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.

- Polynomials Class 10 Ex 2.1
- Polynomials Class 10 Ex 2.2
- Polynomials Class 10 Ex 2.3
- Polynomials Class 10 Ex 2.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

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

**Ex 2.2 Class 10 Maths Question 1.
**Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their 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:**

**(i)**x

^{2}– 2x – 8 = x

^{2}– 4x + 2x – 8

= x(x – 4) + 2(x – 4)

= (x + 2) (x – 4)

Either x + 2 = 0 or x – 4 = 0

⇒ x = -2 or x = 4

Hence, zeroes of this polynomial are -2 and 4.

Verification:

Sum of the zeroes = (-2) + (4) = 2

= \(\frac { (-2) }{ 1 }\) = \(\frac { -b }{ a }\)

Product of zeroes = (-2) (4) = -8 = \(\frac { -8 }{ 1 }\)

= \(\frac { c }{ a }\)

Hence verified.

**(ii)** 4s^{2} – 4s + 1 = (2s – l)^{2} = (2s – l)(2s – 1)

Either 2s – 1 = 0 or 2s – 1 = 0

i.e. , s = \(\frac { 1 }{ 2 }\) , \(\frac { 1 }{ 2 }\)

Hence, the two Zeroes are \(\frac { 1 }{ 2 }\) and \(\frac { 1 }{ 2 }\)

Verification:

Hence verified.

**(iii)** 6x^{2} – 3 – 7x = 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

= 3x (2x – 3) + 1(2x – 3)

= (2x – 3) (3x + 1)

Either 2x – 3 = 0 or 3x+1 = 0

Hence verified.

**(iv)** 4u^{2} + 8u ⇒ 4u(u + 2)

Either 4u = 0 or u + 2 = 0

⇒ u = 0 or u = -2

Hence, the two zeroes are 0 and -2.

Verification:

Sum of the zeroes = 0 + (-2) = -2

= \(\frac { -8 }{ 4 }\) = \(\frac { -b }{ a }\)

Product of zeroes = 0 x (-2) = 0 = \(\frac { c }{ a }\)

Hence verified.

**(v)** t^{2} – 15 = t^{2} – (\( \sqrt{15} \))^{2
}= (t + (\( \sqrt{15} \)) (t- (\( \sqrt{15} \))

Either t + (\( \sqrt{15} \) = 0 or t – (\( \sqrt{15} \) = 0

⇒ t = -(\( \sqrt{15} \) or t = (\( \sqrt{15} \)

Hence, the two zeroes are -(\( \sqrt{15} \) and + \( \sqrt{15} \).

Verification:

Sum of the zeroes = -(\( \sqrt{15} \) + \( \sqrt{15} \) = 0

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

Product of zeroes = –\( \sqrt{15} \) x \( \sqrt{15} \) = -15

= \(\frac { c }{ a }\)

Hence verified.

**(vi)** 3x^{2} – x – 4 = 3x^{2} – 4x + 3x – 4

= x(3x – 4) + l(3x 4)

= (x + 1) (3x – 4)

Either x + 1 = 0 or 3x-4 = 0

⇒ x = -1 or x = \(\frac { 4 }{ 3 }\)

Verification:

Sum of the zeroes = -1 + \(\frac { 4 }{ 3 }\) = \(\frac { 1 }{ 3 }\) = \(\frac { -b }{ a }\)

Product of zeroes = -1 x \(\frac { 4 }{ 3 }\) = \(\frac { -4 }{ 3 }\) = \(\frac { c }{ a }\)

Hence verified.

**Ex 2.2 Class 10 Maths Question 2.
**Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively:

**Solution:**

**(i)**Let the zeroes of polynomial be α and β.

Then, α + β = \(\frac { 1 }{ 4 }\) and αβ = -1

∴ Required polynomial is given by,

x

^{2}– (α + β)x + αβ = x

^{2}– \(\frac { 1 }{ 4 }\)x + (-1)

= x

^{2}– \(\frac { 1 }{ 4 }\)x – 1

= 4x

^{2}– x – 4

**(ii)** Let the zeroes of polynomial be α and β.

Then, α + β= √2 and αβ = \(\frac { 1 }{ 3 }\)

∴ Required polynomial is:

x^{2} – (α + β)x + αβ = x^{2} – √2x + \(\frac { 1 }{ 3 }\)

= 3x^{2} – 3√2x + 1

**(iii)** Let the zeroes of the polynomial be α and β.

Then, α + β = 0 and αβ = √5

∴ Required polynomial

= x^{2} – (α + β)x + αβ

= x^{2}– 0 x x + √5 = x^{2} + √5

**(iv)** Let the zeroes of the polynomial be α and β.

Then, α + β = 1 and αβ = 1.

∴ Required polynomial

= x^{2} – (α + β)x + αβ

= x^{2} – x + 1

**(v)** Let the zeroes of the polynomial be α and β.

Then, α + β = – \(\frac { 1 }{ 4 }\) and αβ = \(\frac { 1 }{ 4 }\)

∴ Required polynomial

= x^{2} – (α + β)x + αβ

= x^{2} – (- \(\frac { 1 }{ 4 }\) ) + \(\frac { 1 }{ 4 }\)

= 4x^{2} + x + 1 = 0

**(vi)** Let the zeroes of the polynomial be α and β.

Then, α + β = 4 and αβ = 1.

∴ Required polynomial = x^{2} -(α + β)x + αβ

= x^{2} – 4x + 1

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