})
Zeroes are: (t - \sqrt{15})(t + \sqrt{15}) = 0
\Rightarrow (t - \sqrt{15}) = 0 \Rightarrow t = \sqrt{15}
\Rightarrow (t + \sqrt{15}) = 0 \Rightarrow t = -\sqrt{15}
Relationship between zeroes and coefficients
Sum of zeroes = \sqrt{15} - \sqrt{15} = 0
Product of zeroes = (\sqrt{15})(-\sqrt{15}) = -15
By Formula: Compare to standard form of polynomial ax^2 + bx + c
Sum of zeroes = -\frac{b}{a} = \frac{0}{1} = 0
Product of zeroes = \frac{c}{a} = \frac{-15}{1} = -15
Example: 1 (vi) 3x^2 - x - 4
Solution: 3x^2 - x - 4
= 3x^2 - 4x + 3x - 4
= x(3x - 4) + 1(3x - 4)
= (x + 1)(3x - 4)
Zeroes are: (x + 1)(3x - 4) = 0
x + 1 = 0 \Rightarrow x = -1
3x - 4 = 0 \Rightarrow x = \frac{4}{3}
Relationship between zeroes and coefficients
Sum of zeroes = -1 + \frac{4}{3} = \frac{-3 + 4}{3} = \frac{1}{3}
Product of zeroes = (-1)\left(\frac{4}{3}\right) = -\frac{4}{3}
By Formula: compare to standard form of polynomial ax^2 + bx + c
Sum of zeroes = -\frac{b}{a} = -\frac{(-1)}{3} = \frac{1}{3}
Product of zeroes = \frac{c}{a} = -\frac{4}{3}

Example: 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
Example: 2 (i) \frac{1}{4}, -1
Solution: \frac{1}{4}, -1
Let the quadratic polynomial be ax^2 + bx + c, and its zeroes be \alpha and \beta, we have
\alpha + \beta = \frac{1}{4} = -\frac{b}{a}, \quad \alpha\beta = -1 = \frac{c}{a}
\Rightarrow \frac{-4}{4} = \frac{c}{a}
\therefore b = -1, \quad a = 4, \quad c = -4
Quadratic polynomial is 4x^2 - x - 4
Example: 2 (ii) \sqrt{2}, \frac{1}{3}
Solution: \sqrt{2}, \frac{1}{3}
Let the quadratic polynomial be ax^2 + bx + c and its zeroes be \alpha and \beta, we have
\alpha + \beta = \sqrt{2} = -\frac{b}{a} \Rightarrow -\frac{b}{a} = \frac{3\sqrt{2}}{3}
\alpha\beta = \frac{1}{3} = \frac{c}{a}
If a = 3, then b = -3\sqrt{2}, c = 1
Quadratic polynomial is
3x^2 - 3\sqrt{2}x + 1
Example: 2 (iii) 0, \sqrt{5}
Solution: 0, \sqrt{5}
Let the quadratic polynomial be ax^2 + bx + c and its zeroes be \alpha and \beta, we have
\alpha + \beta = \frac{0}{1} = -\frac{b}{a}
\alpha\beta = \frac{\sqrt{5}}{1} = \frac{c}{a}
If a = 1 then b = 0, c = \sqrt{5}
Quadratic polynomial is
x^2 + 0 \cdot x + \sqrt{5} \Rightarrow x^2 + \sqrt{5}
Example: 2 (iv) 1, 1
Solution: 1, 1
If the zeroes are \alpha, \beta then
\alpha + \beta = 1, \quad \alpha\beta = 1
Quadratic polynomial is
k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]
\Rightarrow k[x^2 - x + 1]
Assume k = 1
\Rightarrow x^2 - x + 1
Example: 2 (v) -\frac{1}{4}, \frac{1}{4}
Solution: -\frac{1}{4}, \frac{1}{4}
If the zeroes are \alpha, \beta then
\alpha + \beta = -\frac{1}{4} and \alpha\beta = \frac{1}{4}
Quadratic polynomial is
k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]
\Rightarrow k\left[x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4}\right]
\Rightarrow k\left(x^2 + \frac{1}{4}x + \frac{1}{4}\right)
\Rightarrow k\left(\frac{4x^2 + x + 1}{4}\right)
Assume k = 4 then
\Rightarrow \frac{4(4x^2 + x + 1)}{4}
\Rightarrow 4x^2 + x + 1
Example: 2 (vi) 4, 1
Solution: 4, 1
If the zeroes are \alpha, \beta then
\alpha + \beta = 4 and \alpha\beta = 1
Quadratic polynomial is
k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]
\Rightarrow k[x^2 - 4x + 1]
Assume k = 1 then
\Rightarrow x^2 - 4x + 1

4. Practice questions of quadratic polynomials in class 10 for students

(1.) Find the zeroes of the quadratic polynomial 3x^2 + 5x - 2 and verify the relationship between the zeroes and the coefficients
(2.) What will zeroes of polynomial x^2 - 4 be?
Answers. (1.) -1, \frac{2}{7} (2.) \pm 2

Also Read This Article:- Quadratic equation

5.Key Points Practice Questions of Qudratic Polynomials in Class 10

(1.) The general forms of polynomials are called as ax+b linear, ax^2+bx+c quadratic and ax^3+bx^2+cx+d cubic.
(2.) The values of x for polynomial f(x) are said to zeroes of polynomial, if f(x) = 0.
(3.) The number of zeroes of a polynomial is equal to its highest degree. A quadratic polynomial has two zeroes.
(4.) If \alpha, \beta are the zeroes of a quadratic polynomial f(x) = ax^2+bx+c, then \alpha + \beta = -\frac{b}{a} and \alpha\beta = \frac{c}{a}.
(5.) If \alpha, \beta are the zeroes of quadratic polynomials, then it can be written as k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right].
(6.) If f(x) is a polynomial and g(x) is nonzero polynomial, then there exist two polynomials q(x) and r(x) such that f(x) = g(x) \times q(x) + r(x), where g(x) = 0 or degree r(x) degree g(x). This is known as division algorithm.
(7.) If f(x) = ax^2 + bx + c is a quadratic polynomial, then f(x) = 0, a \neq 0 is known as a quadratic equation. The zeroes of polynomial f(x) and the roots of quadratic equation f(x) = 0 are the same.

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6.Frequently Asked Questions Related to Qudratic Polynomials in Class 10: Relationship Between Zeroes and Coefficients

Q:1.What is meant by the zeros of a polynomial?

Ans:In general,we can define the zero of polynomial that a real number ‘a’ is a zero of polynomial f(x),if f(a) = 0

Q:2.What is the relationship between the zeros of a Polynomial and its coefficients?

Ans:Quadratic Polynomial f(x)=ax^2 + bx + c
Sum of zeroes (\alpha + \beta) = -\frac{b}{a}
Product of zeroes = \frac{c}{a}

Q:3.Write the relationship between the zeros and their Coefficents of a cubic Polynomial

Ans: If \alpha, \beta, \gamma are the zeroes of the cubic polynomial ax^3 + bx^2 + cx + d, then
\alpha + \beta + \gamma = -\frac{b}{a}
\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}
and \alpha\beta\gamma = -\frac{d}{a}
By answering the above questions,you can know about the primary terms of Qudratic Polynomials in Class 10: Relationship Between Zeroes and Coefficients.

7.छात्र-छात्राओं से आज प्रश्न (Today’s Question to students):

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*”28 सेमी की भुजा वाले वर्ग के भीतर बनाए जा सकने वाले सबसे बड़े वृत्त का क्षेत्रफल क्या होगा?
(What is the area of the largest circle that cob be formed within a square with sides of 28 cm?)”*
*पिछली प्रश्नोत्तरी का हल*1.6 
100^{\frac{1}{2}} \times (0.001)^{\frac{1}{3}} - (0.0016)^{\frac{1}{4}} \times 3^0 + \left(\frac{5}{4}\right)^{-1}
= (10^2)^{\frac{1}{2}} \times \left(\frac{1}{1000}\right)^{1/3} - \left(\frac{16}{10000}\right)^{1/4} \times 1 + \frac{4}{5}
= 10 \times \left(\frac{1}{10^3}\right)^{\frac{1}{3}} - \left(\frac{2^4}{10^4}\right)^{1/4} + \frac{4}{5}
= 10 \times \frac{1}{10} - \frac{2}{10} + \frac{4}{5}
= 1 - \frac{1}{5} + \frac{4}{5}
= \frac{5 - 1 + 4}{5} = \frac{8}{5} = 1.6
*Previous Quiz Solution*1.6

100^{\frac{1}{2}} \times (0.001)^{1/3} - (0.0016)^{1/4} \times 3^0 + \left(\frac{5}{4}\right)^{-1}
= (10^2)^{\frac{1}{2}} \times \left(\frac{1}{1000}\right)^{1/3} - \left(\frac{16}{10000}\right)^{1/4} \times 1 + \frac{4}{5}
= 10 \times \left(\frac{1}{10^3}\right)^{1/3} - \left(\frac{2^4}{10^4}\right)^{1/4} + \frac{4}{5}
= 10 \times \frac{1}{10} - \frac{2}{10} + \frac{4}{5}
= 1 - \frac{1}{5} + \frac{4}{5}
= \frac{5 - 1 + 4}{5} = \frac{8}{5} = 1.6
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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