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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.2

**9th Maths Algebra Exercise 3.2 Question 1.**

Find the value of the polynomial f(y) = 6y – 3y^{2} + 3 at (i) y = 1 (ii) y = -1 (iii) y = 0

Solution:

(i) At y = 1,

f(1) = 6(1) – 3(1)^{2} + 3 = 6 – 3 + 3 = 6

(ii) At y = -1,

f(-1) = 6(-1) – 3(-1)^{2} + 3 = -6 – 3 + 3 = -6

(iii) At y = 0,

f(0) = 6(0) – 3(0)^{2} + 3 = 0 – 0 + 3 = 3

**9th Maths Exercise 3.2 Question 2.**

If f(x) = x^{2} – \(2 \sqrt{2} x\) + 1, find p (\(2 \sqrt{2}\))

Solution:

p(\(2 \sqrt{2}\)) = (\(2 \sqrt{2}\))^{2} – \(2 \sqrt{2}\)(\(2 \sqrt{2}\)) + 1

= 4 × 2 – 4 × 2 + 1

= 8 – 8 + 1 = 1

**Ex 3.2 Class 9 Algebra Question 3.**

Find the zeros of the polynomial in each of the following :

(i) p(x) = x – 3

(ii) p(x) = 2x + 5

(iii) q(y) = 2y – 3,

(iv) f(z) = 8z

(v) p(x) = ax where a ≠ 0,

(vi) h(x) = ax + b, a ≠ 0, a, b ∈ R

Solution:

(i) x = 3.

p( 3) = 3 – 3 = 0

∴ The zero of the polynomial is x = 3.

(iv) f(z) = 8z,

If 8z = 0

z = \(\frac{0}{8}\) = 0

f(0) = 8(0) = 0

∴ z = 0 is the zero of the given polynomial.

**Class 9 Maths Exercise 3.2 Solutions Question 4.**

Find the roots of the polynomial equations.

(i) 5x – 6 = 0

(ii) x + 3 = 0

(iii) 10x + 9 = 0

(iv) 9x – 4 = 0

Solution:

(i) 5x – 6 = 0

5x = 6

∴ x = \(\frac{6}{5}\)

(ii) x + 3 = 0

∴ x = -3

(iii) 10x + 9 = 0

10x = -9

∴ x = \(\frac{-9}{10}\)

(iv) 9x – 4 = 0

9x = 4

∴ x = \(\frac{4}{9}\)

**Algebra 3.2 Answers Question 5.**

Verify whether the following are zeros of the polynomial indicated against them, or not.

(i) p(x) = 2x – 1, x = \(\frac{1}{2}\)

(ii) p(x) = x^{3} – 1, x = 1,

(iii) p(x) = ax + b, x = \(\frac{-b}{a}\)

(iv) p(x) = (x + 3) (x – 4), x = 4, x = -3

Solution:

(i) p(x) = 2x – 1, x = \(\frac{1}{2}\)

p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\)) – 1 = 1 – 1 = 0

∴ x = \(\frac{1}{2}\) is the zero of the given polynomial.

(ii) p(x) = x^{3} – 1, x = 1

p(1) = 1^{3} – 1 = 1 – 1 = 0

∴ x = 1 is the zero of the given polynomial

(iii) p(x) = ax + b, x = \(\frac{-b}{a}\)

p(\(\frac{-b}{a}\)) = a(\(\frac{-b}{a}\)) + b

= -b + b = 0

∴ x = \(\frac{-b}{a}\) is the zero of the given polynomial.

(iv) p(x) = (x + 3) (x – 4), x = 4, x = -3

p(-3) = (-3 + 3) (-3 – 4) = 0(-7) = 0

p(4) = (4 + 3) (4 – 4) = 7(0) = 0

∴ x = -3, x = 4 are the zeros of the given polynomial.

**Samacheer Kalvi 9th Standard Maths Question 6.**

Find the number of zeros of the following polynomial represented by their graphs.

Solution:

(i) The curve cuts the x-axis at two points. ∴ The equation has 2 zeros.

(ii) Since the curve cuts the x-axis at 3 different points. The number of zeros of the given curve is three.

(iii) Since the curve doesn’t cut the x axis. The number of zeros of the given curve is zero.

(iv) The curve cut the x-axis at one point. ∴ The equation has one zero.

(v) The curve cut the x axis at one point. ∴ The equation has one zero.