Get step-by-step Rbse Solutions for Class 9 Maths Chapter 2 Exercise 2.2. Learn to find the value of $P(x)$ and verify/calculate the zeroes of linear and quadratic polynomials. Essential for CBSE/RBSE exams.
1. Find the value of the polynomial $P(x) = 5x – 4x^2 + 3$ at:
(i) $x = 0$
$$P(0) = 5(0) – 4(0)^2 + 3$$
$$P(0) = 0 – 0 + 3$$
$$\mathbf{P(0) = 3}$$
(ii) $x = –1$
$$P(-1) = 5(-1) – 4(-1)^2 + 3$$
$$P(-1) = -5 – 4(1) + 3$$
$$P(-1) = -5 – 4 + 3$$
$$P(-1) = -9 + 3$$
$$\mathbf{P(-1) = -6}$$
(iii) $x = 2$
$$P(2) = 5(2) – 4(2)^2 + 3$$
$$P(2) = 10 – 4(4) + 3$$
$$P(2) = 10 – 16 + 3$$
$$P(2) = 13 – 16$$
$$\mathbf{P(2) = -3}$$
2. Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials:
(i) $p(y) = y^2 – y + 1$
- $p(0) = (0)^2 – (0) + 1 = \mathbf{1}$
- $p(1) = (1)^2 – (1) + 1 = 1 – 1 + 1 = \mathbf{1}$
- $p(2) = (2)^2 – (2) + 1 = 4 – 2 + 1 = \mathbf{3}$
(ii) $p(t) = 2 + t + 2t^2 – t^3$
- $p(0) = 2 + 0 + 2(0)^2 – (0)^3 = \mathbf{2}$
- $p(1) = 2 + 1 + 2(1)^2 – (1)^3 = 2 + 1 + 2 – 1 = \mathbf{4}$
- $p(2) = 2 + 2 + 2(2)^2 – (2)^3 = 4 + 2(4) – 8 = 4 + 8 – 8 = \mathbf{4}$
(iii) $p(x) = x^3$
- $p(0) = (0)^3 = \mathbf{0}$
- $p(1) = (1)^3 = \mathbf{1}$
- $p(2) = (2)^3 = \mathbf{8}$
(iv) $p(x) = (x – 1) (x + 1)$
- $p(0) = (0 – 1)(0 + 1) = (-1)(1) = \mathbf{-1}$
- $p(1) = (1 – 1)(1 + 1) = (0)(2) = \mathbf{0}$
- $p(2) = (2 – 1)(2 + 1) = (1)(3) = \mathbf{3}$
3. Verify whether the following are zeroes of the polynomial, indicated against them.
A number $k$ is a zero of a polynomial $p(x)$ if $p(k) = 0$.
(i) $p(x) = 3x + 1, x = -\frac{1}{3}$
$$p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1 = -1 + 1 = 0$$
Result: Yes, $-\frac{1}{3}$ is a zero.
(ii) $p(x) = 5x – \pi, x = \frac{4}{5}$
$$p(\frac{4}{5}) = 5(\frac{4}{5}) – \pi = 4 – \pi$$
Result: No, $\frac{4}{5}$ is not a zero (since $4 – \pi \neq 0$).
(iii) $p(x) = x^2 – 1, x = 1, -1$
- For $x=1$: $p(1) = (1)^2 – 1 = 1 – 1 = 0$
- For $x=-1$: $p(-1) = (-1)^2 – 1 = 1 – 1 = 0$Result: Yes, both $1$ and $-1$ are zeroes.
(iv) $p(x) = (x + 1) (x – 2), x = -1, 2$
- For $x=-1$: $p(-1) = (-1 + 1)(-1 – 2) = (0)(-3) = 0$
- For $x=2$: $p(2) = (2 + 1)(2 – 2) = (3)(0) = 0$Result: Yes, both $-1$ and $2$ are zeroes.
(v) $p(x) = x^2, x = 0$
$$p(0) = (0)^2 = 0$$
Result: Yes, $0$ is a zero.
(vi) $p(x) = lx + m, x = -\frac{m}{l}$
$$p(-\frac{m}{l}) = l(-\frac{m}{l}) + m = -m + m = 0$$
Result: Yes, $-\frac{m}{l}$ is a zero.
(vii) $p(x) = 3x^2 – 1, x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
- For $x = -\frac{1}{\sqrt{3}}$:$$p(-\frac{1}{\sqrt{3}}) = 3(-\frac{1}{\sqrt{3}})^2 – 1 = 3(\frac{1}{3}) – 1 = 1 – 1 = 0$$Result: Yes, $-\frac{1}{\sqrt{3}}$ is a zero.
- For $x = \frac{2}{\sqrt{3}}$:$$p(\frac{2}{\sqrt{3}}) = 3(\frac{2}{\sqrt{3}})^2 – 1 = 3(\frac{4}{3}) – 1 = 4 – 1 = 3$$Result: No, $\frac{2}{\sqrt{3}}$ is not a zero (since $3 \neq 0$).
(viii) $p(x) = 2x + 1, x = \frac{1}{2}$
$$p(\frac{1}{2}) = 2(\frac{1}{2}) + 1 = 1 + 1 = 2$$
Result: No, $\frac{1}{2}$ is not a zero (since $2 \neq 0$).
4. Find the zero of the polynomial in each of the following cases:
To find the zero, set the polynomial equal to zero and solve for $x$ ($p(x) = 0$).
(i) $p(x) = x + 5$
$$x + 5 = 0$$
$$\mathbf{x = -5}$$
(ii) $p(x) = x – 5$
$$x – 5 = 0$$
$$\mathbf{x = 5}$$
(iii) $p(x) = 2x + 5$
$$2x + 5 = 0$$
$$2x = -5$$
$$\mathbf{x = -\frac{5}{2}}$$
(iv) $p(x) = 3x – 2$
$$3x – 2 = 0$$
$$3x = 2$$
$$\mathbf{x = \frac{2}{3}}$$
(v) $p(x) = 3x$
$$3x = 0$$
$$\mathbf{x = 0}$$
(vi) $p(x) = ax, a \neq 0$
$$ax = 0$$
Since $a \neq 0$, we must have:
$$\mathbf{x = 0}$$
(vii) $p(x) = cx + d, c \neq 0$
$$cx + d = 0$$
$$cx = -d$$
$$\mathbf{x = -\frac{d}{c}}$$