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Q3(viii):
Verify whether the following are zeroes of the polynomial, indicated against them. (viii) $p(x) = 2x + 1, x = \frac{1}{2}$

Solution :

Initial Setup & Theoretical Foundation

We are tasked with verifying whether the given value of $x$ is a zero (or root) of the provided linear polynomial. The given parameters are:

  • Polynomial: $p(x) = 2x + 1$
  • Value to verify: $x = \frac{1}{2}$

[Theoretical Justification: By the fundamental definition of polynomials, a real number $c$ is considered a "zero" of a polynomial $p(x)$ if and only if evaluating the polynomial at $x = c$ yields a result of zero. Mathematically, this is expressed as $p(c) = 0$.]

Step 1: Substitution of the Given Value

To test the condition $p(c) = 0$, we substitute $x = \frac{1}{2}$ into the polynomial $p(x)$.

$p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) + 1$

Step 2: Algebraic Evaluation

Next, we perform the arithmetic operations following the standard order of operations (PEMDAS/BODMAS).

First, multiply the terms:

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

Now, substitute this product back into the expression:

$p\left(\frac{1}{2}\right) = 1 + 1$

$p\left(\frac{1}{2}\right) = 2$

Step 3: Logical Verification

We must now compare our evaluated result with the required condition for a zero.

$p\left(\frac{1}{2}\right) = 2 \neq 0$

[Since the polynomial evaluates to $2$ rather than $0$, the value $x = \frac{1}{2}$ fails the condition $p(c) = 0$.]

Graphical Representation & Geometric Proof

Geometrically, the zeroes of a polynomial $p(x)$ correspond to the $x$-intercepts of its graph $y = p(x)$. The graph below plots the linear function $y = 2x + 1$. Notice that the true zero occurs where the line crosses the x-axis at $x = -\frac{1}{2}$, whereas evaluating at $x = \frac{1}{2}$ yields a y-coordinate of $2$.

x y 0 1 -1 1 2 y = 2x + 1 True Zero: (-0.5, 0) Tested Point: (0.5, 2)

Final Conclusion

Because the evaluation of the polynomial at the given value does not result in zero, the value is not a root of the equation.

Final Solution: No, $x = \frac{1}{2}$ is not a zero of the polynomial $p(x) = 2x + 1$, because $p\left(\frac{1}{2}\right) = 2 \neq 0$.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.2


CBSE Solutions for Class 9 Mathematics Polynomials


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