# RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1

The basic study of polynomials and their geometrical representation of linear and quadratic polynomials along with the geometrical meaning of their zeros are the main insights of this chapter. The RD Sharma Solutions Class 10 Exercise 2.1, contains problems dealing with finding the zeros of polynomials and verification of the relationship between the zeros and their coefficients. In order for students to excel in the exercise questions and boost their confidence, the RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1 PDF is provided below.

## RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1 Download PDF

### Access RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.1

1. Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

(i) f(x) = x2Â â€“ 2x â€“ 8

Solution:

Given,

f(x) = x2Â â€“ 2x â€“ 8

To find the zeros, we put f(x) = 0

â‡’ x2Â â€“ 2x â€“ 8 = 0

â‡’Â  x2Â – 4x + 2x – 8 = 0

â‡’ x(x – 4) + 2(x – 4) = 0

â‡’ (x – 4)(x + 2) = 0

This gives us 2 zeros, for

x = 4 and x = -2

Hence, the zeros of the quadratic equation are 4 and -2.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

4 + (-2)= – (-2) / 1

2 = 2

Product of roots = constant / coefficient of x2

4 x (-2) = (-8) / 1

-8 = -8

Therefore, the relationship between zeros and their coefficients is verified.

(ii) g(s) = 4s2Â â€“ 4s + 1

Solution:

Given,

g(s) = 4s2Â â€“ 4s + 1

To find the zeros, we put g(s) = 0

â‡’ 4s2Â â€“ 4s + 1 = 0

â‡’Â  4s2Â – 2s – 2s + 1= 0

â‡’ Â 2s(2s – 1) – (2s – 1)Â = 0

â‡’ (2s – 1)(2s â€“ 1) = 0

This gives us 2 zeros, for

s = 1/2 and s = 1/2

Hence, the zeros of the quadratic equation are 1/2 and 1/2.

Now, for verification

Sum of zeros = – coefficient of s / coefficient of s2

1/2 + 1/2 = – (-4) / 4

1 = 1

Product of roots = constant / coefficient of s2

1/2 x 1/2 = 1/4

1/4 = 1/4

Therefore, the relationship between zeros and their coefficients is verified.

(iii) h(t)=t2Â â€“ 15

Solution:

Given,

h(t) = t2Â â€“ 15 = t2Â +(0)t â€“ 15

To find the zeros, we put h(t) = 0

â‡’ t2Â â€“ 15 = 0

â‡’ (t + âˆš15)(t – âˆš15)= 0

This gives us 2 zeros, for

t = âˆš15 and t = -âˆš15

Hence, the zeros of the quadratic equation are âˆš15 and -âˆš15.

Now, for verification

Sum of zeros = – coefficient of t / coefficient of t2

âˆš15 + (-âˆš15) = – (0) / 1

0 = 0

Product of roots = constant / coefficient of t2

âˆš15 x (-âˆš15) = -15/1

-15 = -15

Therefore, the relationship between zeros and their coefficients is verified.

(iv) f(x) = 6x2Â â€“ 3 â€“ 7x

Solution:

Given,

f(x) = 6x2Â â€“ 3 â€“ 7x

To find the zeros, we put f(x) = 0

â‡’ 6x2Â â€“ 3 â€“ 7x = 0

â‡’Â  6x2Â – 9x + 2x – 3 = 0

â‡’ 3x(2x – 3) + 1(2x – 3) = 0

â‡’ (2x – 3)(3x + 1) = 0

This gives us 2 zeros, for

x = 3/2 and x = -1/3

Hence, the zeros of the quadratic equation are 3/2 and -1/3.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

3/2 + (-1/3) = – (-7) / 6

7/6 = 7/6

Product of roots = constant / coefficient of x2

3/2 x (-1/3) = (-3) / 6

-1/2 = -1/2

Therefore, the relationship between zeros and their coefficients is verified.

(v) p(x) = x2Â + 2âˆš2x â€“ 6

Solution:

Given,

p(x) = x2Â + 2âˆš2x â€“ 6

To find the zeros, we put p(x) = 0

â‡’ x2Â + 2âˆš2x â€“ 6 = 0

â‡’Â  x2Â + 3âˆš2x – âˆš2x – 6 = 0

â‡’ x(x + 3âˆš2) – âˆš2 (x + 3âˆš2) = 0

â‡’ (x – âˆš2)(x + 3âˆš2) = 0

This gives us 2 zeros, for

x = âˆš2 and x = -3âˆš2

Hence, the zeros of the quadratic equation are âˆš2 and -3âˆš2.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

âˆš2 + (-3âˆš2) = – (2âˆš2) / 1

-2âˆš2 = -2âˆš2

Product of roots = constant / coefficient of x2

âˆš2 x (-3âˆš2) = (-6) / 2âˆš2

-3 x 2 = -6/1

-6 = -6

Therefore, the relationship between zeros and their coefficients is verified.

(vi) q(x)=âˆš3x2Â + 10x + 7âˆš3

Solution:

Given,

q(x) = âˆš3x2Â + 10x + 7âˆš3

To find the zeros, we put q(x) = 0

â‡’ âˆš3x2Â + 10x + 7âˆš3 = 0

â‡’Â  âˆš3x2Â + 3x +7x + 7âˆš3x = 0

â‡’ âˆš3x(x + âˆš3) + 7 (x + âˆš3) = 0

â‡’ (x + âˆš3)(âˆš3x + 7) = 0

This gives us 2 zeros, for

x = -âˆš3 and x = -7/âˆš3

Hence, the zeros of the quadratic equation are -âˆš3 and -7/âˆš3.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

-âˆš3 + (-7/âˆš3) = – (10) /âˆš3

(-3-7)/ âˆš3 = -10/âˆš3

-10/ âˆš3 = -10/âˆš3

Product of roots = constant / coefficient of x2

(-âˆš3) x (-7/âˆš3) = (7âˆš3) / âˆš3

7 = 7

Therefore, the relationship between zeros and their coefficients is verified.

(vii) f(x) = x2Â â€“ (âˆš3 + 1)x + âˆš3

Solution:

Given,

f(x) = x2Â â€“ (âˆš3 + 1)x + âˆš3

To find the zeros, we put f(x) = 0

â‡’ x2Â â€“ (âˆš3 + 1)x + âˆš3 = 0

â‡’Â  x2Â – âˆš3x – x + âˆš3 = 0

â‡’ x(x – âˆš3) – 1 (x – âˆš3) = 0

â‡’ (x – âˆš3)(x – 1) = 0

This gives us 2 zeros, for

x = âˆš3 and x = 1

Hence, the zeros of the quadratic equation are âˆš3 and 1.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

âˆš3 + 1 = – (-(âˆš3 +1)) / 1

âˆš3 + 1 = âˆš3 +1

Product of roots = constant / coefficient of x2

1 x âˆš3 = âˆš3 / 1

âˆš3 = âˆš3

Therefore, the relationship between zeros and their coefficients is verified.

(viii) g(x)=a(x2+1)â€“x(a2+1)

Solution:

Given,

g(x) = a(x2+1)â€“x(a2+1)

To find the zeros, we put g(x) = 0

â‡’ a(x2+1)â€“x(a2+1) = 0

â‡’ ax2Â + a âˆ’ a2x â€“ x = 0

â‡’ ax2Â âˆ’ a2x â€“ x + a = 0

â‡’ ax(x âˆ’ a) âˆ’ 1(x â€“ a) = 0

â‡’ (x â€“ a)(ax â€“ 1) = 0

This gives us 2 zeros, for

x = a and x = 1/a

Hence, the zeros of the quadratic equation are a and 1/a.

Now, for verification

Sum of zeros = – coefficient of x / coefficient of x2

a + 1/a = – (-(a2 + 1)) / a

(a2 + 1)/a = (a2 + 1)/a

Product of roots = constant / coefficient of x2

a x 1/a = a / a

1 = 1

Therefore, the relationship between zeros and their coefficients is verified.

(ix) h(s) = 2s2Â â€“ (1 + 2âˆš2)s + âˆš2

Solution:

Given,

h(s) = 2s2Â â€“ (1 + 2âˆš2)s + âˆš2

To find the zeros, we put h(s) = 0

â‡’ 2s2Â â€“ (1 + 2âˆš2)s + âˆš2 = 0

â‡’Â  2s2Â â€“ 2âˆš2s â€“ s + âˆš2 = 0

â‡’ 2s(sÂ â€“ âˆš2) -1(s – âˆš2) = 0

â‡’ (2s – 1)(s – âˆš2) = 0

This gives us 2 zeros, for

x = âˆš2 and x = 1/2

Hence, the zeros of the quadratic equation are âˆš3 and 1.

Now, for verification

Sum of zeros = – coefficient of s / coefficient of s2

âˆš2 + 1/2 = – (-(1 + 2âˆš2)) / 2

(2âˆš2 + 1)/2 = (2âˆš2 +1)/2

Product of roots = constant / coefficient of s2

1/2 x âˆš2 = âˆš2 / 2

âˆš2 / 2 = âˆš2 / 2

Therefore, the relationship between zeros and their coefficients is verified.

(x) f(v) = v2Â + 4âˆš3v – 15

Solution:

Given,

f(v) = v2Â + 4âˆš3v â€“ 15

To find the zeros, we put f(v) = 0

â‡’ v2Â + 4âˆš3v â€“ 15 = 0

â‡’Â  v2Â + 5âˆš3v – âˆš3v – 15 = 0

â‡’ v(v + 5âˆš3) – âˆš3 (v + 5âˆš3) = 0

â‡’ (v – âˆš3)(v + 5âˆš3) = 0

This gives us 2 zeros, for

v = âˆš3 and v = -5âˆš3

Hence, the zeros of the quadratic equation are âˆš3 and -5âˆš3.

Now, for verification

Sum of zeros = – coefficient of v / coefficient of v2

âˆš3 + (-5âˆš3) = – (4âˆš3) / 1

-4âˆš3 = -4âˆš3

Product of roots = constant / coefficient of v2

âˆš3 x (-5âˆš3) = (-15) / 1

-5 x 3 = -15

-15 = -15

Therefore, the relationship between zeros and their coefficients is verified.

(xi) p(y) = y2Â + (3âˆš5/2)y â€“ 5

Solution:

Given,

p(y) = y2Â + (3âˆš5/2)y â€“ 5

To find the zeros, we put f(v) = 0

â‡’ y2Â + (3âˆš5/2)y â€“ 5 = 0

â‡’Â  y2Â – âˆš5/2 y + 2âˆš5y – 5 = 0

â‡’ y(y – âˆš5/2) + 2âˆš5 (y – âˆš5/2) = 0

â‡’ (y + 2âˆš5)(y – âˆš5/2) = 0

This gives us 2 zeros, for

y = âˆš5/2 and y = -2âˆš5

Hence, the zeros of the quadratic equation are âˆš5/2 and -2âˆš5.

Now, for verification

Sum of zeros = – coefficient of y / coefficient of y2

âˆš5/2 + (-2âˆš5) = – (3âˆš5/2) / 1

-3âˆš5/2 = -3âˆš5/2

Product of roots = constant / coefficient of y2

âˆš5/2 x (-2âˆš5) = (-5) / 1

– (âˆš5)2 = -5

-5 = -5

Therefore, the relationship between zeros and their coefficients is verified.

(xii) q(y) = 7y2Â – (11/3)y â€“ 2/3

Solution:

Given,

q(y) = 7y2Â – (11/3)y â€“ 2/3

To find the zeros, we put q(y) = 0

â‡’ 7y2Â – (11/3)y â€“ 2/3 = 0

â‡’Â  (21y2 – 11y -2)/3 = 0

â‡’ 21y2 – 11y – 2 = 0

â‡’ 21y2 – 14y + 3y – 2 = 0

â‡’ 7y(3y – 2) â€“ 1(3y + 2) = 0

â‡’ (3y – 2)(7y + 1) = 0

This gives us 2 zeros, for

y = 2/3 and y = -1/7

Hence, the zeros of the quadratic equation are 2/3 and -1/7.

Now, for verification

Sum of zeros = – coefficient of y / coefficient of y2

2/3 + (-1/7) = – (-11/3) / 7

-11/21 = -11/21

Product of roots = constant / coefficient of y2

2/3 x (-1/7) = (-2/3) / 7

– 2/21 = -2/21

Therefore, the relationship between zeros and their coefficients is verified.

2. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by factorization.

(i) -8/3 , 4/3

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

f(x) = x2 + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -8/3 and product of zero= 4/3

Thus,

The required polynomial f(x) is,

â‡’ x2 – (-8/3)x + (4/3)

â‡’ x2 + 8/3x + (4/3)

So, to find the zeros we put f(x) = 0

â‡’ x2 + 8/3x + (4/3) = 0

â‡’ 3x2 + 8x + 4 = 0

â‡’ 3x2 + 6x + 2x + 4 = 0

â‡’ 3x(x + 2) + 2(x + 2) = 0

â‡’ (x + 2) (3x + 2) = 0

â‡’ (x + 2) = 0 and, or (3x + 2) = 0

Therefore, the two zeros are -2 and -2/3.

(ii) 21/8 , 5/16

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

f(x) = x2 + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = 21/8 and product of zero = 5/16

Thus,

The required polynomial f(x) is,

â‡’ x2 – (21/8)x + (5/16)

â‡’ x2 – 21/8x + 5/16

So, to find the zeros we put f(x) = 0

â‡’ x2 – 21/8x + 5/16 = 0

â‡’ 16x2 – 42x + 5 = 0

â‡’ 16x2 – 40x – 2x + 5 = 0

â‡’ 8x(2x – 5) – 1(2x – 5) = 0

â‡’ (2x – 5) (8x – 1) = 0

â‡’ (2x – 5) = 0 and, or (8x – 1) = 0

Therefore, the two zeros are 5/2 and 1/8.

(iii) -2âˆš3, -9

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

f(x) = x2 + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -2âˆš3 and product of zero = -9

Thus,

The required polynomial f(x) is,

â‡’ x2 – (-2âˆš3)x + (-9)

â‡’ x2 + 2âˆš3x â€“ 9

So, to find the zeros we put f(x) = 0

â‡’ x2 + 2âˆš3x â€“ 9 = 0

â‡’ x2 + 3âˆš3x – âˆš3x â€“ 9 = 0

â‡’ x(x + 3âˆš3) – âˆš3(x + 3âˆš3) = 0

â‡’ (x + 3âˆš3) (x – âˆš3) = 0

â‡’ (x + 3âˆš3) = 0 and, or (x – âˆš3) = 0

Therefore, the two zeros are -3âˆš3and âˆš3.

(iv) -3/2âˆš5, -1/2

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

f(x) = x2 + -(sum of zeros) x + (product of roots)

Here, the sum of zeros is = -3/2âˆš5 and product of zero = -1/2

Thus,

The required polynomial f(x) is,

â‡’ x2 – (-3/2âˆš5)x + (-1/2)

â‡’ x2 + 3/2âˆš5x â€“ 1/2

So, to find the zeros we put f(x) = 0

â‡’ x2 + 3/2âˆš5x â€“ 1/2 = 0

â‡’ 2âˆš5x2 + 3x – âˆš5 = 0

â‡’ 2âˆš5x2 + 5x â€“ 2x – âˆš5 = 0

â‡’ âˆš5x(2x + âˆš5) – 1(2x + âˆš5) = 0

â‡’ (2x + âˆš5) (âˆš5x – 1) = 0

â‡’ (2x + âˆš5) = 0 and, or (âˆš5x – 1) = 0

Therefore, the two zeros are -âˆš5/2 and 1/âˆš5.

3. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ f(x) = x2 â€“ 5x + 4, find the value ofÂ 1/Î± + 1/Î² â€“ 2Î±Î².

Solution:

From the question, itâ€™s given that:

Î± and Î²Â are the roots of the quadratic polynomial f(x) where a = 1, b = -5 and c = 4

So, we can find

Sum of the roots =Â Î±+Î²Â = -b/a = – (-5)/1 = 5

Product of the roots =Â Î±Î²Â = c/a = 4/1 = 4

To find, 1/Î± +1/Î² â€“ 2Î±Î²

â‡’ [(Î± +Î²)/ Î±Î²] â€“ 2Î±Î²

â‡’ (5)/ 4 â€“ 2(4) = 5/4 â€“ 8 = -27/ 4

4. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ p(y) = 5y2 â€“ 7y + 1, find the value ofÂ 1/Î±+1/Î².

Solution:

From the question, itâ€™s given that:

Î± and Î²Â are the roots of the quadratic polynomial f(x) where a =5, b = -7 and c = 1

So, we can find

Sum of the roots =Â Î±+Î²Â = -b/a = – (-7)/5 = 7/5

Product of the roots =Â Î±Î²Â = c/a = 1/5

To find, 1/Î± +1/Î²

â‡’ (Î± +Î²)/ Î±Î²

â‡’ (7/5)/ (1/5) = 7

5. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ f(x)=x2 â€“ x â€“ 4, find the value ofÂ 1/Î±+1/Î²â€“Î±Î².

Solution:

From the question, itâ€™s given that:

Î± and Î²Â are the roots of the quadratic polynomial f(x) where a = 1, b = -1 and c = – 4

So, we can find

Sum of the roots =Â Î±+Î²Â = -b/a = – (-1)/1 = 1

Product of the roots =Â Î±Î²Â = c/a = -4 /1 = – 4

To find, 1/Î± +1/Î² â€“ Î±Î²

â‡’ [(Î± +Î²)/ Î±Î²] â€“ Î±Î²

â‡’ [(1)/ (-4)] â€“ (-4) = -1/4 + 4 = 15/ 4

6. IfÂ Î± and Î²Â are the zeroes of the quadratic polynomialÂ f(x) = x2 + x â€“ 2, find the value ofÂ 1/Î± â€“ 1/Î².

Solution:

From the question, itâ€™s given that:

Î± and Î²Â are the roots of the quadratic polynomial f(x) where a = 1, b = 1 and c = – 2

So, we can find

Sum of the roots =Â Î±+Î²Â = -b/a = – (1)/1 = -1

Product of the roots =Â Î±Î²Â = c/a = -2 /1 = – 2

To find, 1/Î± – 1/Î²

â‡’ [(Î² – Î±)/ Î±Î²]

â‡’

7. If one of the zero of the quadratic polynomialÂ f(x) = 4x2 â€“ 8kx â€“ 9Â is negative of the other, then find the value of k.

Solution:

From the question, itâ€™s given that:

The quadratic polynomial f(x) where a = 4, b = -8k and c = – 9

And, for roots to be negative of each other, let the roots be Î± and â€“ Î±.

So, we can find

Sum of the roots =Â Î± – Î±Â = -b/a = – (-8k)/1 = 8k = 0 [âˆµ Î± – Î±Â = 0]

â‡’ k = 0

8. Â If the sum of the zeroes of the quadratic polynomialÂ f(t)=kt2 + 2t + 3kÂ is equal to their product, then find the value of k.

Solution:

Given,

The quadratic polynomial f(t)=kt2 + 2t + 3k,Â where a = k, b = 2 and c = 3k.

And,

Sum of the roots = Product of the roots

â‡’ (-b/a) = (c/a)

â‡’ (-2/k) = (3k/k)

â‡’ (-2/k) = 3

âˆ´ k = -2/3

9. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ p(x) = 4x2 â€“ 5x â€“ 1, find the value ofÂ Î±2Î²+Î±Î²2.

Solution:

From the question, itâ€™s given that:

Î± and Î²Â are the roots of the quadratic polynomial p(x) where a = 4, b = -5 and c = -1

So, we can find

Sum of the roots =Â Î±+Î²Â = -b/a = – (-5)/4 = 5/4

Product of the roots =Â Î±Î²Â = c/a = -1/4

To find, Î±2Î²+Î±Î²2

â‡’ Î±Î²(Î± +Î²)

â‡’ (-1/4)(5/4) = -5/16

10. IfÂ Î± and Î²Â are the zeros of the quadratic polynomialÂ f(t)=t2â€“ 4t + 3, find the value ofÂ Î±4Î²3+Î±3Î²4.

Solution:

From the question, itâ€™s given that:

Î± and Î²Â are the roots of the quadratic polynomial f(t) where a = 1, b = -4 and c = 3

So, we can find

Sum of the roots =Â Î±+Î²Â = -b/a = – (-4)/1 = 4

Product of the roots =Â Î±Î²Â = c/a = 3/1 = 3

To find, Î±4Î²3+Î±3Î²4

â‡’ Î±3Î²3 (Î± +Î²)

â‡’ (Î±Î²)3 (Î± +Î²)

â‡’ (3)3 (4) = 27 x 4 = 108