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Ex.2.2 Q4 Polynomials Solution - NCERT Maths Class 9

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Question

Find the zero of the polynomials in each of the following cases:

(i) \(\begin{align} p(x)=x+5 \end{align}\)

(ii) \(\begin{align}p(x)=x-5\end{align}\)

(iii) \(\begin{align}p(x)=2 x+5\end{align}\)

(iv)\(\begin{align}p(x)=3 x-2\end{align}\)

(v) \(\begin{align}p(x)=3 x\end{align}\)

(vi) \(\begin{align}p(x)=a x, a \neq 0\end{align}\)

(vii) \(\begin{align}p(x)=c x+d, c \neq 0, c, d \end{align}\) are real numbers.

 Video Solution
Polynomials
Ex 2.2 | Question 4

Text Solution

Reasoning:

In general, we say that a zero of a polynomial \(p(x) \) is a number \(c\) such that\(p(c) = 0.\)

Steps:

(i) \(\begin{align} p(x)=x+5 \end{align}\)

\[\begin{align}p(x)&=x+5 \\ \qquad p(x)&=0 \rightarrow x+5=0 \\ \Rightarrow x&=-5 \end{align}\]

\(\therefore-5\) is the zero of \(p(x) \)

(ii) \(\begin{align}p(x)=x-5\end{align}\)

\[\begin{align}{p(x)}&={x-5} \\ p(x)&=0 \rightarrow x-5=0 \\ \Rightarrow x&=5 \end{align}\]

\(\therefore 5\) is the zero of \(p(x) \)

(iii) \(\begin{align}p(x)=2 x+5\end{align}\)

\[\begin{align}p(x)&=2 x+5 \\p(x)&=0 \rightarrow 2 x+5=0 \\ {\Rightarrow 2 x}&={-5} \\ {\Rightarrow x}&={\frac{-5}{2}} \\ \end{align}\]

\(\begin{align}\therefore \frac{-5}{2}\end{align}\) is the zero of \(p(x) \)

(iv)\(\begin{align}p(x)=3 x-2\end{align}\)

\[\begin{align}p(x)& =3 x-2 \\ {p(x)}&={0 \rightarrow 3 x-2=0} \\ {\Rightarrow 3 x}&={2} \\ {\Rightarrow x}&={\frac{2}{3}} \end{align}\]

\(\begin{align}\therefore \frac{2}{3}\end{align}\) is the zero of \(p(x) \)

(v) \(\begin{align}p(x)=3 x\end{align}\)

\[\begin{align} p(x)&=3 x \\ p(x)&=0 \rightarrow 3 x =0 \\ x&=0 \\ \end{align}\]

\(\therefore 0\) is the zero of \(p(x) \)

(vi) \(\begin{align}p(x)=a x, a \neq 0\end{align}\)

\[\begin{align}p(x)&=a x, a \neq 0 \\ {p(x)}&={0 \rightarrow a x=0} \\ {\quad x}&={0} \end{align}\]

\(\therefore 0\) is the zero of \(p(x) \)

(vii) \(\begin{align}p(x)=c x+d, c \neq 0, c, d \end{align}\) are real numbers.

\[\begin{align}{p(x)}&=c x+d, c \neq 0, c, d \quad \\&\text { are real numbers. } \\\\ {p(x)}&={0 \rightarrow c x+d=0} \\ {\Rightarrow c x}&={-d} \\ {\Rightarrow x}&={\frac{-d}{c}}\end{align}\]

\(\begin{align}\therefore \frac{-d}{c}\end{align}\) is the zero of \(p(x) \)

 Video Solution
Polynomials
Ex 2.2 | Question 4
  
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