NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-2-ex-2-2/

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Polynomials |

Exercise |
Ex 2.2 |

Number of Questions Solved |
2 |

Category |
NCERT Solutions |

### NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

**Question 1.**

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

**(i) x ^{2} – 2x – 8**

**(ii) 4s**

^{2}– 4s + 1**(iii) 6x**

^{2}– 3 – 7x**(iv) 4u**

^{2}+ 8u**(v) t**

^{2}-15**(vi) 3x**

^{2}– x – 4**Solution:**

**Question 2.**

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

**Solution:**

**(i)** Zeroes of polynomial are not given, sum of zeroes = \(\frac { 1 }{ 4 }\) and product of zeroes = -1

If ax^{2} + bx + c is a quadratic polynomial, then

α + β = sum of zeroes = \(\frac { -b }{ a }\) = \(\frac { 1 }{ 4 }\) and αβ = product of zeroes = \(\frac { c }{ a }\) = -1

Quadratic polynomial is ax^{2} + bx + c

Let a = k, ∴ b = \(\frac { -k }{ 4 }\) and c = -k

Putting these values, we get

For different values of k, we can have quadratic polynomials all having sum of zeroes as \(\frac { 1 }{ 4 }\) and product of zeroes as -1.

**(ii)** Sum of zeroes = α + β = √2 = \(\frac { -b }{ a }\); product of zeroes = αβ = \(\frac { 1 }{ 3 }\) = \(\frac { c }{ a }\)

Quadratic polynomial is ax^{2} + bx + c

Let a = k,b = -√2k and c = \(\frac { k }{ 3 }\)

Putting these values we get

For all different real values of k, we can have different quadratic polynomials of the form 3×2 – 3√2x +1 having sum of zeroes = √2 and product of zeroes = \(\frac { 1 }{ 3 }\)

**(iii)** Sum of zeroes = α + β = 0 = \(\frac { -b }{ a }\); product of zeroes = αβ = √5 = \(\frac { c }{ a }\)

Let quadratic polynomial is ax^{2} + bx + c

Let a = k,b = 0, c = √5 k

Putting these values, we get

k[x^{2} – 0x + √5 ] = k(x^{2} + √5).

For different real values of k, we can have different quadratic polynomials of the form

x^{2} + √5, having sum of zeroes = 0 and product of zeroes = √5

**(iv)** Sum of zeroes = α + β = 1= \(\frac { -b }{ a }\); product of zeroes = αβ = 1 = \(\frac { c }{ a }\)

Let quadratic polynomial is ax^{2} + bx + c.

Let a=k, c = k, b = -k

Putting these values, we get k[x^{2} -x +1]

Quadratic polynomial is of the form x^{2} -x + 1 for different values of k.

**(v)** Sum of zeroes = α + β = \(\frac { -1 }{ 4 }\)= \(\frac { -b }{ a }\); product of zeroes = αβ = \(\frac { 1 }{ 4 }\) = \(\frac { c }{ a }\)

Let quadratic polynomial is ax^{2} + bx + c

Let a=k, b= \(\frac { k }{ 4 }\), c= \(\frac { k }{ 4 }\)

Putting these values, we get k

Quadratic polynomial is of the form 4x^{2} +x + 1 for different values of k.

**(vi)** Sum of zeroes = α + β = 4 = \(\frac { -b }{ a }\); product of zeroes = αβ = 1 = \(\frac { c }{ a }\)

Let quadratic polynomial is ax^{2} + bx + c

Let a = k,b = -4k and c = k

Putting these values, we get

k[x^{2} – 4x + 1]

Quadratic polynomial is of the form x^{2} – 4x + 1 for different values of k.

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