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Q4(i):
Find the zero of the polynomial in each of the following cases:
(i) $p(x) = x + 5$
Solution :
Initial Setup & Theoretical Foundation
We are given the linear polynomial:
$p(x) = x + 5$
Step 1: Applying the Definition of a Zero
The zero (or root) of a polynomial $p(x)$ is defined as the real number $c$ such that $p(c) = 0$. [Per the Fundamental Theorem of Algebra and basic polynomial theory, a linear polynomial of degree 1 will have exactly one real zero].
Step 2: Algebraic Evaluation
To find the zero, we equate the polynomial to zero:
$x + 5 = 0$
Subtracting $5$ from both sides of the equation [By the Subtraction Property of Equality]:
$x = 0 - 5$
$x = -5$
Step 3: Verification
Substitute $x = -5$ back into the original polynomial to ensure the condition $p(x) = 0$ is rigorously satisfied:
$p(-5) = (-5) + 5$
$p(-5) = 0$
The result is verified.
Step 4: Geometric Interpretation
Geometrically, the zero of a polynomial represents the x-coordinate of the point where the graph of the function $y = p(x)$ intersects the x-axis (the line where $y = 0$). As shown in the Cartesian plane below, the line $y = x + 5$ crosses the x-axis exactly at the coordinate point $(-5, 0)$.
Final Solution: The zero of the polynomial $p(x) = x + 5$ is $x = -5$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.2
- Q1(i): Find the value of the polynomial $5x – 4x^2 + 3$ at (i) $x = 0$
- Q1(ii): Find the value of the polynomial $5x – 4x^2 + 3$ at (ii) $x = –1$
- Q1(iii): Find the value of the polynomial $5x – 4x^2 + 3$ at (iii) $x = 2$
- Q2(i): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (i) $p(y) = y^2 – y + 1$
- Q2(ii): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (ii) $p(t) = 2 + t + 2t^2 – t^3$
- Q2(iii): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (iii) $p(x) = x^3$
- Q2(iv): Find $p(0)$, $p(1)$ and $p(2)$ for each of the following polynomials: (iv) $p(x) = (x – 1) (x + 1)$
- Q3(i): Verify whether the following are zeroes of the polynomial, indicated against them. (i) $p(x) = 3x + 1, x = –\frac{1}{3}$
- Q3(ii): Verify whether the following are zeroes of the polynomial, indicated against them. (ii) $p(x) = 5x – \pi, x = \frac{4}{5}$
- Q3(iii): Verify whether the following are zeroes of the polynomial, indicated against them. (iii) $p(x) = x^2 – 1, x = 1, –1$
- Q3(iv): Verify whether the following are zeroes of the polynomial, indicated against them. (iv) $p(x) = (x + 1) (x – 2), x = – 1, 2$
- Q3(v): Verify whether the following are zeroes of the polynomial, indicated against them. (v) $p(x) = x^2, x = 0$
- Q3(vi): Verify whether the following are zeroes of the polynomial, indicated against them. (vi) $p(x) = lx + m, x = –\frac{m}{l}$
- Q3(vii): Verify whether the following are zeroes of the polynomial, indicated against them. (vii) $p(x) = 3x^2 – 1, x = –\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$
- Q3(viii): Verify whether the following are zeroes of the polynomial, indicated against them. (viii) $p(x) = 2x + 1, x = \frac{1}{2}$
- Q4(ii): Find the zero of the polynomial in each of the following cases: (ii) $p(x) = x – 5$
- Q4(iii): Find the zero of the polynomial in each of the following cases: (iii) $p(x) = 2x + 5$
- Q4(iv): Find the zero of the polynomial in each of the following cases: (iv) $p(x) = 3x – 2$
- Q4(v): Find the zero of the polynomial in each of the following cases: (v) $p(x) = 3x$
- Q4(vi): Find the zero of the polynomial in each of the following cases: (vi) $p(x) = ax, a \neq 0$
- Q4(vii): Find the zero of the polynomial in each of the following cases: (vii) $p(x) = cx + d, c \neq 0, c, d$ are real numbers.
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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