NCERT Solutions for Class 9 Maths Exercise 2.2 Question 3

NCERT Solutions for Class 9 Maths Exercise 2.2 Question 3

NCERT Solutions for Class 9 Maths Exercise 2.2 Question 3

ncert-solutions-for-class-9-maths-exercise-2-2-question-3

Understanding the Question 🧐

This question asks us to verify if a given value of a variable is a “zero of the polynomial”. A value is a “zero” if, when we substitute it into the polynomial, the entire expression simplifies to 0. If the result is any other number, it is not a zero.

Verify whether the following are zeroes of the polynomial, indicated against them.


Solutions and Verifications 📝

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

Substitute &&x = -\frac{1}{3}&&: &&p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1 = -1 + 1 = 0&&
Since the result is 0, Yes, &&-\frac{1}{3}&& is a zero of the polynomial.

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

Substitute &&x = \frac{4}{5}&&: &&p(\frac{4}{5}) = 5(\frac{4}{5}) – \pi = 4 – \pi&&
Since &&4 – \pi \neq 0&&, No, &&\frac{4}{5}&& is not a zero of the polynomial.

(iii) &&p(x) = x^2 – 1, x = 1, -1&&

We need to check both values.
For &&x = 1&&: &&p(1) = (1)^2 – 1 = 1 – 1 = 0&&.
For &&x = -1&&: &&p(-1) = (-1)^2 – 1 = 1 – 1 = 0&&.
Since both results are 0, Yes, both 1 and -1 are zeroes of the polynomial.

(iv) &&p(x) = (x + 1)(x – 2), x = -1, 2&&

Check both values.
For &&x = -1&&: &&p(-1) = (-1 + 1)(-1 – 2) = (0)(-3) = 0&&.
For &&x = 2&&: &&p(2) = (2 + 1)(2 – 2) = (3)(0) = 0&&.
Since both results are 0, Yes, both -1 and 2 are zeroes of the polynomial.

(v) &&p(x) = x^2, x = 0&&

Substitute &&x = 0&&: &&p(0) = (0)^2 = 0&&
Since the result is 0, Yes, 0 is a zero of the polynomial.

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

Substitute &&x = -\frac{m}{l}&&: &&p(-\frac{m}{l}) = l(-\frac{m}{l}) + m = -m + m = 0&&
Since the result is 0, Yes, &&-\frac{m}{l}&& is a zero of the polynomial.

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

Check both values.
For &&x = -\frac{1}{\sqrt{3}}&&: &&p(-\frac{1}{\sqrt{3}}) = 3(-\frac{1}{\sqrt{3}})^2 – 1 = 3(\frac{1}{3}) – 1 = 1 – 1 = 0&&.
For &&x = \frac{2}{\sqrt{3}}&&: &&p(\frac{2}{\sqrt{3}}) = 3(\frac{2}{\sqrt{3}})^2 – 1 = 3(\frac{4}{3}) – 1 = 4 – 1 = 3&&.
The first value gives 0, but the second gives 3.
Conclusion: &&-\frac{1}{\sqrt{3}}&& is a zero, but &&\frac{2}{\sqrt{3}}&& is not a zero.

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

Substitute &&x = \frac{1}{2}&&: &&p(\frac{1}{2}) = 2(\frac{1}{2}) + 1 = 1 + 1 = 2&&
Since the result is not 0, No, &&\frac{1}{2}&& is not a zero of the polynomial.

📝 Key Points:
  • A zero of a polynomial is a value that makes the polynomial equal to 0.
  • It is also called a root of the polynomial equation &&p(x) = 0&&.
  • To check if a value is a zero, simply substitute it into the polynomial. If the result is 0, it’s a zero.

FAQ ❓

Q: What is a ‘zero of a polynomial’?

A: A zero of a polynomial is a special number that, when you substitute it for the variable, makes the whole polynomial equal to 0.

Q: Can a polynomial have two zeroes?

A: Yes. A polynomial can have multiple zeroes. For example, a quadratic polynomial (degree 2) like &&x^2 – 1&& can have up to two zeroes, which we found to be 1 and -1.

Further Reading 📖

To learn more about finding the zeroes of polynomials, you can refer to the official NCERT textbook for Class 9 Maths, Chapter 2. More resources are available on the NCERT website at https://ncert.nic.in/.

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