NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2#
Welcome to our comprehensive NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2. This exercise focuses on the relationship between the zeroes and coefficients of a polynomial. Students find the zeroes of quadratic polynomials and verify the relationship, and also form quadratic polynomials when the sum and product of zeroes are given.
Key Relationships for a Quadratic Polynomial $p(x) = ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$:
- Sum of zeroes: $\alpha + \beta = -\dfrac{b}{a}$
- Product of zeroes: $\alpha \beta = \dfrac{c}{a}$
- The quadratic polynomial: $x^2 - (\alpha + \beta)x + \alpha\beta$
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. #
(i) $x^2 - 2x - 8$
The parabola opens upward and cuts the x-axis at the zeroes $x = -2$ and $x = 4$.
We factorise the polynomial $x^2 - 2x - 8$:
$$= x^2 - 4x + 2x - 8$$$$= x(x - 4) + 2(x - 4)$$
$$= (x - 4)(x + 2)$$
The zeroes are obtained by setting each factor to zero:
$$x - 4 = 0$$$$\Rightarrow x = 4$$
$$x + 2 = 0$$
$$\Rightarrow x = -2$$
So the zeroes are $\alpha = 4$ and $\beta = -2$.
Verification:
Here $a = 1$, $b = -2$, $c = -8$.
$\alpha + \beta = 4 + (-2)$
$\alpha + \beta = 2$
$\alpha + \beta = -\dfrac{(-2)}{1}$
$\alpha + \beta = -\dfrac{b}{a}$ ✓
$\alpha\beta = 4 \times (-2)$
$\alpha\beta = -8$
$\alpha\beta = \dfrac{-8}{1}$
$\alpha\beta = \dfrac{c}{a}$ ✓
(ii) $4s^2 - 4s + 1$
The parabola just touches the x-axis at $s = \tfrac{1}{2}$ (a repeated zero).
We factorise the polynomial $4s^2 - 4s + 1$:
$$= 4s^2 - 2s - 2s + 1$$$$= 2s(2s - 1) - 1(2s - 1)$$
$$= (2s - 1)(2s - 1)$$
$$= (2s - 1)^2$$
Setting to zero:
$$2s - 1 = 0$$$$\Rightarrow s = \dfrac{1}{2}$$
So both zeroes are equal: $\alpha = \dfrac{1}{2}$ and $\beta = \dfrac{1}{2}$.
Verification:
Here $a = 4$, $b = -4$, $c = 1$.
$\alpha + \beta = \dfrac{1}{2} + \dfrac{1}{2}$
$\alpha + \beta = 1$
$\alpha + \beta = -\dfrac{(-4)}{4}$
$\alpha + \beta = -\dfrac{b}{a}$ ✓
$\alpha\beta = \dfrac{1}{2} \times \dfrac{1}{2}$
$\alpha\beta = \dfrac{1}{4}$
$\alpha\beta = \dfrac{c}{a}$ ✓
(iii) $6x^2 - 3 - 7x$
Rewriting: $6x^2 - 7x - 3$
We factorise the polynomial $6x^2 - 7x - 3$:
$$= 6x^2 - 9x + 2x - 3$$$$= 3x(2x - 3) + 1(2x - 3)$$
$$= (2x - 3)(3x + 1)$$
Zeroes:
$$2x - 3 = 0$$$$\Rightarrow x = \dfrac{3}{2}$$
and
$$3x + 1 = 0$$$$\Rightarrow x = -\dfrac{1}{3}$$
So $\alpha = \dfrac{3}{2}$, $\beta = -\dfrac{1}{3}$.
Verification:
Here $a = 6$, $b = -7$, $c = -3$.
$\alpha + \beta = \dfrac{3}{2} - \dfrac{1}{3}$
$\alpha + \beta = \dfrac{9 - 2}{6}$
$\alpha + \beta = \dfrac{7}{6}$
$\alpha + \beta = -\dfrac{(-7)}{6}$
$\alpha + \beta = -\dfrac{b}{a}$ ✓
$\alpha\beta = \dfrac{3}{2} \times \left(-\dfrac{1}{3}\right)$
$\alpha\beta = -\dfrac{1}{2}$
$\alpha\beta = \dfrac{-3}{6}$
$\alpha\beta = \dfrac{c}{a}$ ✓
(iv) $4u^2 + 8u$
We factorise the polynomial $4u^2 + 8u$:
$$= 4u(u + 2)$$Zeroes:
$$4u = 0$$$$\Rightarrow u = 0$$
and
$$u + 2 = 0$$$$\Rightarrow u = -2$$
So $\alpha = 0$, $\beta = -2$.
Verification:
Here $a = 4$, $b = 8$, $c = 0$.
$\alpha + \beta = 0 + (-2)$
$\alpha + \beta = -2$
$\alpha + \beta = -\dfrac{8}{4}$
$\alpha + \beta = -\dfrac{b}{a}$ ✓
$\alpha\beta = 0 \times (-2)$
$\alpha\beta = 0$
$\alpha\beta = \dfrac{0}{4}$
$\alpha\beta = \dfrac{c}{a}$ ✓
(v) $t^2 - 15$
We factorise the polynomial $t^2 - 15$:
$$= (t - \sqrt{15})(t + \sqrt{15})$$Zeroes: $t = \sqrt{15}$ and $t = -\sqrt{15}$
So $\alpha = \sqrt{15}$, $\beta = -\sqrt{15}$.
Verification:
Here $a = 1$, $b = 0$, $c = -15$.
$\alpha + \beta = \sqrt{15} + (-\sqrt{15})$
$\alpha + \beta = 0$
$\alpha + \beta = -\dfrac{0}{1}$
$\alpha + \beta = -\dfrac{b}{a}$ ✓
$\alpha\beta = \sqrt{15} \times (-\sqrt{15})$
$\alpha\beta = -15$
$\alpha\beta = \dfrac{-15}{1}$
$\alpha\beta = \dfrac{c}{a}$ ✓
(vi) $3x^2 - x - 4$
We factorise the polynomial $3x^2 - x - 4$:
$$= 3x^2 - 4x + 3x - 4$$$$= x(3x - 4) + 1(3x - 4)$$
$$= (3x - 4)(x + 1)$$
Zeroes:
$$3x - 4 = 0$$$$\Rightarrow x = \dfrac{4}{3}$$
and
$$x + 1 = 0$$$$\Rightarrow x = -1$$
So $\alpha = \dfrac{4}{3}$, $\beta = -1$.
Verification:
Here $a = 3$, $b = -1$, $c = -4$.
$\alpha + \beta = \dfrac{4}{3} + (-1)$
$\alpha + \beta = \dfrac{4 - 3}{3}$
$\alpha + \beta = \dfrac{1}{3}$
$\alpha + \beta = -\dfrac{(-1)}{3}$
$\alpha + \beta = -\dfrac{b}{a}$ ✓
$\alpha\beta = \dfrac{4}{3} \times (-1)$
$\alpha\beta = -\dfrac{4}{3}$
$\alpha\beta = \dfrac{-4}{3}$
$\alpha\beta = \dfrac{c}{a}$ ✓
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. #
(i) $\dfrac{1}{4},\; -1$
Sum of zeroes $\alpha + \beta = \dfrac{1}{4}$
Product of zeroes $\alpha\beta = -1$
The required quadratic polynomial is:
$$p(x) = k\left[x^2 - (\alpha + \beta)x + \alpha\beta\right]$$$$p(x) = k\left[x^2 - \dfrac{1}{4}x - 1\right]$$
Taking $k = 4$:
$$p(x) = 4x^2 - x - 4$$∴ The quadratic polynomial is $\mathbf{4x^2 - x - 4}$.
(ii) $\sqrt{2},\; \dfrac{1}{3}$
Sum of zeroes $\alpha + \beta = \sqrt{2}$
Product of zeroes $\alpha\beta = \dfrac{1}{3}$
The required polynomial:
$$p(x) = k\left[x^2 - \sqrt{2}\,x + \dfrac{1}{3}\right]$$Taking $k = 3$:
$$p(x) = 3x^2 - 3\sqrt{2}\,x + 1$$∴ The quadratic polynomial is $\mathbf{3x^2 - 3\sqrt{2}\,x + 1}$.
(iii) $0,\; \sqrt{5}$
Sum of zeroes $\alpha + \beta = 0$
Product of zeroes $\alpha\beta = \sqrt{5}$
The required polynomial:
$$p(x) = k\left[x^2 - 0 \cdot x + \sqrt{5}\right]$$$$p(x) = k\left[x^2 + \sqrt{5}\right]$$
Taking $k = 1$:
∴ The quadratic polynomial is $\mathbf{x^2 + \sqrt{5}}$.
(iv) $1,\; 1$
Sum of zeroes $\alpha + \beta = 1$
Product of zeroes $\alpha\beta = 1$
The required polynomial:
$$p(x) = k\left[x^2 - x + 1\right]$$Taking $k = 1$:
∴ The quadratic polynomial is $\mathbf{x^2 - x + 1}$.
(v) $-\dfrac{1}{4},\; \dfrac{1}{4}$
Sum of zeroes $\alpha + \beta = -\dfrac{1}{4}$
Product of zeroes $\alpha\beta = \dfrac{1}{4}$
The required polynomial:
$$p(x) = k\left[x^2 + \dfrac{1}{4}x + \dfrac{1}{4}\right]$$Taking $k = 4$:
$$p(x) = 4x^2 + x + 1$$∴ The quadratic polynomial is $\mathbf{4x^2 + x + 1}$.
(vi) $4,\; 1$
Sum of zeroes $\alpha + \beta = 4$
Product of zeroes $\alpha\beta = 1$
The required polynomial:
$$p(x) = k\left[x^2 - 4x + 1\right]$$Taking $k = 1$:
∴ The quadratic polynomial is $\mathbf{x^2 - 4x + 1}$.