IX Std Term I - Chapter 3 - Ex 3.2

Exercise 3.2

 1. Find the zeros of the following polynomials.

 (i)                 p(x) = 4x – 1

Solution:

Given that,

p(x) = 4x – 1 = 4(x -1/4)  

p(1/4) = 4(1/4 - 1/4) = 0

Hence, 1/4  is the zero of p(x).

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(ii)               p(x) = 3x + 5

Solution:

Given that,

p(x) = 3x + 5 = 3(x + 5/3)

p(-5/3) = 3(-5/3 + 5/3) = 0

Hence, - 5/3 is the zero of p(x).

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(iii)             p(x) = 2x

Solution:

Given that,

p(x) = 2x

p(0) = 2 × 0 = 0

Hence, 0 is the zero of p(x).

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(iv)             p(x) = x + 9

Solution:

Given that,

p(x) = x + 9

p(-9) = -9 + 9 = 0

Hence, -9 is the zero of p(x).

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2. Find the roots of the following polynomial equations.

 

(i)                 x - 3 = 0

Solution:

x – 3 = 0

ð  x = 3

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(ii)               5x - 6 = 0

Solution:

5x – 6 = 0

ð  5x = 6

ð  x = 6/5 

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(iii)             11x + 1 = 0

Solution:

11x + 1 = 0

ð  11x = -1

ð     x  = - 1/11

 

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(iv)             -9x = 0

Solution:

-9x = 0

ð  x = 0/-9    

ð  x = 0

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3. Verify whether the following are roots of the polynomial equations indicated against them.

(i)                 x² -  5x + 6 = 0; x = 2, 3

Solution:

Let p(x) = x² -  5x + 6 = 0

P(2) = 2² – (5 × 2) + 6 = 4 – 10 +6 = 0

Hence, x = 2 is the root of p(x).

P(3) = 3² – (5× 3) + 6 = 9 – 15 + 6 = 0

Hence, x = 3 is the root of p(x).  

 
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(ii)               x² + 4x + 3 = 0; x = -1, 2

Solution:

Let p(x) = x² + 4x + 3 = 0

P(-1) = (-1)² + (4× (-1)) + 3 = 1 – 4 + 3 = 0

Hence, x = -1 is the root of p(x).

P(2) = 2² + (4 × 2) + 3 = 4 +8 + 3 = 15 ≠ 0

Hence, x = 2 is not the root of p(x).

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(iii)             x³ - 2x² - 5x + 6 = 0; x = 1, -2, 3

Solution:

Let p(x) = x³ - 2x² - 5x + 6 = 0

P(1) = 1³ – (2 × 1²) – (5 × 1) + 6 = 1 – 2 – 5 + 6 = 0

Hence, x = 1 is the root of p(x).

P(-2) = (-2)³ – (2 × (-2)²) – (5 × (-2)) + 6 = -8 – 8 + 10 + 6 = 0

Hence, x = -2 is the root of p(x).

P(3) = 3³ – (2× (3)²) – (5 × 3) + 6 = 27 – 18 – 15 + 6 = 0

Hence, x = 3 is the root of p(x).

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(iv)             x³ -  2x² – x + 2 = 0; x = -1, 2, 3

Solution:

Let p(x) = x³ - 2x² – x + 2 = 0

 

P(-1) = (-1)³ – (2 × (-1)²) – (-1) + 2 = -1 - 2 + 1 + 2 = 0

Hence, x = -1 is the root of p(x).

P(2) = (2)³ – (2 × 2²) – 2 + 2 = 8 – 8 - 2 + 2 = 0

Hence, x = 2 is the root of p(x).

P(3) = (3)³ – (2 × 3²) – 3 + 2 = 27 – 18 – 3 + 2 = 8 ≠ 0

Hence, x = 3 is not the root of p(x).

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