NCERT Solutions for Class 10 Mathematics Chapter 2: Polynomials (Complete Guide)
Polynomials serve as a fundamental structural bridge in secondary mathematics, connecting basic linear arithmetic to higher-order functional analysis, coordinate geometry, and algebraic calculus. In the CBSE Class 10 curriculum, Chapter 2: Polynomials develops the theoretical principles of single-variable algebraic functions over the real field ($\mathbb{R}[x]$). Mastering this chapter requires a solid grasp of how the degree of an algebraic expression governs its roots, the geometric meaning of intersections on the Cartesian plane, and the algebraic relationship between polynomial coefficients and their zeroes.
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| POLYNOMIAL CLASSIFICATION ARCHITECTURE |
| |
| Polynomial Form: P(x) = a_n*x^n + a_(n-1)*x^(n-1) + ... + a_1*x + a_0 (where a_n != 0, n in W) |
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| +-----------------------+----------------------------------+----------------------------------+ |
| | Linear Polynomial | Quadratic Polynomial | Cubic Polynomial | |
| | Degree: n = 1 | Degree: n = 2 | Degree: n = 3 | |
| | Form: ax + b (a != 0) | Form: ax^2 + bx + c (a != 0) | Form: ax^3 + bx^2 + cx + d | |
| | Max Zeroes: 1 | Max Zeroes: 2 | Max Zeroes: 3 | |
| | Graph: Straight Line | Graph: Parabola (Upward/Downward)| Graph: Continuous S-Curve | |
| +-----------------------+----------------------------------+----------------------------------+ |
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The geometric interpretation of zeroes is a key conceptual focus of the latest CBSE curriculum. For any polynomial function $y = p(x)$, the real roots correspond directly to the $x$-coordinates of the points where the graph intersects or touches the $x$-axis. A polynomial of degree $n$ can intersect the $x$-axis at a maximum of $n$ distinct points, meaning it possesses at most $n$ real zeroes. For a quadratic polynomial $p(x) = ax^2 + bx + c$ ($a \neq 0$), the graph forms a symmetric curve known as a parabola, opening upward when the leading coefficient is positive ($a > 0$) and downward when it is negative ($a < 0$).
The core analytical foundation of this chapter lies in the Relations Between Zeroes and Coefficients. Derived from the factor theorem, if $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = ax^2 + bx + c$, then $(x – \alpha)$ and $(x – \beta)$ are its linear factors. Equating coefficients yields two central invariant identities:
$$\text{Sum of Zeroes: } \alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$$
$$\text{Product of Zeroes: } \alpha\beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$$
According to official CBSE board evaluation criteria, solutions to polynomial problems require a clear, stepwise presentation. When verifying relationships, students must first explicitly state the coefficients ($a, b, c$), determine the zeroes using systematic methods (such as middle-term splitting), and verify both the sum and product relationships independently. When constructing a quadratic polynomial from given conditions, students should write the general standard form $k[x^2 – (\alpha + \beta)x + \alpha\beta]$ (where $k$ is a non-zero real constant) to ensure full marks.
| Polynomial Type | Standard Form | Degree | Maximum Zeroes | Geometric Graph Shape | Invariant Relations Between Zeroes & Coefficients |
| Linear | $p(x) = ax + b$ ($a \neq 0$) | $1$ | $1$ | Straight Line | Zero is given by $x = -\frac{b}{a} = -\frac{\text{Constant term}}{\text{Coefficient of } x}$ |
| Quadratic | $p(x) = ax^2 + bx + c$ ($a \neq 0$) | $2$ | $2$ | Parabola (Opens upward if $a > 0$, downward if $a < 0$) | $\alpha + \beta = -\frac{b}{a}$ $\alpha\beta = \frac{c}{a}$ |
| Cubic | $p(x) = ax^3 + bx^2 + cx + d$ ($a \neq 0$) | $3$ | $3$ | Continuous Cubic Inflection Curve | $\alpha + \beta + \gamma = -\frac{b}{a}$ $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$ $\alpha\beta\gamma = -\frac{d}{a}$ |
| Quadratic Construction | $k[x^2 – Sx + P]$ ($k \neq 0$) | $2$ | $2$ | Family of parabolas scaled by real non-zero factor $k$ | $S = \text{Sum of zeroes} = \alpha + \beta$ $P = \text{Product of zeroes} = \alpha\beta$ |
Chapter-End Exercises – Exercise 2.1 (Page No. 28)
Question 1. The graphs of $y = p(x)$ are given in the figures below, for some polynomials $p(x)$. Find the number of zeroes of $p(x)$, in each case.
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GEOMETRICAL GRAPH SCHEMATICS (EXERCISE 2.1)
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(i) (ii) (iii)
Y Y Y
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-----+----- Line -----*-------+---- -----*-------+---*----
| (Parallel to X) | * | *| *
| | * | * | *
-+- -+- -+-
X X X
Zero Intersections One Intersection Three Intersections
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(iv) (v) (vi)
Y Y Y
| * | * * | * * *
-----*---+-*-- -----*-----+-*-+-*-- -----*----+-*-+-*---
| * * | * |* * | * * * * *
| * | | |
-+- -+- -+-
X X X
Two Intersections Four Intersections Three Intersections (1 Cross + 2 Touches)
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(i)
Answer:
The graph of $y = p(x)$ is a horizontal straight line parallel to the $x$-axis.
Because the graph does not intersect or touch the $x$-axis at any point, the number of real points of intersection is zero.
Therefore, the number of zeroes of $p(x)$ is $0$.
(ii)
Answer:
The graph of $y = p(x)$ is a continuous curve that intersects the axes on the Cartesian plane.
Inspecting the curve shows that it crosses the horizontal $x$-axis at exactly one point.
Therefore, the number of zeroes of $p(x)$ is $1$.
(iii)
Answer:
The graph of $y = p(x)$ is a continuous cubic-type curve.
The curve intersects the horizontal $x$-axis at three distinct points.
Therefore, the number of zeroes of $p(x)$ is $3$.
(iv)
Answer:
The graph of $y = p(x)$ is an upward-opening parabola representing a quadratic polynomial.
The curve crosses the horizontal $x$-axis at two distinct points.
Therefore, the number of zeroes of $p(x)$ is $2$.
(v)
Answer:
The graph of $y = p(x)$ is a continuous wave-like polynomial curve.
The curve intersects the horizontal $x$-axis at four distinct points.
Therefore, the number of zeroes of $p(x)$ is $4$.
(vi)
Answer:
The graph of $y = p(x)$ intersects the $x$-axis at one point and touches the $x$-axis at two other distinct points without crossing below it.
Each point where the graph touches the axis corresponds to a real zero (a root of even multiplicity). Counting all contact points gives:
$$1 \text{ (cross)} + 2 \text{ (tangent touch points)} = 3 \text{ real intersection points}$$
Therefore, the number of zeroes of $p(x)$ is $3$.
Chapter-End Exercises – Exercise 2.2 (Page No. 33)
Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
(i) $x^2 – 2x – 8$
(ii) $4s^2 – 4s + 1$
(iii) $6x^2 – 3 – 7x$
(iv) $4u^2 + 8u$
(v) $t^2 – 15$
(vi) $3x^2 – x – 4$
(i) $x^2 – 2x – 8$ [BOARD EXAM FAVORITE / CBSE 2020, 2023]
Answer:
Let the given quadratic polynomial be:
$$p(x) = x^2 – 2x – 8$$
Step 1: Finding the Zeroes by Factorisation
To find the zeroes, we set $p(x) = 0$ and split the middle term.
We need two numbers whose sum is $-2$ and whose product is $1 \times (-8) = -8$. These numbers are $-4$ and $+2$:
$$\begin{aligned} x^2 – 2x – 8 &= 0 \\ x^2 – 4x + 2x – 8 &= 0 \\ x(x – 4) + 2(x – 4) &= 0 \\ (x – 4)(x + 2) &= 0 \end{aligned}$$
Setting each linear factor to zero gives:
$$\begin{aligned} x – 4 = 0 &\implies x = 4 \\ x + 2 = 0 &\implies x = -2 \end{aligned}$$
Thus, the zeroes of the polynomial are $\alpha = 4$ and $\beta = -2$.
Step 2: Verification of Relationships with Coefficients
Comparing $p(x) = x^2 – 2x – 8$ with the standard form $ax^2 + bx + c$:
$$a = 1, \quad b = -2, \quad c = -8$$
- Verification of Sum of Zeroes:$$\begin{aligned} \text{Sum of zeroes } (\alpha + \beta) &= 4 + (-2) = 2 \\ -\frac{b}{a} &= -\frac{-2}{1} = 2 \end{aligned}$$$$\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \quad \text{[Verified]}$$
- Verification of Product of Zeroes:$$\begin{aligned} \text{Product of zeroes } (\alpha \cdot \beta) &= 4 \times (-2) = -8 \\ \frac{c}{a} &= \frac{-8}{1} = -8 \end{aligned}$$$$\alpha \cdot \beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \quad \text{[Verified]}$$
(ii) $4s^2 – 4s + 1$ [BOARD EXAM FAVORITE / CBSE 2019, 2022]
Answer:
Let the given polynomial in variable $s$ be:
$$p(s) = 4s^2 – 4s + 1$$
Step 1: Finding the Zeroes by Factorisation
We set $p(s) = 0$. We split the middle term using two numbers whose sum is $-4$ and whose product is $4 \times 1 = 4$. These numbers are $-2$ and $-2$:
$$\begin{aligned} 4s^2 – 4s + 1 &= 0 \\ 4s^2 – 2s – 2s + 1 &= 0 \\ 2s(2s – 1) – 1(2s – 1) &= 0 \\ (2s – 1)(2s – 1) &= 0 \\ (2s – 1)^2 &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} 2s – 1 = 0 &\implies s = \frac{1}{2} \\ 2s – 1 = 0 &\implies s = \frac{1}{2} \end{aligned}$$
Thus, the polynomial has two equal zeroes: $\alpha = \frac{1}{2}$ and $\beta = \frac{1}{2}$.
Step 2: Verification of Relationships with Coefficients
Comparing $p(s) = 4s^2 – 4s + 1$ with the standard form $as^2 + bs + c$:
$$a = 4, \quad b = -4, \quad c = 1$$
- Verification of Sum of Zeroes:$$\begin{aligned} \text{Sum of zeroes } (\alpha + \beta) &= \frac{1}{2} + \frac{1}{2} = 1 \\ -\frac{b}{a} &= -\frac{-4}{4} = 1 \end{aligned}$$$$\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } s}{\text{Coefficient of } s^2} \quad \text{[Verified]}$$
- Verification of Product of Zeroes:$$\begin{aligned} \text{Product of zeroes } (\alpha \cdot \beta) &= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \\ \frac{c}{a} &= \frac{1}{4} \end{aligned}$$$$\alpha \cdot \beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } s^2} \quad \text{[Verified]}$$
(iii) $6x^2 – 3 – 7x$ [BOARD EXAM FAVORITE / CBSE 2018, 2021, 2023]
Answer:
First, rearrange the given polynomial in standard descending order of powers of $x$:
$$p(x) = 6x^2 – 7x – 3$$
Step 1: Finding the Zeroes by Factorisation
We set $p(x) = 0$. We look for two numbers whose sum is $-7$ and whose product is $6 \times (-3) = -18$. These numbers are $-9$ and $+2$:
$$\begin{aligned} 6x^2 – 7x – 3 &= 0 \\ 6x^2 – 9x + 2x – 3 &= 0 \\ 3x(2x – 3) + 1(2x – 3) &= 0 \\ (2x – 3)(3x + 1) &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} 2x – 3 = 0 &\implies x = \frac{3}{2} \\ 3x + 1 = 0 &\implies x = -\frac{1}{3} \end{aligned}$$
Thus, the zeroes are $\alpha = \frac{3}{2}$ and $\beta = -\frac{1}{3}$.
Step 2: Verification of Relationships with Coefficients
Comparing $p(x) = 6x^2 – 7x – 3$ with standard form $ax^2 + bx + c$:
$$a = 6, \quad b = -7, \quad c = -3$$
- Verification of Sum of Zeroes:$$\begin{aligned} \text{Sum of zeroes } (\alpha + \beta) &= \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9 – 2}{6} = \frac{7}{6} \\ -\frac{b}{a} &= -\frac{-7}{6} = \frac{7}{6} \end{aligned}$$$$\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \quad \text{[Verified]}$$
- Verification of Product of Zeroes:$$\begin{aligned} \text{Product of zeroes } (\alpha \cdot \beta) &= \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{3}{6} = -\frac{1}{2} \\ \frac{c}{a} &= \frac{-3}{6} = -\frac{1}{2} \end{aligned}$$$$\alpha \cdot \beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \quad \text{[Verified]}$$
(iv) $4u^2 + 8u$ [BOARD EXAM FAVORITE / CBSE 2020]
Answer:
Let the given polynomial in variable $u$ be:
$$p(u) = 4u^2 + 8u$$
Step 1: Finding the Zeroes by Factorisation
We set $p(u) = 0$ and factor out the common algebraic term $4u$:
$$\begin{aligned} 4u^2 + 8u &= 0 \\ 4u(u + 2) &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} 4u = 0 &\implies u = 0 \\ u + 2 = 0 &\implies u = -2 \end{aligned}$$
Thus, the zeroes of the polynomial are $\alpha = 0$ and $\beta = -2$.
Step 2: Verification of Relationships with Coefficients
Comparing $p(u) = 4u^2 + 8u + 0$ with the standard form $au^2 + bu + c$:
$$a = 4, \quad b = 8, \quad c = 0$$
- Verification of Sum of Zeroes:$$\begin{aligned} \text{Sum of zeroes } (\alpha + \beta) &= 0 + (-2) = -2 \\ -\frac{b}{a} &= -\frac{8}{4} = -2 \end{aligned}$$$$\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } u}{\text{Coefficient of } u^2} \quad \text{[Verified]}$$
- Verification of Product of Zeroes:$$\begin{aligned} \text{Product of zeroes } (\alpha \cdot \beta) &= 0 \times (-2) = 0 \\ \frac{c}{a} &= \frac{0}{4} = 0 \end{aligned}$$$$\alpha \cdot \beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } u^2} \quad \text{[Verified]}$$
(v) $t^2 – 15$ [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
Let the given polynomial in variable $t$ be:
$$p(t) = t^2 – 15$$
Step 1: Finding the Zeroes by Factorisation
Using the difference-of-squares identity $a^2 – b^2 = (a – b)(a + b)$, we set $p(t) = 0$:
$$\begin{aligned} t^2 – 15 &= 0 \\ t^2 – (\sqrt{15})^2 &= 0 \\ (t – \sqrt{15})(t + \sqrt{15}) &= 0 \end{aligned}$$
Setting each linear factor to zero:
$$\begin{aligned} t – \sqrt{15} = 0 &\implies t = \sqrt{15} \\ t + \sqrt{15} = 0 &\implies t = -\sqrt{15} \end{aligned}$$
Thus, the zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.
Step 2: Verification of Relationships with Coefficients
Comparing $p(t) = 1t^2 + 0t – 15$ with the standard form $at^2 + bt + c$:
$$a = 1, \quad b = 0, \quad c = -15$$
- Verification of Sum of Zeroes:$$\begin{aligned} \text{Sum of zeroes } (\alpha + \beta) &= \sqrt{15} + (-\sqrt{15}) = 0 \\ -\frac{b}{a} &= -\frac{0}{1} = 0 \end{aligned}$$$$\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } t}{\text{Coefficient of } t^2} \quad \text{[Verified]}$$
- Verification of Product of Zeroes:$$\begin{aligned} \text{Product of zeroes } (\alpha \cdot \beta) &= (\sqrt{15}) \times (-\sqrt{15}) = -15 \\ \frac{c}{a} &= \frac{-15}{1} = -15 \end{aligned}$$$$\alpha \cdot \beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } t^2} \quad \text{[Verified]}$$
(vi) $3x^2 – x – 4$ [BOARD EXAM FAVORITE / CBSE 2020, 2022]
Answer:
Let the given quadratic polynomial be:
$$p(x) = 3x^2 – x – 4$$
Step 1: Finding the Zeroes by Factorisation
We set $p(x) = 0$. We split the middle term using two numbers whose sum is $-1$ and whose product is $3 \times (-4) = -12$. These numbers are $-4$ and $+3$:
$$\begin{aligned} 3x^2 – x – 4 &= 0 \\ 3x^2 – 4x + 3x – 4 &= 0 \\ x(3x – 4) + 1(3x – 4) &= 0 \\ (3x – 4)(x + 1) &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} 3x – 4 = 0 &\implies x = \frac{4}{3} \\ x + 1 = 0 &\implies x = -1 \end{aligned}$$
Thus, the zeroes are $\alpha = \frac{4}{3}$ and $\beta = -1$.
Step 2: Verification of Relationships with Coefficients
Comparing $p(x) = 3x^2 – x – 4$ with the standard form $ax^2 + bx + c$:
$$a = 3, \quad b = -1, \quad c = -4$$
- Verification of Sum of Zeroes:$$\begin{aligned} \text{Sum of zeroes } (\alpha + \beta) &= \frac{4}{3} + (-1) = \frac{4 – 3}{3} = \frac{1}{3} \\ -\frac{b}{a} &= -\frac{-1}{3} = \frac{1}{3} \end{aligned}$$$$\alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \quad \text{[Verified]}$$
- Verification of Product of Zeroes:$$\begin{aligned} \text{Product of zeroes } (\alpha \cdot \beta) &= \frac{4}{3} \times (-1) = -\frac{4}{3} \\ \frac{c}{a} &= \frac{-4}{3} = -\frac{4}{3} \end{aligned}$$$$\alpha \cdot \beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \quad \text{[Verified]}$$
Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
(i) $\frac{1}{4}, -1$
(ii) $\sqrt{2}, \frac{1}{3}$
(iii) $0, \sqrt{5}$
(iv) $1, 1$
(v) $-\frac{1}{4}, \frac{1}{4}$
(vi) $4, 1$
(i) $\frac{1}{4}, -1$ [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
Given Data:
- Sum of zeroes, $S = \alpha + \beta = \frac{1}{4}$
- Product of zeroes, $P = \alpha\beta = -1$
Formula Required:
A quadratic polynomial with sum of zeroes $S$ and product of zeroes $P$ is given by:
$$p(x) = k\left[x^2 – Sx + P\right] \quad \text{where } k \text{ is any non-zero real constant.}$$
Step-by-Step Calculation:
Substituting $S = \frac{1}{4}$ and $P = -1$:
$$\begin{aligned} p(x) &= k\left[x^2 – \left(\frac{1}{4}\right)x + (-1)\right] \\ &= k\left[x^2 – \frac{1}{4}x – 1\right] \\ &= \frac{k}{4}\left[4x^2 – x – 4\right] \end{aligned}$$
Choosing the non-zero constant $k = 4$ to clear the fractional denominator:
$$p(x) = 4x^2 – x – 4$$
Final Statement:
A required quadratic polynomial is $4x^2 – x – 4$.
(ii) $\sqrt{2}, \frac{1}{3}$ [BOARD EXAM FAVORITE / CBSE 2020, 2022]
Answer:
Given Data:
- Sum of zeroes, $S = \alpha + \beta = \sqrt{2}$
- Product of zeroes, $P = \alpha\beta = \frac{1}{3}$
Formula Required:
$$p(x) = k\left[x^2 – Sx + P\right] \quad (k \neq 0)$$
Step-by-Step Calculation:
Substituting $S = \sqrt{2}$ and $P = \frac{1}{3}$:
$$\begin{aligned} p(x) &= k\left[x^2 – \sqrt{2}x + \frac{1}{3}\right] \\ &= \frac{k}{3}\left[3x^2 – 3\sqrt{2}x + 1\right] \end{aligned}$$
Choosing $k = 3$:
$$p(x) = 3x^2 – 3\sqrt{2}x + 1$$
Final Statement:
A required quadratic polynomial is $3x^2 – 3\sqrt{2}x + 1$.
(iii) $0, \sqrt{5}$
Answer:
Given Data:
- Sum of zeroes, $S = \alpha + \beta = 0$
- Product of zeroes, $P = \alpha\beta = \sqrt{5}$
Formula Required:
$$p(x) = k\left[x^2 – Sx + P\right] \quad (k \neq 0)$$
Step-by-Step Calculation:
Substituting $S = 0$ and $P = \sqrt{5}$:
$$\begin{aligned} p(x) &= k\left[x^2 – (0)x + \sqrt{5}\right] \\ &= k\left[x^2 + \sqrt{5}\right] \end{aligned}$$
Choosing $k = 1$:
$$p(x) = x^2 + \sqrt{5}$$
Final Statement:
A required quadratic polynomial is $x^2 + \sqrt{5}$.
(iv) $1, 1$
Answer:
Given Data:
- Sum of zeroes, $S = \alpha + \beta = 1$
- Product of zeroes, $P = \alpha\beta = 1$
Formula Required:
$$p(x) = k\left[x^2 – Sx + P\right] \quad (k \neq 0)$$
Step-by-Step Calculation:
Substituting $S = 1$ and $P = 1$:
$$\begin{aligned} p(x) &= k\left[x^2 – (1)x + 1\right] \\ &= k\left[x^2 – x + 1\right] \end{aligned}$$
Choosing $k = 1$:
$$p(x) = x^2 – x + 1$$
Final Statement:
A required quadratic polynomial is $x^2 – x + 1$.
(v) $-\frac{1}{4}, \frac{1}{4}$ [BOARD EXAM FAVORITE / CBSE 2021, 2023]
Answer:
Given Data:
- Sum of zeroes, $S = \alpha + \beta = -\frac{1}{4}$
- Product of zeroes, $P = \alpha\beta = \frac{1}{4}$
Formula Required:
$$p(x) = k\left[x^2 – Sx + P\right] \quad (k \neq 0)$$
Step-by-Step Calculation:
Substituting $S = -\frac{1}{4}$ and $P = \frac{1}{4}$:
$$\begin{aligned} p(x) &= k\left[x^2 – \left(-\frac{1}{4}\right)x + \frac{1}{4}\right] \\ &= k\left[x^2 + \frac{1}{4}x + \frac{1}{4}\right] \\ &= \frac{k}{4}\left[4x^2 + x + 1\right] \end{aligned}$$
Choosing $k = 4$:
$$p(x) = 4x^2 + x + 1$$
Final Statement:
A required quadratic polynomial is $4x^2 + x + 1$.
(vi) $4, 1$
Answer:
Given Data:
- Sum of zeroes, $S = \alpha + \beta = 4$
- Product of zeroes, $P = \alpha\beta = 1$
Formula Required:
$$p(x) = k\left[x^2 – Sx + P\right] \quad (k \neq 0)$$
Step-by-Step Calculation:
Substituting $S = 4$ and $P = 1$:
$$\begin{aligned} p(x) &= k\left[x^2 – (4)x + 1\right] \\ &= k\left[x^2 – 4x + 1\right] \end{aligned}$$
Choosing $k = 1$:
$$p(x) = x^2 – 4x + 1$$
Final Statement:
A required quadratic polynomial is $x^2 – 4x + 1$.
Advanced Algebraic Identities & Symmetric Functions of Zeroes
To solve Higher Order Thinking Skills (HOTS) problems in quadratic polynomials, symmetric expressions of the roots $\alpha$ and $\beta$ must be rewritten purely in terms of the elementary symmetric polynomials: Sum ($S = \alpha + \beta = -\frac{b}{a}$) and Product ($P = \alpha\beta = \frac{c}{a}$).
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STANDARD ALGEBRAIC REDUCTION TOOLKIT FOR ROOTS (a, b)
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1. Sum of Squares: a^2 + b^2 = (a + b)^2 - 2ab
2. Difference of Roots: |a - b| = sqrt[(a + b)^2 - 4ab]
3. Sum of Cubes: a^3 + b^3 = (a + b)^3 - 3ab(a + b) = (a + b)(a^2 - ab + b^2)
4. Difference of Cubes: a^3 - b^3 = (a - b)(a^2 + ab + b^2) = (a - b)[(a + b)^2 - ab]
5. Sum of Reciprocals: 1/a + 1/b = (a + b) / (ab)
6. Sum of Inverse Squares: 1/a^2 + 1/b^2 = (a^2 + b^2) / (a^2 * b^2) = [(a + b)^2 - 2ab] / (ab)^2
7. Biquadratic Root Sum: a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 = [(a + b)^2 - 2ab]^2 - 2(ab)^2
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15 High-Yield Board Exam FAQs
FAQ 1 (HOTS). If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 – 5x + k$ such that $\alpha – \beta = 1$, find the value of $k$. [CBSE 2020, 2023]
Answer:
Comparing $p(x) = x^2 – 5x + k$ with $ax^2 + bx + c$:
$$a = 1, \quad b = -5, \quad c = k$$
Using the coefficient relations:
$$\alpha + \beta = -\frac{b}{a} = -\frac{-5}{1} = 5$$
$$\alpha\beta = \frac{c}{a} = \frac{k}{1} = k$$
We are given $\alpha – \beta = 1$. Using the algebraic identity:
$$(\alpha – \beta)^2 = (\alpha + \beta)^2 – 4\alpha\beta$$
Substituting the known values:
$$\begin{aligned} (1)^2 &= (5)^2 – 4(k) \\ 1 &= 25 – 4k \\ 4k &= 25 – 1 \\ 4k &= 24 \\ k &= 6 \end{aligned}$$
Therefore, the value of $k$ is $6$.
FAQ 2 (HOTS). If $\alpha$ and $\beta$ are the zeroes of the polynomial $f(x) = 2x^2 + 5x + k$, find the value of $k$ such that $\alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4}$. [CBSE 2019, 2022]
Answer:
For the polynomial $f(x) = 2x^2 + 5x + k$, the coefficients are $a = 2, b = 5, c = k$.
$$\alpha + \beta = -\frac{5}{2}, \quad \alpha\beta = \frac{k}{2}$$
Rewrite the given condition using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta$:
$$\begin{aligned} (\alpha^2 + \beta^2) + \alpha\beta &= \frac{21}{4} \\ \left[(\alpha + \beta)^2 – 2\alpha\beta\right] + \alpha\beta &= \frac{21}{4} \\ (\alpha + \beta)^2 – \alpha\beta &= \frac{21}{4} \end{aligned}$$
Substituting the values of $(\alpha + \beta)$ and $\alpha\beta$:
$$\begin{aligned} \left(-\frac{5}{2}\right)^2 – \frac{k}{2} &= \frac{21}{4} \\ \frac{25}{4} – \frac{k}{2} &= \frac{21}{4} \\ \frac{25}{4} – \frac{21}{4} &= \frac{k}{2} \\ \frac{4}{4} &= \frac{k}{2} \\ 1 &= \frac{k}{2} \implies k = 2 \end{aligned}$$
Therefore, the value of $k$ is $2$.
FAQ 3 (Algebraic Transformation). If $\alpha$ and $\beta$ are the zeroes of $p(x) = 3x^2 – 4x + 1$, form a quadratic polynomial whose zeroes are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$. [CBSE 2021]
Answer:
From $p(x) = 3x^2 – 4x + 1$, we have:
$$\alpha + \beta = \frac{4}{3}, \quad \alpha\beta = \frac{1}{3}$$
Let the new zeroes be $\alpha’ = \frac{\alpha}{\beta}$ and $\beta’ = \frac{\beta}{\alpha}$.
- Sum of new zeroes ($S’$):
$$\begin{aligned} S’ &= \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 – 2\alpha\beta}{\alpha\beta} \\ S’ &= \frac{\left(\frac{4}{3}\right)^2 – 2\left(\frac{1}{3}\right)}{\frac{1}{3}} = \frac{\frac{16}{9} – \frac{2}{3}}{\frac{1}{3}} = \frac{\frac{16 – 6}{9}}{\frac{1}{3}} = \frac{\frac{10}{9}}{\frac{1}{3}} = \frac{10}{3} \end{aligned}$$
- Product of new zeroes ($P’$):
$$P’ = \left(\frac{\alpha}{\beta}\right) \times \left(\frac{\beta}{\alpha}\right) = 1$$
The required quadratic polynomial is:
$$q(x) = k[x^2 – S’x + P’] = k\left[x^2 – \frac{10}{3}x + 1\right]$$
Setting $k = 3$:
$$q(x) = \underline{3x^2 – 10x + 3}$$
FAQ 4 (Reciprocal Roots). If one zero of the polynomial $(a^2 + 9)x^2 + 13x + 6a$ is the reciprocal of the other, find the value of $a$. [CBSE 2018, 2020]
Answer:
Let one zero be $\alpha$. Then the other zero is $\beta = \frac{1}{\alpha}$.
For the polynomial $p(x) = (a^2 + 9)x^2 + 13x + 6a$:
- Coefficient of $x^2$: $A = a^2 + 9$
- Coefficient of $x$: $B = 13$
- Constant term: $C = 6a$
Using the product of zeroes:
$$\begin{aligned} \alpha \times \beta &= \frac{C}{A} \\ \alpha \times \frac{1}{\alpha} &= \frac{6a}{a^2 + 9} \\ 1 &= \frac{6a}{a^2 + 9} \\ a^2 + 9 &= 6a \\ a^2 – 6a + 9 &= 0 \\ (a – 3)^2 &= 0 \implies a = 3 \end{aligned}$$
Therefore, the value of $a$ is $3$.
FAQ 5 (Equal Magnitude, Opposite Sign). Find the value of $m$ if the zeroes of the polynomial $p(x) = 2x^2 – (m^2 – 4m)x – 7$ are equal in magnitude but opposite in sign.
Answer:
Let the zeroes be $\alpha$ and $-\alpha$.
The sum of zeroes is:
$$\text{Sum} = \alpha + (-\alpha) = 0$$
Using the coefficient relation:
$$\begin{aligned} -\frac{B}{A} &= 0 \implies -\frac{-(m^2 – 4m)}{2} = 0 \\ m^2 – 4m &= 0 \\ m(m – 4) &= 0 \end{aligned}$$
This gives $m = 0$ or $m = 4$.
If $m = 0$, $p(x) = 2x^2 – 7$, which gives zeroes $\pm \sqrt{\frac{7}{2}}$.
If $m = 4$, $p(x) = 2x^2 – 7$, which also gives zeroes $\pm \sqrt{\frac{7}{2}}$.
Therefore, $m = 0$ or $m = 4$.
FAQ 6 (Assertion-Reasoning).
- Assertion (A): The polynomial $p(x) = x^2 + 4x + 5$ has no real zeroes.
- Reason (R): A quadratic polynomial $ax^2 + bx + c$ has no real zeroes if its discriminant $D = b^2 – 4ac < 0$.
Answer:
For the polynomial $p(x) = x^2 + 4x + 5$, the discriminant is:
$$D = b^2 – 4ac = (4)^2 – 4(1)(5) = 16 – 20 = -4 < 0$$
Since $D < 0$, the parabola does not cross or touch the $x$-axis, meaning $p(x)$ has no real zeroes.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).
FAQ 7 (Geometric Analysis). If the graph of a quadratic polynomial $y = ax^2 + bx + c$ opens downward and does not touch or cross the $x$-axis, determine the signs of $a$ and $D = b^2 – 4ac$.
Answer:
- Because the parabola opens downward, the leading coefficient must be negative: $a < 0$.
- Because the graph does not intersect or touch the $x$-axis, it has no real zeroes, which means the discriminant must be strictly negative: $D = b^2 – 4ac < 0$.
FAQ 8 (HOTS Ratio Problem). If the zeroes of the polynomial $p(x) = ax^2 + bx + c$ are in the ratio $m : n$, prove that $mnb^2 = (m + n)^2 ac$. [CBSE 2019]
Answer:
Let the zeroes be $\alpha = mk$ and $\beta = nk$, where $k \neq 0$.
$$\alpha + \beta = mk + nk = (m + n)k = -\frac{b}{a} \implies k = -\frac{b}{a(m + n)} \quad \text{— (Equation 1)}$$
$$\alpha\beta = (mk)(nk) = mnk^2 = \frac{c}{a} \quad \text{— (Equation 2)}$$
Substitute Equation 1 into Equation 2:
$$\begin{aligned} mn \left[-\frac{b}{a(m + n)}\right]^2 &= \frac{c}{a} \\ mn \frac{b^2}{a^2 (m + n)^2} &= \frac{c}{a} \end{aligned}$$
Multiply both sides by $a^2(m + n)^2$:
$$mnb^2 = ac(m + n)^2$$
Hence, $mnb^2 = (m + n)^2 ac$ is proved.
FAQ 9 (Cubic Identity). If $\alpha, \beta, \gamma$ are the zeroes of the cubic polynomial $p(x) = 2x^3 – 5x^2 – 14x + 8$, find the value of $\alpha^2 + \beta^2 + \gamma^2$.
Answer:
For the cubic polynomial $p(x) = 2x^3 – 5x^2 – 14x + 8$, the coefficients are $a = 2, b = -5, c = -14, d = 8$.
$$\sum \alpha = \alpha + \beta + \gamma = -\frac{-5}{2} = \frac{5}{2}$$
$$\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{-14}{2} = -7$$
Using the algebraic expansion identity:
$$(\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)$$
Rearranging gives:
$$\begin{aligned} \alpha^2 + \beta^2 + \gamma^2 &= (\alpha + \beta + \gamma)^2 – 2(\alpha\beta + \beta\gamma + \gamma\alpha) \\ &= \left(\frac{5}{2}\right)^2 – 2(-7) \\ &= \frac{25}{4} + 14 = \frac{25 + 56}{4} = \frac{81}{4} \end{aligned}$$
Therefore, $\alpha^2 + \beta^2 + \gamma^2 = \frac{81}{4}$.
FAQ 10 (Zero Multiplicity). Explain how the graph of a quadratic polynomial behaves at the $x$-axis when it has two equal real zeroes.
Answer:
When a quadratic polynomial $p(x) = ax^2 + bx + c$ has two equal real zeroes ($\alpha = \beta$), its discriminant is zero ($D = b^2 – 4ac = 0$).
Geometrically, the parabola touches the $x$-axis at exactly one point $(x = \alpha)$ without crossing through it. The $x$-axis acts as a tangent to the curve at the parabola’s vertex $\left(-\frac{b}{2a}, 0\right)$.
FAQ 11 (Linear Transformation). If $\alpha$ and $\beta$ are the zeroes of $p(x) = x^2 – p(x + 1) – c$, show that $(\alpha + 1)(\beta + 1) = 1 – c$. [CBSE 2018, 2020]
Answer:
First, rewrite $p(x)$ in standard form:
$$p(x) = x^2 – px – (p + c)$$
Here, $a = 1, b = -p, c’ = -(p + c)$.
$$\alpha + \beta = -(-p) = p, \quad \alpha\beta = -(p + c)$$
Expanding the expression:
$$\begin{aligned} (\alpha + 1)(\beta + 1) &= \alpha\beta + \alpha + \beta + 1 \\ &= \alpha\beta + (\alpha + \beta) + 1 \end{aligned}$$
Substituting $\alpha + \beta = p$ and $\alpha\beta = -(p + c)$:
$$\begin{aligned} (\alpha + 1)(\beta + 1) &= -(p + c) + p + 1 \\ &= -p – c + p + 1 \\ &= 1 – c \end{aligned}$$
Hence, $(\alpha + 1)(\beta + 1) = 1 – c$ is verified.
FAQ 12 (Case-Study Application). An architect designs an arched entrance to a bridge modelled by the quadratic equation $h(x) = -\frac{1}{2}x^2 + 2x + 6$, where $h(x)$ is the height in metres and $x$ is the horizontal distance in metres from the origin. Find the horizontal width of the arch at ground level ($h(x) = 0$).
Answer:
To find the span at ground level, set $h(x) = 0$:
$$\begin{aligned} -\frac{1}{2}x^2 + 2x + 6 &= 0 \\ x^2 – 4x – 12 &= 0 \\ (x – 6)(x + 2) &= 0 \end{aligned}$$
This yields $x = 6$ and $x = -2$.
The horizontal width is the absolute distance between the two $x$-intercepts:
$$\text{Width} = |x_2 – x_1| = |6 – (-2)| = 6 + 2 = 8\text{ metres}$$
Therefore, the width of the entrance at ground level is $8\text{ metres}$.
FAQ 13 (HOTS Difference of Squares). If the sum of the squares of the zeroes of the quadratic polynomial $p(x) = x^2 – 8x + k$ is $40$, find the value of $k$. [CBSE 2020]
Answer:
For $p(x) = x^2 – 8x + k$, the coefficients are $a = 1, b = -8, c = k$.
$$\alpha + \beta = -\frac{-8}{1} = 8, \quad \alpha\beta = k$$
We are given $\alpha^2 + \beta^2 = 40$.
$$\begin{aligned} \alpha^2 + \beta^2 &= (\alpha + \beta)^2 – 2\alpha\beta \\ 40 &= (8)^2 – 2k \\ 40 &= 64 – 2k \\ 2k &= 64 – 40 \\ 2k &= 24 \\ k &= 12 \end{aligned}$$
Therefore, the value of $k$ is $12$.
FAQ 14 (Polynomial Sign Properties). Can a quadratic polynomial $ax^2 + bx + c$ with positive coefficients ($a > 0, b > 0, c > 0$) have positive zeroes? Explain.
Answer:
No. Let $\alpha$ and $\beta$ be the zeroes of $ax^2 + bx + c$.
Using the coefficient relations:
- $\alpha + \beta = -\frac{b}{a} < 0$ (since $a, b > 0$)
- $\alpha\beta = \frac{c}{a} > 0$ (since $a, c > 0$)
A positive product ($\alpha\beta > 0$) means both zeroes must share the same sign (both positive or both negative). Since their sum is negative ($\alpha + \beta < 0$), both zeroes must be negative.
Therefore, a quadratic polynomial with all positive coefficients cannot have positive zeroes.
FAQ 15 (HOTS Reciprocal Transformation). If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = ax^2 + bx + c$, find a quadratic polynomial whose zeroes are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
Answer:
For $f(x) = ax^2 + bx + c$, we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
Let the new zeroes be $\alpha’ = \frac{1}{\alpha}$ and $\beta’ = \frac{1}{\beta}$.
- Sum of new zeroes ($S’$):
$$S’ = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-\frac{b}{a}}{\frac{c}{a}} = -\frac{b}{c}$$
- Product of new zeroes ($P’$):
$$P’ = \frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{\frac{c}{a}} = \frac{a}{c}$$
The required quadratic polynomial is:
$$g(x) = k[x^2 – S’x + P’] = k\left[x^2 – \left(-\frac{b}{c}\right)x + \frac{a}{c}\right] = \frac{k}{c}\left[cx^2 + bx + a\right]$$
Choosing $k = c$:
$$g(x) = \underline{cx^2 + bx + a}$$
To score full marks on Polynomials questions in the CBSE Class 10 Board Examinations, present every step systematically. When factorising quadratics, clearly show the middle-term split before writing down the linear factors. When verifying relations, always compute and state $\alpha + \beta$ alongside $-\frac{b}{a}$, and $\alpha\beta$ alongside $\frac{c}{a}$ with explicit variable labels. When forming a quadratic polynomial from given numbers, write the complete general expression $k[x^2 – Sx + P]$ and explicitly define $k$ as a non-zero real number to satisfy full-credit marking criteria. For graph-based questions, state the exact number of contact points with the $x$-axis and provide a clear geometric justification for each answer.