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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 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) When y = 1

f(y) = 6y – 3y^{2} + 3

f(1) = 6(1) – 3(1)^{2} + 3

= 6 – 3 + 3 = 6

(ii) When y = – 1

f(y) = 6y – 3y^{2} + 3

f(-1) = 6(-1) – 3(-1)^{2} + 3

= – 6 – 3 + 3

= – 6

(iii) When y = 0

f(y) = 6y – 3y^{2} + 3

f(0) = 6(0) – 3(0)^{2} + 3

= 0 – 0 + 3

= 3

Question 2.

If p(x) = x^{2} – 2√2x + 1, find p(2√2).

Solution:

p(x) = x^{2} – 2√2x + 1

p(2√2) = (2√2)^{2} – 2√2 (2√2) + 1

= 8 – 8 + 1

= 0 + 1

= 1

Question 3.

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

(i) P(x) = x – 3

Solution:

p( 3) = 3 – 3

= 0

p(3) is the zero of p(x)

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

Solution:

= -5 + 5

= 2(0)

= 0

Hence –\(\frac{5}{2}\) is the zero of p(x).

(iii) q(y) = 2y – 3

Solution:

= 2 × 0

= 0

Hence \(\frac{3}{2}\) is the zero of q(y).

(iv) f(z) = 8z

Solution:

f(0) = 8 × 0

= 0

Hence 0 is the zero of f(z)

(v) p(x) = ax when a ≠ 0

Solution:

p(0) = a(0)

= 0

Hence, 0 is the zero of p(x)

(vi) h(x) = ax + b, a ≠ 0, a, b∈R

Solution:

Hence –\(\frac{b}{a}\) is the zero of h(x).

Question 4.

Find the roots of the polynomial equations.

(i) 5x – 6 = 0

Solution:

5x = 6

x = \(\frac{6}{5}\)

\(\frac{6}{5}\) is the root of the polynomial.

(ii) x + 3 = 0

Solution:

x = -3

-3 is the root of the polynomial.

(iii) 10x + 9 = 0

Solution:

10x = -9

x = –\(\frac{9}{10}\)

–\(\frac{9}{10}\) is the root of the polynomial.

(iv) 9x – 4 = 0

Solution:

9x = 4

x = \(\frac{4}{9}\)

\(\frac{4}{9}\) is the root of the polynomial.

Question 5.

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

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

Solution:

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

= 1 – 1

= 0

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

(ii) p(x) = x^{3} – 1, x = 1

Solution:

p(1) = 1^{3} – 1

= 1 – 1

= 0

∴ 1 is the zero of the polynomial

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

Solution:

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

= -b + b

= 0

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

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

Solution:

P(-3) = (-3 + 3) (-3 – 4)

= (0) (-7)

= 0

P( 4) = (4 + 3) (4 – 4)

= (7) (0)

= 0

∴ -3 and 4 are the zeros of the polynomial.

Question 6.

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

Solution:

(i) Number of zeros = 2 (The curve is intersecting the x-axis at 2 points)

(ii) Number of zeros = 3 (The curve is intersecting the x-axis at 3 points)

(iii) Number of zeros = 0 (The curve is not intersecting the x-axis)

(iv) Number of zeros = 1 (The curve is intersecting at the origin)

(v) Number of zeros = 1 (The curve is intersecting the x-axis at one point)