Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $ p(x)=3 x+1, x=-\frac{1}{3} $
(ii) $ p(x)=5 x-\pi, x=\frac{4}{5} $
(iii) $ p(x)=x^{2}-1, x=1,-1 $
(iv) $ p(x)=(x+1)(x-2), x=-1,2 $
(v) $ p(x)=x^{2}, x=0 $
(vi) $ p(x)=l x+m, x=-\frac{m}{l} $
(vii) $ p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} $
(viii) $ p(x)=2 x+1, x=\frac{1}{2} $

AcademicMathematicsNCERTClass 9

To do:

We have to verify whether the given values are zeroes of the given polynomials. 

Solution:

The zero of the polynomial is defined as any real value of $x$, for which the value of the polynomial becomes zero.

Therefore,

(i) \( p(x)=3 x+1, x=-\frac{1}{3} \)

$p(-\frac{1}{3})=3(-\frac{1}{3})+1$

$=-1+1$

$=0$

Therefore, $x=-\frac{1}{3}$ is the zero of the polynomial $p(x)=3 x+1$.

(ii) \( p(x)=5 x-\pi, x=\frac{4}{5} \)

$p(\frac{4}{5})=5(\frac{4}{5})-\pi$

$=4-\pi$

$≠0$

Therefore, $x=\frac{4}{5}$ is not the zero of the polynomial $p(x)=5x-\pi$.

(iii) \( p(x)=x^{2}-1, x=1,-1 \)

$p(1)=(1)^2-1$

$=1-1$

$=0$

$p(-1)=(-1)^2-1$

$=1-1$

$=0$

Therefore, $x=1,-1$ are the zeroes of the polynomial $p(x)=x^2-1$.

(iv) \( p(x)=(x+1)(x-2), x=-1,2 \)

$p(-1)=(-1+1)(-1-2)$

$=(0)(-3)$

$=0$

$p(2)=(2+1)(2-2)$

$=(3)(0)$

$=0$

Therefore, $x=-1,2$ are the zeroes of the polynomial $p(x)=(x+1)(x-2)$.

(v) \( p(x)=x^{2}, x=0 \)

$p(0)=(0)^2$

$=0$

Therefore, $x=0$ is the zero of the polynomial $p(x)=x^2$.

(vi) \( p(x)=l x+m, x=-\frac{m}{l} \)

$p(-\frac{m}{l})=l(-\frac{m}{l})+m$

$=-m+m$

$=0$

Therefore, $x=-\frac{m}{l}$ is the zero of the polynomial $p(x)=l x+m$.

(vii) \( p(x)=3 x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \)

$p(-\frac{1}{\sqrt3})=3(-\frac{1}{\sqrt3})^2-1$

$=3(\frac{1}{3})-1$

$=1-1$

$=0$

$p(-\frac{2}{\sqrt3})=3(-\frac{2}{\sqrt3})^2-1$

$=3(\frac{4}{3})-1$

$=4-1$

$=3$

$≠0$

Therefore, $x=-\frac{1}{\sqrt3}$ is the zero of the polynomial $p(x)=3x^2-1$ and $x=-\frac{2}{\sqrt3}$ is not the zero of the polynomial $p(x)=3x^2-1$.

(viii) \( p(x)=2 x+1, x=\frac{1}{2} \)

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

$=1+1$

$=2$

$≠0$

Therefore, $x=\frac{1}{2}$ is not the zero of the polynomial $p(x)=2 x+1$.

raja
Updated on 10-Oct-2022 13:39:07

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