SEBA Class 10 Maths Exercise 2.2 Solutions: Polynomials | New Book 2026

Get free solutions to SEBA Class 10 Maths Exercise 2.2 Solutions from the new SCERT textbook 2026. In this article, we have solved all questions—including finding zeroes of quadratic polynomials, verifying the relationship between zeroes and coefficients, forming quadratic polynomials, and multiple-choice questions—in a simple, step-by-step manner.

If you are a Class 10 student under the SEBA preparing for your upcoming HSLC examinations, then these reliable and easy-to-understand solutions will help you clear your concepts and score 90+ marks in the matric exam.

See More:

• SEBA Class 10 Maths Chapter 1.1 Solutions
• SEBA Class 10 Maths Chapter 7.1 Solutions

Seba Class 10 Maths EXERCISE 2.2 Solution

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) $x^2 – 2x – 8$

Solution:-

$x^2 – 2x – 8$

$= x^2 – (4 – 2)x – 8$

$= x^2 – 4x + 2x – 8$

$= x(x – 4) + 2(x – 4)$

$= (x + 2)(x – 4)$

The value of $x^2 – 2x – 8$ is zero when,

$(x + 2) = 0 \quad\text{or}\quad (x – 4) = 0$

$\Rightarrow x = -2 \quad\quad\quad \Rightarrow x = 4$

$x = -2$ and $x = 4$ are the zeroes of the polynomial.

Verification:

Comparing $x^2 – 2x – 8$ with $ax^2 + bx + c$, we have $a = 1$, $b = -2$, $c = -8$.

And $\alpha = -2$, $\beta = 4$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow -2 + 4 = -\frac{-2}{1}$

$\Rightarrow 2 = 2$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow -2 \times 4 = \frac{-8}{1}$

$\Rightarrow -8 = -8$

Hence verified.

(ii) $4s^2 – 4s + 1$

Solution:-

Given, $4s^2 – 4s + 1$

$= 4s^2 – (2 + 2)s + 1$

$= 4s^2 – 2s – 2s + 1$

$= 2s(2s – 1) – 1(2s – 1)$

$= (2s – 1)(2s – 1)$

The value of $4s^2 – 4s + 1$ is zero when,

$(2s – 1) = 0 \quad\text{or}\quad (2s – 1) = 0$

$\Rightarrow 2s = 1 \quad\quad\quad \Rightarrow 2s = 1$

$\Rightarrow s = \frac{1}{2} \quad\quad\quad \Rightarrow s = \frac{1}{2}$

$s = \frac{1}{2}$ and $s = \frac{1}{2}$ are the zeroes of the polynomial.

Verification:

(here $a = 4$, $b = -4$, $c = 1$)

And $\alpha = \frac{1}{2}$, $\beta = \frac{1}{2}$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow \frac{1}{2} + \frac{1}{2} = -\frac{-4}{4}$

$\Rightarrow 1 = 1$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$\Rightarrow \frac{1}{4} = \frac{1}{4}$

Hence verified.

(iii) $6x^2 – 3 – 7x$

Solution:-

$6x^2 – 7x – 3$

$= 6x^2 – (9 – 2)x – 3$

$= 6x^2 – 9x + 2x – 3$

$= 3x(2x – 3) + 1(2x – 3)$

$= (3x + 1)(2x – 3)$

The value of $6x^2 – 7x – 3$ is zero when,

$(3x + 1) = 0 \quad\text{or}\quad (2x – 3) = 0$

$\Rightarrow 3x = -1 \quad\quad\quad \Rightarrow 2x = 3$

$\Rightarrow x = \frac{-1}{3} \quad\quad\quad \Rightarrow x = \frac{3}{2}$

$x = \frac{-1}{3}$ and $x = \frac{3}{2}$ are the zeroes of the polynomial.

Verification:

(here $a = 6$, $b = -7$, $c = -3$)

And $\alpha = \frac{-1}{3}$, $\beta = \frac{3}{2}$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow \frac{-1}{3} + \frac{3}{2} = -\frac{-7}{6}$

$\Rightarrow \frac{-2 + 9}{6} = \frac{7}{6}$

$\Rightarrow \frac{7}{6} = \frac{7}{6}$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow \frac{-1}{3} \times \frac{3}{2} = \frac{-3}{6}$

$\Rightarrow \frac{-1}{2} = \frac{-1}{2}$

Hence verified.

(iv) $4u^2 + 8u$

Solution:-

$4u^2 + 8u$

$= 4u(u + 2)$

The value of $4u^2 + 8u$ is zero when,

$4u = 0 \quad\text{or}\quad (u + 2) = 0$

$\Rightarrow u = 0 \quad\quad\quad \Rightarrow u = -2$

$u = 0$ and $u = -2$ are the zeroes of the polynomial.

Verification:

(here $a = 4$, $b = 8$, $c = 0$)

And $\alpha = 0$, $\beta = -2$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow 0 + (-2) = -\frac{8}{4}$

$\Rightarrow -2 = -2$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow 0 \times (-2) = \frac{0}{4}$

$\Rightarrow 0 = 0$

Hence verified.

(v) $t^2 – 15$

Solution:-

$t^2 – 15$

$= t^2 – (\sqrt{15})^2$

$= (t + \sqrt{15})(t – \sqrt{15})$

The value of $t^2 – 15$ is zero when,

$(t + \sqrt{15}) = 0 \quad\text{or}\quad (t – \sqrt{15}) = 0$

$\Rightarrow t = -\sqrt{15} \quad\quad\quad \Rightarrow t = \sqrt{15}$

$t = -\sqrt{15}$ and $t = \sqrt{15}$ are the zeroes of the polynomial.

Verification:

(here $a = 1$, $b = 0$, $c = -15$)

And $\alpha = -\sqrt{15}$, $\beta = \sqrt{15}$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow -\sqrt{15} + \sqrt{15} = -\frac{0}{1}$

$\Rightarrow 0 = 0$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow (-\sqrt{15}) \times (\sqrt{15}) = \frac{-15}{1}$

$\Rightarrow -15 = -15$

Hence verified.

(vi) $3x^2 – x – 4$

Solution:-

$3x^2 – x – 4$

$= 3x^2 – (4 – 3)x – 4$

$= 3x^2 – 4x + 3x – 4$

$= x(3x – 4) + 1(3x – 4)$

$= (x + 1)(3x – 4)$

The value of $3x^2 – x – 4$ is zero when,

$(x + 1) = 0 \quad\text{or}\quad (3x – 4) = 0$

$\Rightarrow x = -1 \quad\quad\quad \Rightarrow 3x = 4$

$\quad\quad\quad\quad\quad\quad\quad\quad \Rightarrow x = \frac{4}{3}$

$x = -1$ and $x = \frac{4}{3}$ are the zeroes of the polynomial.

Verification:

(here $a = 3$, $b = -1$, $c = -4$)

And $\alpha = -1$, $\beta = \frac{4}{3}$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow -1 + \frac{4}{3} = -\frac{-1}{3}$

$\Rightarrow \frac{-3 + 4}{3} = \frac{1}{3}$

$\Rightarrow \frac{1}{3} = \frac{1}{3}$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow -1 \times \frac{4}{3} = \frac{-4}{3}$

$\Rightarrow \frac{-4}{3} = \frac{-4}{3}$

Hence verified.

(vii) $x^2 + 7x + 12$

Solution:-

$x^2 + 7x + 12$

$= x^2 + (4 + 3)x + 12$

$= x^2 + 4x + 3x + 12$

$= x(x + 4) + 3(x + 4)$

$= (x + 3)(x + 4)$

The value of $x^2 + 7x + 12$ is zero when,

$(x + 3) = 0 \quad\text{or}\quad (x + 4) = 0$

$\Rightarrow x = -3 \quad\quad\quad \Rightarrow x = -4$

$x = -3$ and $x = -4$ are the zeroes of the polynomial.

Verification:

(here $a = 1$, $b = 7$, $c = 12$)

And $\alpha = -3$, $\beta = -4$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow -3 + (-4) = -\frac{7}{1}$

$\Rightarrow -7 = -7$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow -3 \times (-4) = \frac{12}{1}$

$\Rightarrow 12 = 12$

Hence verified.

(viii) $x^2 – 4x + 3$

Solution:-

$x^2 – 4x + 3$

$= x^2 – (3 + 1)x + 3$

$= x^2 – 3x – x + 3$

$= x(x – 3) – 1(x – 3)$

$= (x – 1)(x – 3)$

The value of $x^2 – 4x + 3$ is zero when,

$(x – 1) = 0 \quad\text{or}\quad (x – 3) = 0$

$\Rightarrow x = 1 \quad\quad\quad \Rightarrow x = 3$

$x = 1$ and $x = 3$ are the zeroes of the polynomial.

Verification:

(here $a = 1$, $b = -4$, $c = 3$)

And $\alpha = 1$, $\beta = 3$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow 1 + 3 = -\frac{-4}{1}$

$\Rightarrow 4 = 4$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow 1 \times 3 = \frac{3}{1}$

$\Rightarrow 3 = 3$

Hence verified.

(ix) $x^2 – 6x – 7$

Solution:-

$x^2 – 6x – 7$

$= x^2 – (7 – 1)x – 7$

$= x^2 – 7x + x – 7$

$= x(x – 7) + 1(x – 7)$

$= (x + 1)(x – 7)$

The value of $x^2 – 6x – 7$ is zero when,

$(x + 1) = 0 \quad\text{or}\quad (x – 7) = 0$

$\Rightarrow x = -1 \quad\quad\quad \Rightarrow x = 7$

$x = -1$ and $x = 7$ are the zeroes of the polynomial.

Verification:

(here $a = 1$, $b = -6$, $c = -7$)

And $\alpha = -1$, $\beta = 7$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow -1 + 7 = -\frac{-6}{1}$

$\Rightarrow 6 = 6$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow -1 \times 7 = \frac{-7}{1}$

$\Rightarrow -7 = -7$

Hence verified.

(x) $2x^2 – 5x – 7$

Solution:-

$2x^2 – 5x – 7$

$= 2x^2 – (7 – 2)x – 7$

$= 2x^2 – 7x + 2x – 7$

$= x(2x – 7) + 1(2x – 7)$

$= (x + 1)(2x – 7)$

The value of $2x^2 – 5x – 7$ is zero when,

$(x + 1) = 0 \quad\text{or}\quad (2x – 7) = 0$

$\Rightarrow x = -1 \quad\quad\quad \Rightarrow 2x = 7$

$\quad\quad\quad\quad\quad\quad\quad\quad \Rightarrow x = \frac{7}{2}$

$x = -1$ and $x = \frac{7}{2}$ are the zeroes of the polynomial.

Verification:

(here $a = 2$, $b = -5$, $c = -7$)

And $\alpha = -1$, $\beta = \frac{7}{2}$

We know,

$\text{Sum of zeroes: } \alpha + \beta = -\frac{b}{a}$

$\Rightarrow -1 + \frac{7}{2} = -\frac{-5}{2}$

$\Rightarrow \frac{-2 + 7}{2} = \frac{5}{2}$

$\Rightarrow \frac{5}{2} = \frac{5}{2}$

$\text{Product of zeroes, } \alpha.\beta = \frac{c}{a}$

$\Rightarrow -1 \times \frac{7}{2} = \frac{-7}{2}$

$\Rightarrow \frac{-7}{2} = \frac{-7}{2}$

Hence verified.

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) $\frac{1}{4}, -1$

Solution:-

$\text{Sum of the zeroes} = \frac{1}{4}$

$\text{Product of the zeroes} = -1$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – \left(\frac{1}{4}\right)x + (-1)$

$= x^2 – \frac{1}{4}x – 1$

$= 4x^2 – x – 4 \quad [\text{Multiply by } 4]$

(ii) $\sqrt{2}, \frac{1}{3}$

Solution:-

$\text{Sum of the zeroes} = \sqrt{2}$

$\text{Product of the zeroes} = \frac{1}{3}$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – (\sqrt{2})x + \left(\frac{1}{3}\right)$

$= x^2 – \sqrt{2}x + \frac{1}{3}$

$= 3x^2 – 3\sqrt{2}x + 1 \quad [\text{Multiply by } 3]$

(iii) $0, \sqrt{5}$

Solution:-

$\text{Sum of the zeroes} = 0$

$\text{Product of the zeroes} = \sqrt{5}$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – (0)x + (\sqrt{5})$

$= x^2 + \sqrt{5}$

(iv) $1, 1$

Solution:-

$\text{Sum of the zeroes} = 1$

$\text{Product of the zeroes} = 1$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – (1)x + (1)$

$= x^2 – x + 1$

(v) $-\frac{1}{4}, \frac{1}{4}$

Solution:-

$\text{Sum of the zeroes} = -\frac{1}{4}$

$\text{Product of the zeroes} = \frac{1}{4}$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – \left(-\frac{1}{4}\right)x + \left(\frac{1}{4}\right)$

$= x^2 + \frac{1}{4}x + \frac{1}{4}$

$= 4x^2 + x + 1 \quad [\text{Multiply by } 4]$

(vi) $4, 1$

Solution:-

$\text{Sum of the zeroes} = 4$

$\text{Product of the zeroes} = 1$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – (4)x + (1)$

$= x^2 – 4x + 1$

3: Find quadratic polynomials whose zeroes are given:

(i) $-4$ and $\frac{3}{2}$

Solution:-

$\text{Sum of the zeroes} = -4 + \frac{3}{2} = \frac{-8 + 3}{2} = -\frac{5}{2}$

$\text{Product of the zeroes} = -4 \times \frac{3}{2} = -6$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – \left(-\frac{5}{2}\right)x + (-6)$

$= x^2 + \frac{5}{2}x – 6$

$= 2x^2 + 5x – 12 \quad [\text{Multiply by } 2]$

(ii) $5$ and $2$

Solution:-

$\text{Sum of the zeroes} = 5 + 2 = 7$

$\text{Product of the zeroes} = 5 \times 2 = 10$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – (7)x + (10)$

$= x^2 – 7x + 10$

(iii) $\frac{1}{3}$ and $-1$

Solution:-

$\text{Sum of the zeroes} = \frac{1}{3} + (-1) = \frac{1 – 3}{3} = -\frac{2}{3}$

$\text{Product of the zeroes} = \frac{1}{3} \times (-1) = -\frac{1}{3}$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – \left(-\frac{2}{3}\right)x + \left(-\frac{1}{3}\right)$

$= x^2 + \frac{2}{3}x – \frac{1}{3}$

$= 3x^2 + 2x – 1 \quad [\text{Multiply by } 3]$

(iv) $\frac{3}{2}$ and $-2$

Solution:-

$\text{Sum of the zeroes} = \frac{3}{2} + (-2) = \frac{3 – 4}{2} = -\frac{1}{2}$

$\text{Product of the zeroes} = \frac{3}{2} \times (-2) = -3$

$\therefore \text{Required quadratic polynomial: } x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})$

$= x^2 – \left(-\frac{1}{2}\right)x + (-3)$

$= x^2 + \frac{1}{2}x – 3$

$= 2x^2 + x – 6 \quad [\text{Multiply by } 2]$

4. If $\alpha$ and $\beta$ are the two zeroes of the polynomial $x^2 – p(x+1) + c$ such that $(\alpha+1)(\beta+1) = 0$, then the value of $c$ is:

(a) 1

(b) 2

(c) -1

(d) -2

Solⁿ: (c) -1

Given polynomial: $x^2 – p(x+1) + c = x^2 – px + (c – p)$

• $\alpha + \beta = -\frac{-p}{1} = p$

• $\alpha\beta = \frac{c – p}{1} = c – p$

Given:

$(\alpha + 1)(\beta + 1) = 0$

$\implies \alpha\beta + (\alpha + \beta) + 1 = 0$

$\implies (c – p) + p + 1 = 0$

$\implies c + 1 = 0$

$\implies c = -1$

Correct Option: (c) -1

5. If $a-b$, $a$ and $a+b$ are the zeroes of the polynomial $p(x) = x^3 – 3x^2 – 6x + 8$, then the values of $a$ and $b$ are:

(i) $a = 1$

(ii) $b = \pm 3$

(iii) $a = -1$

(iv) $b = a$

Choose the correct answer:

(a) (i) and (iv)

(b) (i) and (ii)

(c) (iii) and (ii)

(d) (iii) and (iv)

Solⁿ: (b) (i) and (ii)

For $p(x) = x^3 – 3x^2 – 6x + 8$, the coefficients are $A = 1, B = -3, C = -6, D = 8$.

Sum of zeroes:

$(a – b) + a + (a + b) = -\frac{B}{A}$

$\implies 3a = -\frac{-3}{1} = 3 $

$\implies a = 1$

Product of zeroes:

$(a – b) \times a \times (a + b) = -\frac{D}{A}$

$\implies a(a^2 – b^2) = -\frac{8}{1} = -8$

Substituting $a = 1$, we get

$1(1 – b^2) = -8 $

$\implies 1 – b^2 = -8 $

$\implies b^2 = 9 $

$\implies b = \pm 3$

Thus, statements (i) and (ii) are true.

Correct Option: (b) (i) and (ii)

6. Assertion (A): If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 – 6x + p$ such that $(\alpha+\beta)^2 – 2\alpha\beta = 40$, then the value of $p$ is -1.

Reason (R): The sum and the product of the zeroes of the quadratic polynomial $ax^2 + bx + c$ are $-\frac{b}{a}$ and $\frac{c}{a}$ respectively.

Choose the correct option:

(a) Both assertion (A) and reason (R) are true and R is the correct explanation of assertion (A).

(b) Both assertion (A) and reason (R) are true but R is not the correct explanation of assertion (A).

(c) Assertion (A) is true but reason (R) is false.

(d) Assertion (A) is false but reason (R) is true.

Solⁿ: (d) Assertion (A) is false but reason (R) is true.

For Assertion (A):

From $x^2 – 6x + p$, we have $\alpha + \beta = 6$ and $\alpha\beta = p$.

Given:

$(\alpha + \beta)^2 – 2\alpha\beta = 40$

$\implies (6)^2 – 2p = 40$

$\implies 36 – 2p = 40$

$\implies -2p = 4 \implies p = -2$

Since Assertion states $p = -1$, Assertion (A) is false.

Reason (R) states the standard relationship between zeroes and coefficients, which is true.

Correct Option: (d) Assertion (A) is false but reason (R) is true.

7. Match the polynomial in Column I with the sum and product of its zeroes in Column II:

Choose the correct answer:

(a) $A \rightarrow Q,\; B \rightarrow P,\; C \rightarrow R$

(b) $A \rightarrow Q,\; B \rightarrow R,\; C \rightarrow P$

(c) $A \rightarrow P,\; B \rightarrow Q,\; C \rightarrow R$

(d) $A \rightarrow P,\; B \rightarrow R,\; C \rightarrow Q$

Solⁿ: (b) $A \rightarrow Q,\; B \rightarrow R,\; C \rightarrow P$

• For (A) $x^2 – 7x + 12$: Sum $= 7$, Product $= 12 \rightarrow (Q)$

• For (B) $x^2 + 7x + 12$: Sum $= -7$, Product $= 12 \rightarrow (R)$

• For (C) $x^2 – 7x – 12$: Sum $= 7$, Product $= -12 \rightarrow (P)$

  Correct Option: (b) $A \rightarrow Q,\; B \rightarrow R,\; C \rightarrow P$

8. Two statements are given below:

Statement (i): The graph of a linear polynomial is a straight line.

Statement (ii): A polynomial cannot have a variable in the denominator.

Choose the correct alternative:

(a) Both (i) and (ii) are true

(b) Both (i) and (ii) are false

(c) (i) is true but (ii) is false

(d) (i) is false but (ii) is true

Solⁿ: (a) Both (i) and (ii) are true

• The graph of a linear polynomial $y = ax + b$ is always a straight line. Hence, (i) is true.

• By definition, in a polynomial, all exponents of variables must be whole numbers; expressions with a variable in the denominator (like $\frac{1}{x-1}$) are not polynomials. Hence, (ii) is true.

  Correct Option: (a) Both (i) and (ii) are true

9. If $p^2 = \frac{32}{50}$, then the value of $p$ is:

(i) An integer

(ii) A rational number

(iii) An irrational number

(iv) A real number

Choose the correct option:

(a) Both (ii) and (iv) are true

(b) Both (i) and (iv) are true

(c) (i) is true but (ii) is false

(d) (ii) is false but (iii) is true

Solⁿ: (a) Both (ii) and (iv) are true

$p^2 = \frac{32}{50} = \frac{16}{25}$

$\implies p = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$

Since $\pm \frac{4}{5}$ is expressed in the form $\frac{p}{q}$ ($q \ne 0$), it is a rational number (ii), and every rational number is also a real number (iv).

Correct Option: (a) Both (ii) and (iv) are true

10. If one of the zeroes of the polynomial $x^3 + ax^2 + bx + c$ is -1, then the product of the other two zeroes is:

(a) $b – a + 1$

(b) $b – a – 1$

(c) $a – b + 1$

(d) $a – b – 1$

Solⁿ: (a) $b – a + 1$

Let the zeroes of $p(x) = x^3 + ax^2 + bx + c$ be $\alpha, \beta$, and $\gamma = -1$.

Sum of zeroes:

$\alpha + \beta + (-1) = -a $

$\implies \alpha + \beta = 1 – a$

Sum of products taken two at a time:

$\alpha\beta + \beta(-1) + (-1)\alpha = b$

$\implies \alpha\beta – (\alpha + \beta) = b$

$\implies \alpha\beta – (1 – a) = b$

$\implies \alpha\beta = b + (1 – a) = b – a + 1$

Correct Option: (a) $b – a + 1$

11. $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $x^2 – 6x + a$. If $3\alpha + 2\beta = 20$, then the value of $a$ is:

(a) 5

(b) -7

(c) 12

(d) -16

Solⁿ: (d) -16

From $x^2 – 6x + a$:

$\alpha + \beta = -\frac{-6}{1} = 6 $

$\implies \beta = 6 – \alpha \quad \text{— (1)}$

$\alpha\beta = a \quad \text{— (2)}$

Given:

$3\alpha + 2\beta = 20$

$3\alpha + 2(6 – \alpha) = 20$ [Substituting (1) into the given equation]

$\implies 3\alpha + 12 – 2\alpha = 20$

$\implies \alpha = 20 – 12 = 8$

From (1): $\beta = 6 – 8 = -2$

From (2): $a = \alpha\beta = 8 \times (-2) = -16$

Correct Option: (d) -16

12. If $\alpha$ and $\beta$ are the zeroes of the polynomial $2x^2 – 5x + 7$, then find another polynomial whose zeroes are $2\alpha + 3\beta$ and $3\alpha + 2\beta$.

Solⁿ:

For $2x^2 – 5x + 7$:

$\alpha + \beta = \frac{5}{2}$

$\alpha\beta = \frac{7}{2}$

Let the new zeroes be $S_1 = 2\alpha + 3\beta$ and $S_2 = 3\alpha + 2\beta$.

Sum of new zeroes:

$S_1 + S_2 = (2\alpha + 3\beta) + (3\alpha + 2\beta) $

$= 5(\alpha + \beta) = 5\left(\frac{5}{2}\right) = \frac{25}{2}$

Product of new zeroes:

$S_1 \times S_2 = (2\alpha + 3\beta)(3\alpha + 2\beta)$

$= 6\alpha^2 + 4\alpha\beta + 9\alpha\beta + 6\beta^2$

$= 6(\alpha^2 + \beta^2) + 13\alpha\beta$

$= 6[(\alpha + \beta)^2 – 2\alpha\beta] + 13\alpha\beta$

$= 6\left[\left(\frac{5}{2}\right)^2 – 2\left(\frac{7}{2}\right)\right] + 13\left(\frac{7}{2}\right)$

$= 6\left[\frac{25}{4} – 7\right] + \frac{91}{2}$

$= 6\left(-\frac{3}{4}\right) + \frac{91}{2} $

$= -\frac{9}{2} + \frac{91}{2} = \frac{82}{2} = 41$

The required quadratic polynomial is:

$[x^2 – (\text{Sum})x + \text{Product}] $

$= x^2 – \frac{25}{2}x + 41$

$= 2x^2 – 25x + 82$ [Multiply by 2]

13. If $\alpha$ and $\beta$ are the zeroes of the polynomial $ax^2 + bx + c$, then find the values of the following:

(i) $\alpha^2 + \beta^2$

Solution:

From $ax^2 + bx + c$, we have:

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

$\alpha\beta = \frac{c}{a}$

Now, $\alpha^2 + \beta^2$

$= (\alpha + \beta)^2 – 2\alpha\beta $

$= \left(-\frac{b}{a}\right)^2 – 2\left(\frac{c}{a}\right) $

$= \frac{b^2}{a^2} – \frac{2c}{a} $

$= \frac{b^2 – 2ac}{a^2}$

(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} $

$= \frac{b^2 – ac}{a^2}$

(iii) $\alpha^2\beta + \alpha\beta^2$

$= \alpha\beta(\alpha + \beta) $

$= \left(\frac{c}{a}\right)\left(-\frac{b}{a}\right) $

$= -\frac{bc}{a^2}$

(iv) $\alpha – \beta$

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

$(\alpha – \beta)^2 =  \left(-\frac{b}{a}\right)^2 – 4\left(\frac{c}{a}\right) $

$(\alpha – \beta)^2 = \frac{b^2 – 4ac}{a^2}$

$\implies \alpha – \beta = \pm\frac{\sqrt{b^2 – 4ac}}{a}$

(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} $

$= \frac{b^2 – 3ac}{a^2}$

14. If the sum and product of the two zeroes of the polynomial $kx^2 + 2x + 3k$ are equal, then find the value of $k$.

Solⁿ:

For $kx^2 + 2x + 3k$:

Here $a = k$, $b = 2$, $c = 3k$.

$\text{Sum of zeroes } (\alpha + \beta) = -\frac{b}{a} = -\frac{2}{k}$

$\text{Product of zeroes } (\alpha\beta) = \frac{c}{a} = \frac{3k}{k} = 3$

According to the question:

$\text{Sum of zeroes} = \text{Product of zeroes}$

$\implies -\frac{2}{k} = 3 $

$\implies 3k = -2 $

$\implies k = -\frac{2}{3}$

15. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 – 5x + 4$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta} – 2\alpha\beta$.

Solⁿ:

For $f(x) = x^2 – 5x + 4$:

$\alpha + \beta = -\frac{-5}{1} = 5$

$\alpha\beta = \frac{4}{1} = 4$

Now:

$\frac{1}{\alpha} + \frac{1}{\beta} – 2\alpha\beta $

$= \frac{\alpha + \beta}{\alpha\beta} – 2\alpha\beta$

$= \frac{5}{4} – 2(4) $

$= \frac{5}{4} – 8 $

$= \frac{5 – 32}{4} $

$= -\frac{27}{4}$

16. Find the cubic polynomial with the sum of the zeroes, sum of the product of its zeroes taken two at a time and product of its zeroes are 2, -7, -14 respectively.

Solⁿ:

A cubic polynomial with zeroes $\alpha, \beta, \gamma$ is given by:

$p(x) = x^3 – (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x – \alpha\beta\gamma$

Given:

$\alpha + \beta + \gamma = 2$

$\alpha\beta + \beta\gamma + \gamma\alpha = -7$

$\alpha\beta\gamma = -14$

Substituting the values:

$p(x) = x^3 – 2x^2 + (-7)x – (-14) = x^3 – 2x^2 – 7x + 14$

17. The adjacent figure shows a mathematical shape of a bridge with hanging wires. Answer the following questions:

(i) Name the shape of the hanging wire?

(A) Linear

(B) Spiral

(C) Parabola

(D) Ellipse

Solⁿ: The curved hanging wire forms a U-shaped curve which represents a parabola.

Correct Option: (C) Parabola

(ii) What will be the expression of the polynomial representing the figure?

(a) $y = ax + b$

(b) $y = ax^2 + bx + c$

(c) $y = ax^3 + bx^2 + cx + d$

(d) $y = ax^4 + cx + d$

Solⁿ: The shape is parabolic, which corresponds to a quadratic polynomial.

Correct Option: (b) $y = ax^2 + bx + c$

(iii) Zeroes of the polynomial can be expressed graphically. Number of zeroes of the polynomial is equal to number of points where the graph of polynomial:

(a) Intersects x-axis

(b) Intersects y-axis

(c) Intersects y-axis and x-axis

(d) None of the above

Solⁿ: By definition, the zeroes of $y = p(x)$ are the x-coordinates of the points where its graph intersects the x-axis.

Correct Option: (a) Intersects x-axis

(iv) The representation of hanging wire on the bridge whose sum of the zeroes is -3 and product of zeroes is 5 is:

(a) $x^3 – 3x – 5$

(b) $x^2 – 3x + 5$

(c) $x^2 + 3x – 5$

(d) $x^2 + 3x + 5$

Solⁿ: $p(x) = x^2 – (\text{Sum})x + \text{Product} $

$= x^2 – (-3)x + 5 = x^2 + 3x + 5$.

Correct Option: (d) $x^2 + 3x + 5$

(v) Graph of a quadratic polynomial is:

(a) Straight line

(b) Circle

(c) Parabola

(d) Any curve

Solⁿ: The graph of any standard quadratic function $y = ax^2 + bx + c$ ($a \ne 0$) is a parabola.

Correct Option: (c) Parabola