This page consists of the **solutions** of the **Exercise 2.2**, **Chapter 2** of **Class 10 Mathematics** under **NCERT** Syllabus. The Chapter 2 of Class 10 Mathematics is about the Polynomials. The Exercise 2.2 deals with the questions based on the Relationship between Zeros and Coefficients of a Polynomial. The Exercise 2.2 consists of 2 questions only.

Description | NCERT Maths Solutions |

Subject | Mathematics |

Class | 10 (X) |

Chapter | 2 |

Exercise | 2.2 |

Topic | The Relationship between Zeros and the Coefficients of a Polynomial |

Syllabus | NCERT |

The solution of all the questions of Exercise 2.2 of Class 10 Mathematics is given below.

**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

**Solution:**

Given, the quadratic polynomial x^2-2x-8

Factorising,

x^2-2x-8 =x^2-(4-2)x-8 =x^2-4x+2x-8 =x(x-4)+2(x-4) =(x-4)(x+2)The zeros of the quadratic polynomial are,

x-4=0 \Rightarrow x=4And x+2=0

\Rightarrow x=-2Sum of the zeros=4+(-2)=4-2=2

Product of the zeros=4\times (-2)=-8

**Verification:**

Here, a=1,\, b=-2 and c=-8

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

Product of zeros=\frac{c}{a}=\frac{-8}{1}=-8

Hence verified.

**ii. **4s^2-4s+1

**Solution:**

Given, the quadratic polynomial 4s^2-4s+1

Factorising,

4s^2-4s+1 =4s^2-(2+2)s+1 =4s^2-2s-2s+1 =2s(2s-1)-1(2s-1) =(2s-1)(2s-1) =(2s-1)^2The zeros of the quadratic polynomial are,

(2s-1)^2=02s-1=0 and 2s-1=0

s=\frac{1}{2},\,\frac{1}{2}Sum of the zeros=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1

Product of the zeros=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}

**Verification:**

Here, a=4,\, b=-4 and c=1

Sum of zeros=\frac{-b}{a}=\frac{-(-4)}{4}=1

Product of zeros=\frac{c}{a}=\frac{1}{4}=\frac{1}{4}

Hence verified.

**iii. **6x^2-3-7x

**Solution:**

Given, the quadratic polynomial 6x^2-3-7x=6x^2-7x-3

Factorising,

6x^2-7x-3 =6x^2-(9-2)x-3 =6x^2-9x+2x-3 =3x(2x-3)+1(2x-3) =(2x-3)(3x+1)The zeros of the quadratic polynomial are,

2x-3=0 and 3x+1=0

2x=3 and 3x=-1

x=\frac{3}{2} and x=\frac{-1}{3}

Sum of the zeros=\frac{3}{2}+\frac{-1}{3}=\frac{7}{6}

Product of the zeros=\frac{3}{2}\times \frac{-1}{3}=-\frac{1}{2}

**Verification:**

Here, a=6,\, b=-7 and c=-3

Sum of zeros=\frac{-b}{a}=\frac{-(-7)}{6}=\frac{7}{6}

Product of zeros=\frac{c}{a}=\frac{-3}{6}=-\frac{1}{2}

Hence verified.

**iv. **4u^2+8u

**Solution:**

Given, the quadratic polynomial 4u^2+8u

Equating the polynomial to zero,

4u^2+8u=0 \Rightarrow 4u(u+2)=0The zeros of the quadratic polynomial are,

4u=0 and u+2=0

u=0 and u=-2

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

Product of the zeros=0\times (-2)=0

**Verification:**

Here, a=4,\, b=8 and c=0

Sum of zeros=\frac{-b}{a}=\frac{-8}{4}=-2

Product of zeros=\frac{c}{a}=\frac{0}{4}=0

Hence verified.

**v. **t^2-15

**Solution:**

Given, the quadratic polynomial t^2-15

Equating the polynomial to zero,

t^2-15=0 \Rightarrow t^2=15 \Rightarrow t=\pm \sqrt{15} \Rightarrow t=+\sqrt{15},\,-\sqrt{15}The zeros of the quadratic polynomial are +\sqrt{15} and -\sqrt{15}

Sum of the zeros=+\sqrt{15}-\sqrt{15}=0

Product of the zeros=+\sqrt{15}\times -\sqrt{15}=-15

**Verification:**

Here, a=1,\, b=0 and c=-15

Sum of zeros=\frac{-b}{a}=\frac{-0}{1}=0

Product of zeros=\frac{c}{a}=\frac{-15}{1}=-15

Hence verified.

**vi. **3x^2-x-4

**Solution:**

Given, the quadratic polynomial 3x^2-x-4

Factorising,

3x^2-x-4 =3x^2-(4-3)x-4 =3x^2-4x+3x-4 =x(3x-4)+1(3x-4) =(3x-4)(x+1)The zeros of the quadratic polynomial are,

3x-4=0 and x+1=0

3x=4 and x=-1

x=\frac{4}{3} and x=-1

Sum of the zeros=\frac{4}{3}+(-1)=\frac{1}{3}

Product of the zeros=\frac{4}{3}\times (-1)=-\frac{4}{3}

**Verification:**

Here, a=3,\, b=-1 and c=-4

Sum of zeros=\frac{-b}{a}=\frac{-(-1)}{3}=\frac{1}{3}

Product of zeros=\frac{c}{a}=\frac{-4}{3}=-\frac{4}{3}

Hence verified.

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

**i.** \frac{1}{4},\, -1

**Solution:**

Given,

α+β=\frac{1}{4}\,\,\,\,\,……(i) and

α\times β=-1\,\,\,\,\,……(ii)We know that,

α+β=\frac{-b}{a}\,\,\,\,\,……(iii) and

α\times β=\frac{c}{a}\,\,\,\,\,……(iv)Comparing equation (i) & (iii) and equation (ii) & (iv)

a=4,\, b=-1 and c=-4

Therefore, the required quadratic polynomial will be,

ax^2+bx+c 4x^2+(-1)x+(-4) 4x^2-x-4**ii. **\sqrt{2},\,\frac{1}{3}

**Solution:**

Given,

α+β=\sqrt{2}\,\,\,\,\,……(i) and

α\times β=\frac{1}{3}\,\,\,\,\,……(ii)We know that,

α+β=\frac{-b}{a}\,\,\,\,\,……(iii) and

α\times β=\frac{c}{a}\,\,\,\,\,……(iv)Comparing equation (i) & (iii) and equation (ii) & (iv)

a=3,\, b=-3\sqrt{2} and c=1

Therefore, the required quadratic polynomial will be,

ax^2+bx+c 3x^2+(-3\sqrt{2})x+(1) 3x^2-2\sqrt{2}x+1**iii. **0,\,\sqrt{5}

**Solution:**

Given,

α+β=0\,\,\,\,\,……(i) and

α\times β=\sqrt{5}\,\,\,\,\,……(ii)We know that,

α+β=\frac{-b}{a}\,\,\,\,\,……(iii) and

α\times β=\frac{c}{a}\,\,\,\,\,……(iv)Comparing equation (i) & (iii) and equation (ii) & (iv)

a=1,\, b=0 and c=\sqrt{5}

Therefore, the required quadratic polynomial will be,

ax^2+bx+c 1x^2+(0)x+\sqrt{5} x^2+\sqrt{5}**iv. **1,\, 1

**Solution:**

Given,

α+β=1\,\,\,\,\,……(i) and

α\times β=1\,\,\,\,\,……(ii)We know that,

α+β=\frac{-b}{a}\,\,\,\,\,……(iii) and

α\times β=\frac{c}{a}\,\,\,\,\,……(iv)Comparing equation (i) & (iii) and equation (ii) & (iv)

a=1,\, b=-1 and c=1

Therefore, the required quadratic polynomial will be,

ax^2+bx+c 1x^2+(-1)x+1 x^2-x+1**v. **-\frac{1}{4},\,\frac{1}{4}

**Solution:**

Given,

α+β=-\frac{1}{4}\,\,\,\,\,……(i) and

α\times β=\frac{1}{4}\,\,\,\,\,……(ii)We know that,

α+β=\frac{-b}{a}\,\,\,\,\,……(iii) and

α\times β=\frac{c}{a}\,\,\,\,\,……(iv)Comparing equation (i) & (iii) and equation (ii) & (iv)

a=4,\, b=1 and c=1

Therefore, the required quadratic polynomial will be,

ax^2+bx+c 4x^2+1x+1 4x^2-x+1**vi. **4,\,1

**Solution:**

Given,

α+β=4\,\,\,\,\,……(i) and

α\times β=1\,\,\,\,\,……(ii)We know that,

α+β=\frac{-b}{a}\,\,\,\,\,……(iii) and

α\times β=\frac{c}{a}\,\,\,\,\,……(iv)Comparing equation (i) & (iii) and equation (ii) & (iv)

a=1,\, b=-4 and c=1

Therefore, the required quadratic polynomial will be,

ax^2+bx+c 1x^2+(-4)x+1 x^2-4x+1I hope the solutions of Class 10, Exercise 2.2, Chapter 2 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 2.2, feel free to contact me at [email protected] or [email protected] or fill the form here.

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