Polynomials Exercise 2.2 Class 10 Solution.

 

Polynomials Exercise 2.2 Class 10 Solution.

Introduction:

यह अभ्यास (Exercise 2.2) कक्षा 10 गणित के Polynomials अध्याय का एक महत्वपूर्ण भाग है। इसमें बहुपदों (polynomials) के zeroes और उनके coefficients के बीच संबंध को समझाया गया है। इस अभ्यास में विद्यार्थियों को quadratic polynomials के zeroes निकालना और उनके योग (sum) तथा गुणनफल (product) को verify करना सिखाया जाता है। साथ ही, दिए गए zeroes के आधार पर polynomial बनाना भी सिखाया गया है। सभी प्रश्नों को सरल और step-by-step तरीके से हल किया गया है, ताकि विद्यार्थियों को यह concept आसानी से समझ में आ सके।

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Important Note:

To find zeroes and relation between zeroes and coefficients.

            Formula:
                    For $ax^2 + bx + c$

• Sum of zeroes = $-\frac{b}{a}$
• Product of zeroes = $\frac{c}{a}$(Coefficient of x²)

(i) $x^2 - 2x - 8$

Solution:

Factor of given quadratic polynomials:
$(x - 4)(x + 2) = 0$

Zeroes = 4, -2

Relationship between the zeroes and the coefficients.

• Sum = $4 + (-2) = 2$
  
• $-b/a = -(-2)/1 = 2$
  
• Product = $4 × (-2) = -8$
  
• $c/a = -8/1 = -8$
  

(ii) $4s^2 - 4s + 1$

Solution:

Factor of given quadratic polynomials:

$(2s - 1)^2 = 0$

Zeroes = 1/2, 1/2

Relationship between the zeroes and the coefficients.

• Sum = $1/2 + 1/2 = 1$
  
• $-b/a = -(-4)/4 = 1$
  
• Product = 1/2 × 1/2 =  $1/4$
  
• $c/a = 1/4$
  

(iii) $6x^2 - 7x - 3$

Solution:

Factor of given quadratic polynomials:

$(3x + 1)(2x - 3) = 0$

Zeroes = -1/3, 3/2

Relationship between the zeroes and the coefficients.

• Sum = $-1/3 + 3/2 = 7/6$
  
• $-b/a = -(-7)/6 = 7/6$
  
• Product = $-1/2$
  
• $c/a = -3/6 = -1/2$
  

(iv) $4u^2 + 8u$

Solution:

Factor of given quadratic polynomials:

$4u(u + 2) = 0$

Zeroes = 0, -2

Relationship between the zeroes and the coefficients.

• Sum = $-2$
  
• $-b/a = -8/4 = -2$
  
• Product = 0
• $c/a = 0$
  

(v) $t^2 - 15$

Solution:

Factor of given quadratic polynomials:

$t = ±\sqrt{15}$

Zeroes = √15, -√15

Relationship between the zeroes and the coefficients.

• Sum = 0
• $-b/a = 0$
  
• Product = $-15$
  
• c/a =  $-15$
  

(vi) $3x^2 - x - 4$

Solution:

Factor of given quadratic polynomials:

$(3x - 4)(x + 1) = 0$

Zeroes = 4/3, -1

Relationship between the zeroes and the coefficients.

• Sum = $4/3 - 1 = 1/3$
  
• −b/a=1/3
• Product = $-4/3$
  
• c/a = −4/3

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Important Note:

To find polynomial from given zeroes

Formula:
            If zeroes = $α, β$

Polynomial:
$x^2-(\alpha +\beta )x+(\alpha \beta )$

(i) $1/4, -1$

Solution:

Here, α = 1/4 and  β = -1

• Sum = (α + β) = $-3/4$
  
• Product = (α × β) = −1/4
  

By comparing with general polynomial 

$x^2-(\alpha +\beta )x+(\alpha \beta )$

The required quadratic polynomial:
$x^2 + \frac{3}{4}x - \frac{1}{4}$

Multiply by 4:
4x² + 3x - 1

(ii) $\sqrt{2}, 1/3$

$Solution:$

Here, α and β are given

• Sum = $\sqrt{2} + 1/3$
  
• Product = $\sqrt{2}/3$
  

The required quadratic polynomial:

$x^2 - (\sqrt{2} + 1/3)x + \sqrt{2}/3$

(iii) $0, \sqrt{5}$

$Solution:$

• Sum = $\sqrt{5}$
• Product = 0

The required quadratic polynomial:
x² - √5x

(iv) $1, 1$

$Solution:$

• Sum = 2
• Product = 1

The required quadratic polynomial:
x² - 2x + 1

(v) $-1/4, 1/4$

$Solution:$

• Sum = 0
• Product = $-1/16$
  

The required quadratic polynomial:
$x^2 - 1/16$

Multiply by 16:
16x² - 1

(vi) $4, 1$

$Solution:$

• Sum = 5
• Product = 4

The required quadratic polynomial:
x² - 5x + 4

Conclusion:

इस अभ्यास में हमने zeroes निकालना और coefficients के साथ उनका संबंध verify करना सीखा। साथ ही, दिए गए zeroes से polynomial बनाना भी समझा। यह chapter algebra की foundation को मजबूत करता है और exam के लिए बहुत महत्वपूर्ण है।

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