📘 SCERT Assam Class 10 Maths Chapter 2 Polynomials Exercise 2.2 Solutions (2026–2027) English Medium
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📚 Chapter 2: Polynomials – Quick Revision
Before solving Exercise 2.2, remember:
A polynomial is an algebraic expression
Zeros of polynomial = values of x where p(x) = 0
✍️ Exercise 2.2 Solutions (Full Explanation)
Question 1: Find the zeros of the following quadratic polynomials and verify the relationship between zeros and coefficients.
- (i) \(x^2 - 2x - 8\)
- (ii) \(4s^2 - 4s + 1\)
- (iii) \(6x^2 - 7x - 3\)
- (iv) \(4u^2 + 8u\)
- (v) \(t^2 - 15\)
- (vi) \(3x^2 - x - 4\)
- (vii) \(x^2 + 7x + 12\)
- (viii) \(x^2 - 4x + 3\)
- (ix) \(x^2 - 6x - 7\)
- (x) \(2x^2 - 5x - 7\)
\[ x^2 - 2x - 8 = x^2 - (4 - 2)x - 8 \] \[ = x^2 - 4x + 2x - 8 \] \[ = x(x - 4) + 2(x - 4) \] \[ = (x - 4)(x + 2) \]
The value of the polynomial becomes zero when: \[ x - 4 = 0 \quad \text{and} \quad x + 2 = 0 \]
\[ \therefore x = 4 \quad \text{and} \quad x = -2 \]
Verification:
Sum of zeroes: \[ = 4 + (-2) = 2 = -\frac{-2}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = 4 \times (-2) = -8 = \frac{-8}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ 4s^2 - 4s + 1 = 4s^2 - (2 + 2)s + 1 \] \[ = 4s^2 - 2s - 2s + 1 \] \[ = 2s(2s - 1) - 1(2s - 1) \] \[ = (2s - 1)(2s - 1) \]
The value of the polynomial becomes zero when: \[ 2s - 1 = 0 \]
\[ \therefore s = \frac{1}{2}, \frac{1}{2} \]
Verification:
Sum of zeroes: \[ = \frac{1}{2} + \frac{1}{2} = 1 = -\frac{-4}{4} = -\frac{\text{Coefficient of } s}{\text{Coefficient of } s^2} \]
Product of zeroes: \[ = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } s^2} \]
\[ 6x^2 - 3 - 7x = 6x^2 - 7x - 3 \] \[ = 6x^2 - (9 - 2)x - 3 \] \[ = 6x^2 - 9x + 2x - 3 \] \[ = 3x(2x - 3) + 1(2x - 3) \] \[ = (2x - 3)(3x + 1) \]
The value of the polynomial becomes zero when: \[ 2x - 3 = 0 \quad \text{and} \quad 3x + 1 = 0 \]
\[ \therefore x = \frac{3}{2} \quad \text{and} \quad x = -\frac{1}{3} \]
Verification:
Sum of zeroes: \[ = \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9 - 2}{6} = \frac{7}{6} = -\frac{-7}{6} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{1}{2} = \frac{-3}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ 4u^2 + 8u = 4u^2 + 8u + 0 \] \[ = 4u(u + 2) \]
The value of the polynomial becomes zero when: \[ 4u = 0 \quad \text{and} \quad u + 2 = 0 \]
\[ \therefore u = 0 \quad \text{and} \quad u = -2 \]
Verification:
Sum of zeroes: \[ = 0 + (-2) = -2 = -\frac{8}{4} = -\frac{\text{Coefficient of } u}{\text{Coefficient of } u^2} \]
Product of zeroes: \[ = 0 \times (-2) = 0 = \frac{0}{4} = \frac{\text{Constant term}}{\text{Coefficient of } u^2} \]
\[ t^2 - 15 = t^2 - (\sqrt{15})^2 \] \[ = (t - \sqrt{15})(t + \sqrt{15}) \]
The value of the polynomial becomes zero when: \[ t - \sqrt{15} = 0 \quad \text{and} \quad t + \sqrt{15} = 0 \]
\[ \therefore t = \sqrt{15} \quad \text{and} \quad t = -\sqrt{15} \]
Verification:
Sum of zeroes: \[ = \sqrt{15} + (-\sqrt{15}) = 0 = -\frac{0}{1} = -\frac{\text{Coefficient of } t}{\text{Coefficient of } t^2} \]
Product of zeroes: \[ = \sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{\text{Constant term}}{\text{Coefficient of } t^2} \]
\[ 3x^2 - x - 4 = 3x^2 - (4 - 3)x - 4 \] \[ = 3x^2 - 4x + 3x - 4 \] \[ = x(3x - 4) + 1(3x - 4) \] \[ = (3x - 4)(x + 1) \]
The value of the polynomial becomes zero when: \[ 3x - 4 = 0 \quad \text{and} \quad x + 1 = 0 \]
\[ \therefore x = \frac{4}{3} \quad \text{and} \quad x = -1 \]
Verification:
Sum of zeroes: \[ = \frac{4}{3} + (-1) = \frac{4 - 3}{3} = \frac{1}{3} = -\frac{-1}{3} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = \frac{4}{3} \times (-1) = -\frac{4}{3} = \frac{-4}{3} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ x^2 + 7x + 12 = x^2 + (3 + 4)x + 12 \] \[ = x^2 + 3x + 4x + 12 \] \[ = x(x + 3) + 4(x + 3) \] \[ = (x + 3)(x + 4) \]
The value of the polynomial becomes zero when: \[ x + 3 = 0 \quad \text{and} \quad x + 4 = 0 \]
\[ \therefore x = -3 \quad \text{and} \quad x = -4 \]
Verification:
Sum of zeroes: \[ = (-3) + (-4) = -7 = -\frac{7}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = (-3) \times (-4) = 12 = \frac{12}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ x^2 - 4x + 3 = x^2 - (1 + 3)x + 3 \] \[ = x^2 - x - 3x + 3 \] \[ = x(x - 1) - 3(x - 1) \] \[ = (x - 1)(x - 3) \]
The value of the polynomial becomes zero when: \[ x - 1 = 0 \quad \text{and} \quad x - 3 = 0 \]
\[ \therefore x = 1 \quad \text{and} \quad x = 3 \]
Verification:
Sum of zeroes: \[ = 1 + 3 = 4 = -\frac{-4}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = 1 \times 3 = 3 = \frac{3}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ x^2 - 6x - 7 = x^2 - (7 - 1)x - 7 \] \[ = x^2 - 7x + x - 7 \] \[ = x(x - 7) + 1(x - 7) \] \[ = (x - 7)(x + 1) \]
The value of the polynomial becomes zero when: \[ x - 7 = 0 \quad \text{and} \quad x + 1 = 0 \]
\[ \therefore x = 7 \quad \text{and} \quad x = -1 \]
Verification:
Sum of zeroes: \[ = 7 + (-1) = 6 = -\frac{-6}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = 7 \times (-1) = -7 = \frac{-7}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
\[ 2x^2 - 5x - 7 = 2x^2 - (7 - 2)x - 7 \] \[ = 2x^2 - 7x + 2x - 7 \] \[ = x(2x - 7) + 1(2x - 7) \] \[ = (2x - 7)(x + 1) \]
The value of the polynomial becomes zero when: \[ 2x - 7 = 0 \quad \text{and} \quad x + 1 = 0 \]
\[ \therefore x = \frac{7}{2} \quad \text{and} \quad x = -1 \]
Verification:
Sum of zeroes: \[ = \frac{7}{2} + (-1) = \frac{7 - 2}{2} = \frac{5}{2} = -\frac{-5}{2} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \]
Product of zeroes: \[ = \frac{7}{2} \times (-1) = -\frac{7}{2} = \frac{-7}{2} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \]
(i) \( \frac{1}{4}, -1 \)
Solution:
From the formulas of sum and product of zeroes, we know:
Sum of zeroes \(= \alpha + \beta\)
Product of zeroes \(= \alpha \beta\)
\[ \alpha + \beta = \frac{1}{4} \] \[ \alpha \beta = -1 \]
If \( \alpha \) and \( \beta \) are zeroes of a quadratic polynomial, then:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - \frac{1}{4}x - 1 = 0 \]
Multiplying by 4: \[ 4x^2 - x - 4 = 0 \]
(ii) \( \sqrt{2}, \frac{1}{3} \)
Solution:
Sum of zeroes \(= \alpha + \beta = \sqrt{2}\)
Product of zeroes \(= \alpha \beta = \frac{1}{3}\)
If \( \alpha \) and \( \beta \) are zeroes of a quadratic polynomial, then:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - \sqrt{2}x + \frac{1}{3} = 0 \]
Multiplying by 3: \[ 3x^2 - 3\sqrt{2}x + 1 = 0 \]
(iii) \( 0, \sqrt{5} \)
Solution:
Given,
Sum of zeroes \(= \alpha + \beta = 0\)
Product of zeroes \(= \alpha \beta = \sqrt{5}\)
If \( \alpha \) and \( \beta \) are zeroes of a quadratic polynomial, then:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - (0)x + \sqrt{5} = 0 \]
\[ x^2 + \sqrt{5} = 0 \]
(iv) \( 1, 1 \)
Solution:
Given,
Sum of zeroes \(= \alpha + \beta = 1\)
Product of zeroes \(= \alpha \beta = 1\)
If \( \alpha \) and \( \beta \) are zeroes of a quadratic polynomial, then:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - x + 1 = 0 \]
(v) \( -\frac{1}{4}, \frac{1}{4} \)
Solution:
Given,
Sum of zeroes \(= \alpha + \beta = -\frac{1}{4}\)
Product of zeroes \(= \alpha \beta = \frac{1}{4}\)
If \( \alpha \) and \( \beta \) are zeroes of a quadratic polynomial, then:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4} = 0 \]
Multiplying by 4: \[ 4x^2 + x + 1 = 0 \]
(vi) \( 4, 1 \)
Solution:
Given,
Sum of zeroes \(= \alpha + \beta = 4\)
Product of zeroes \(= \alpha \beta = 1\)
If \( \alpha \) and \( \beta \) are zeroes of a quadratic polynomial, then:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - 4x + 1 = 0 \]
(i) \( -4 \) and \( \frac{3}{2} \)
Solution:
Let the zeroes be \( \alpha = -4 \) and \( \beta = \frac{3}{2} \)
\[ \text{Sum of zeroes } (\alpha + \beta) = -4 + \frac{3}{2} = \frac{-8 + 3}{2} = -\frac{5}{2} \]
\[ \text{Product of zeroes } (\alpha \beta) = -4 \times \frac{3}{2} = -2 \times 3 = -6 \]
The required quadratic polynomial is given by:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - \left(-\frac{5}{2}\right)x + (-6) = 0 \]
\[ x^2 + \frac{5}{2}x - 6 = 0 \]
Multiplying the equation by 2: \[ 2x^2 + 5x - 12 = 0 \]
(ii) \( 5 \) and \( -2 \)
Solution:
Let the zeroes be \( \alpha = 5 \) and \( \beta = -2 \)
\[ \text{Sum of zeroes } (\alpha + \beta) = 5 + (-2) = 3 \]
\[ \text{Product of zeroes } (\alpha \beta) = 5 \times (-2) = -10 \]
The required quadratic polynomial is:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - 3x - 10 = 0 \]
(iii) \( \frac{1}{3} \) and \( -1 \)
Solution:
Let the zeroes be \( \alpha = \frac{1}{3} \) and \( \beta = -1 \)
\[ \text{Sum of zeroes } (\alpha + \beta) = \frac{1}{3} - 1 = \frac{1 - 3}{3} = -\frac{2}{3} \]
\[ \text{Product of zeroes } (\alpha \beta) = \frac{1}{3} \times (-1) = -\frac{1}{3} \]
The required quadratic polynomial is:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - \left(-\frac{2}{3}\right)x + \left(-\frac{1}{3}\right) = 0 \]
\[ x^2 + \frac{2}{3}x - \frac{1}{3} = 0 \]
Multiplying the equation by 3: \[ 3x^2 + 2x - 1 = 0 \]
(iv) \( \frac{3}{2} \) and \( -2 \)
Solution:
Let the zeroes be \( \alpha = \frac{3}{2} \) and \( \beta = -2 \)
\[ \text{Sum of zeroes } (\alpha + \beta) = \frac{3}{2} - 2 = \frac{3 - 4}{2} = -\frac{1}{2} \]
\[ \text{Product of zeroes } (\alpha \beta) = \frac{3}{2} \times (-2) = -3 \]
The required quadratic polynomial is:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
\[ x^2 - \left(-\frac{1}{2}\right)x - 3 = 0 \]
\[ x^2 + \frac{1}{2}x - 3 = 0 \]
Multiplying the equation by 2: \[ 2x^2 + x - 6 = 0 \]
- \(a = 1\)
- \(b = -p\)
- Constant term \(= -p + c\)
- \(A = 1, B = -3, C = -6, D = 8\)
- Sum of zeroes \((\alpha + \beta) = -\frac{b}{a}\)
- Product of zeroes \((\alpha\beta) = \frac{c}{a}\)
- \(\alpha + \beta = -\frac{-6}{1} = 6\)
- \(\alpha\beta = \frac{p}{1} = p\)
| Column I | Column II |
|---|---|
| (A) \(x^2 - 7x + 12\) | (P) \(7, -12\) |
| (B) \(x^2 + 7x + 12\) | (Q) \(7, 12\) |
| (C) \(x^2 - 7x - 12\) | (R) \(-7, 12\) |
To match the columns, we use the formulas for a quadratic polynomial \(ax^2 + bx + c\):
- Sum of zeroes \((\alpha + \beta) = -\frac{b}{a}\)
- Product of zeroes \((\alpha\beta) = \frac{c}{a}\)
| Polynomial | Sum \((-\frac{b}{a})\) | Product \((\frac{c}{a})\) | Result |
|---|---|---|---|
| (A) \(x^2 - 7x + 12\) | \(-\frac{-7}{1} = 7\) | \(\frac{12}{1} = 12\) | (Q) 7, 12 |
| (B) \(x^2 + 7x + 12\) | \(-\frac{7}{1} = -7\) | \(\frac{12}{1} = 12\) | (R) -7, 12 |
| (C) \(x^2 - 7x - 12\) | \(-\frac{-7}{1} = 7\) | \(\frac{-12}{1} = -12\) | (P) 7, -12 |
- A matches with Q
- B matches with R
- C matches with P
Therefore, Statement (i) is True. Step 2: Evaluate Statement (ii) By definition, a polynomial is an algebraic expression consisting of variables and coefficients, where the exponents of the variables must be non-negative integers (0, 1, 2, ...). If a variable appears in the denominator (e.g., \( \frac{1}{x} \)), its exponent becomes negative (\( x^{-1} \)), which violates the definition of a polynomial.
Therefore, Statement (ii) is True. Step 3: Conclusion Since both statements are mathematically correct:
- Not an integer: Since it is a fraction.
- A rational number: Because it is expressed in the form \(\frac{a}{b}\).
- Not an irrational number: Because it can be written as a simple fraction.
- A real number: All rational numbers are real numbers.
Solution:
Step 1: Sum and product of given zeroes Given polynomial: \[ 2x^2 - 5x + 7 \] \[ \alpha + \beta = \frac{5}{2}, \quad \alpha\beta = \frac{7}{2} \] Step 2: New zeroes New zeroes are: \[ 2\alpha + 3\beta \quad \text{and} \quad 3\alpha + 2\beta \] Step 3: Sum of new zeroes \[ (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5\alpha + 5\beta = 5(\alpha + \beta) \] \[ = 5 \times \frac{5}{2} = \frac{25}{2} \] Step 4: Product of new zeroes \[ (2\alpha + 3\beta)(3\alpha + 2\beta) \] \[ = 6\alpha^2 + 13\alpha\beta + 6\beta^2 \] Use identity: \[ \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \] \[ = \left(\frac{5}{2}\right)^2 - 2\cdot\frac{7}{2} = \frac{25}{4} - 7 = -\frac{3}{4} \] Now, \[ 6(\alpha^2 + \beta^2) + 13\alpha\beta \] \[ = 6\left(-\frac{3}{4}\right) + 13\cdot\frac{7}{2} = -\frac{18}{4} + \frac{91}{2} \] \[ = -\frac{9}{2} + \frac{91}{2} = \frac{82}{2} = 41 \] Step 5: Required polynomial \[ x^2 - \left(\frac{25}{2}\right)x + 41 = 0 \] Multiply by 2: \[ 2x^2 - 25x + 82 = 0 \]Solution:
Given: For polynomial \(ax^2 + bx + c\), \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \] (i) \(\alpha^2 + \beta^2\) \[ \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \] \[ = \left(-\frac{b}{a}\right)^2 - 2\cdot\frac{c}{a} = \frac{b^2}{a^2} - \frac{2c}{a} \] \[ = \frac{b^2 - 2ac}{a^2} \]Answer: \(\frac{b^2 - 2ac}{a^2}\)
(ii) \(\alpha^2 + \alpha\beta + \beta^2\) \[ = (\alpha^2 + \beta^2) + \alpha\beta \] \[ = \frac{b^2 - 2ac}{a^2} + \frac{c}{a} = \frac{b^2 - 2ac + ac}{a^2} \] \[ = \frac{b^2 - ac}{a^2} \]Answer: \(\frac{b^2 - ac}{a^2}\)
(iii) \(\alpha^2\beta + \alpha\beta^2\) \[ = \alpha\beta(\alpha + \beta) \] \[ = \frac{c}{a} \cdot \left(-\frac{b}{a}\right) = -\frac{bc}{a^2} \]Answer: \(-\frac{bc}{a^2}\)
(iv) \(\alpha - \beta\) \[ (\alpha - \beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta \] \[ = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2} \] \[ \Rightarrow \alpha - \beta = \pm \frac{\sqrt{b^2 - 4ac}}{a} \]Answer: \(\pm \frac{\sqrt{b^2 - 4ac}}{a}\)
(v) \(\alpha^2 + \beta^2 - \alpha\beta\) \[ = (\alpha^2 + \beta^2) - \alpha\beta \] \[ = \frac{b^2 - 2ac}{a^2} - \frac{c}{a} = \frac{b^2 - 2ac - ac}{a^2} \] \[ = \frac{b^2 - 3ac}{a^2} \]Answer: \(\frac{b^2 - 3ac}{a^2}\)
- \(a = k\)
- \(b = 2\)
- \(c = 3k\)
- Sum of zeroes \((\alpha + \beta) = -\frac{b}{a} = -\frac{2}{k}\)
- Product of zeroes \((\alpha\beta) = \frac{c}{a} = \frac{3k}{k} = 3\)
Solution:
Step 1: Sum and Product of zeroes Given polynomial: \[ f(x) = x^2 - 5x + 4 \] \[ \alpha + \beta = 5, \quad \alpha\beta = 4 \] Step 2: Find \( \frac{1}{\alpha} + \frac{1}{\beta} \) \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5}{4} \] Step 3: Substitute in the given expression \[ \frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta \] \[ = \frac{5}{4} - 2(4) = \frac{5}{4} - 8 \] \[ = \frac{5 - 32}{4} = -\frac{27}{4} \]Solution:
Step 1: Standard form of cubic polynomial If \( \alpha, \beta, \gamma \) are zeroes, then the cubic polynomial is: \[ x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma \] Step 2: Substitute given values Given: \[ \alpha+\beta+\gamma = 2 \] \[ \alpha\beta+\beta\gamma+\gamma\alpha = -7 \] \[ \alpha\beta\gamma = -14 \] Substitute into formula: \[ x^3 - (2)x^2 + (-7)x - (-14) \] Step 3: Simplify \[ x^3 - 2x^2 - 7x + 14 \]
Explanation: The curved shape formed by the hanging wires of a suspension bridge is mathematically known as a Parabola.
Correct Option: (C) Parabola
Explanation: Since the shape is a parabola, it represents a quadratic polynomial. The standard form of a quadratic polynomial is \(ax^2 + bx + c\).
Correct Option: (b) \(y = ax^2 + bx + c\)
Explanation: Geometrically, the zeroes of any polynomial are the points where its graph crosses or touches the x-axis.
Correct Option: (a) Intersects the x-axis
Explanation: The formula for a quadratic polynomial is \(x^2 - (\text{Sum})x + (\text{Product})\).
Substituting the values: \(x^2 - (-3)x + 5 = x^2 + 3x + 5\).
Correct Option: (d) \(x^2 + 3x + 5\)
Explanation: Every quadratic polynomial, when plotted on a graph, always results in a Parabola.
Correct Option: (c) Parabola
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