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Q3(ii):
Verify whether the following are zeroes of the polynomial, indicated against them.
(ii) $p(x) = 5x – \pi, x = \frac{4}{5}$
Solution :
Given Variables & Initial Setup
We are given the following linear polynomial in one variable:
$p(x) = 5x - \pi$
We are tasked with verifying whether the following specific value of $x$ is a zero (root) of the polynomial:
$x = \frac{4}{5}$
Step 1: Theoretical Foundation
By the fundamental definition of the Zero of a Polynomial [Factor Theorem corollary], a real number $a$ is considered a zero of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = a$ yields exactly zero. Mathematically, this is expressed as:
$p(a) = 0$
Therefore, to verify if $x = \frac{4}{5}$ is a zero, we must substitute $x = \frac{4}{5}$ into $p(x)$ and determine if the resulting value is equal to $0$.
Step 2: Substitution and Algebraic Evaluation
Substitute $x = \frac{4}{5}$ into the polynomial $p(x)$:
$p\left(\frac{4}{5}\right) = 5\left(\frac{4}{5}\right) - \pi$
Perform the multiplication. The factor of $5$ in the numerator and the denominator cancel each other out:
$p\left(\frac{4}{5}\right) = 4 - \pi$
Step 3: Analytical Verification
We must now evaluate the expression $4 - \pi$.
- The number $4$ is a rational integer.
- The number $\pi$ is an irrational mathematical constant representing the ratio of a circle's circumference to its diameter, where $\pi \approx 3.14159...$
Because $\pi$ is strictly less than $4$ (and is irrational), the difference $4 - \pi$ evaluates to a non-zero irrational number:
$4 - \pi \approx 4 - 3.14159 = 0.85841 \neq 0$
| Evaluated Point ($x$) | Polynomial Expression $p(x)$ | Resulting Value | Is $p(x) = 0$? |
|---|---|---|---|
| $x = \frac{\pi}{5}$ (Actual Zero) | $5\left(\frac{\pi}{5}\right) - \pi$ | $0$ | Yes |
| $x = \frac{4}{5}$ (Tested Point) | $5\left(\frac{4}{5}\right) - \pi$ | $4 - \pi \approx 0.858$ | No |
Step 4: Graphical Representation (Visual Verification)
Graphically, the zeroes of a polynomial $p(x)$ correspond to the $x$-intercepts of the graph $y = p(x)$. The graph below plots the linear function $y = 5x - \pi$. The true zero is located at $x = \frac{\pi}{5} \approx 0.628$, while our tested point $x = \frac{4}{5} = 0.8$ clearly yields a positive $y$-value, proving it does not intersect the $x$-axis at this coordinate.
Final Solution: Since $p\left(\frac{4}{5}\right) = 4 - \pi \neq 0$, the value $x = \frac{4}{5}$ is NOT a zero of the polynomial $p(x) = 5x - \pi$.
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(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(i): Find the zero of the polynomial in each of the following cases: (i) $p(x) = x + 5$
- 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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