Ex-2.1 Polynomials, Class 10, Maths RD Sharma Solutions PDF Download

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

(i) f(x) = x² – 2x – 8

(ii) g(s) = 4s² – 4s + 1

(iii) 6x² – 3 – 7x

(iv) h(t) = t² – 15

(v) p(x) = x² + 2√2x – 6

(vi) q(x) = √3x² + 10x + 7√3

(vii) f(x) = x² – (√3 + 1)x + √3

(viii) g(x) = a(x² + 1) – x(a² + 1)

Solution: (i) f(x) = x² – 2x – 8

We have,

f(x) = x² – 2x – 8

=  x² – 4x + 2x – 8

=  x(x – 4) + 2(x – 4)

=  (x + 2)(x – 4)

Zeroes of the polynomials are -2 and 4.

Now,

Sum of the zeroes =  

-2 + 4 =  −(−2)/1

2 = 2

Product of the zeroes =  

-8 =  −8/1

-8 = -8

Hence, the relationship is verified.

(ii) g(s) = 4s² – 4s + 1

We have,

g(s) = 4s² – 4s + 1

=  4s² – 2s – 2s + 1

=  2s(2s – 1)−1(2s – 1)

=  (2s – 1)(2s – 1)

Zeroes of the polynomials are 1/2  and  1/2.

Sum of zeroes =  

1/2 + 1/2 = −(−4)/4

1 = 1

Product of zeroes =   

1/2 × 1/2 = 1/4

1/4 = 1/4

Hence, the relationship is verified.

(iii) 6s²−3−7x

=  6s²−7x−3  = (3x + 11)(2x – 3)

Zeros of the polynomials are 3/2 and −1/3

Sum of the zeros =  

−1/3 + 3/2 = −(−7)/6

7/6 = 7/6

Product of the zeroes =  

−1/3 × 3/2 = −3/6

−3/6 = −3/6

Hence, the relationship is verified.

(iv) h(t) = t² – 15

We have,

h(t) = t² – 15

=  t² –

=  (t + )(t – )

Zeroes of the polynomials are − and

Sum of the zeroes = 0

− +   = 0

0 = 0

Product of zeroes =  

−× = −15

-15 = -15

Hence, the relationship verified.

(v) p(x) = x² + 2√2x – 6

We have,

p(x) = x² + 2√2x – 6

=  x² + 3√2x + 3√2x – 6

=  x(x + 3√2) – √2(x + 3√2)

=  (x + 3√2)(x – √2)

Zeroes of the polynomials are 3√2  and  −3√2

Sum of the zeroes =  −2√2/1

√2 – 3√2 = −2√2

– 2√2 = −2√2

Product of the zeroes =

√2×−3√2 = −6/1

-6 = -6

Hence, the relationship is verified.

(vi) q(x) =  √3x² + 10x + 7√3

=  √3x² + 7x + 3x + 7√3

=  √3x(x + √3)7(x + sqrt3)

=  (x + √3)(7 + sqrt3)

Zeros of the polynomials are −√3 and −7/3√

Sum of zeros =  −10√3

−√3−7/√3 = −10/√3

−10/√3 = −10/√3

Product of the polynomials are −√3,−7/√3

7 = 7

Hence, the relationship is verified.

(vii)  h(x) =  x²−(√3 + 1)x + √3

=  x²−√3−x + √3

= x(x – √3)−1(x−√3)

=  (x – √3)(x−1)

Zeros of the polynomials are 1 and  √3

Sum of zeros =  = −[−√3−1]

1 + √3 = √3 + 1

Product of zeros = = √3

√3 = √3

Hence, the relationship is verified

(viii) g(x) =  a[(x² + 1) – x(a² + 1)]²

=  ax² + a−a²x−x

=  ax²−[(a²x + 1)] + a

=  ax²−a²x – x + a

=  ax(x−a)−1(x – a)  =  (x – a)(ax – 1)

Zeros of the polynomials are 1/a and 1

Sum of the zeros =  

Product of zeros = a/a

1/a×a = a/a

1 = 1

Hence, the relationship is verified.

Q.2: If α and β are the zeroes of the quadratic polynomial f(x) = x² – 5x + 4, find the value of 1/α + 1/β – 2αβ.

Solution: We have,

α and β are the roots of the quadratic polynomial.

f(x) = x² – 5x + 4

Sum of the roots =  α + β  = 5

Product of the roots =  αβ  = 4

So,

Q.3: If α and β are the zeroes of the quadratic polynomial f(x) = x² – 5x + 4, find the value of 1/α + 1/β – 2αβ.

Solution: Since, α and β are the zeroes of the quadratic polynomial.

p(y) = x² – 5x + 4

Sum of the zeroes =  α + β  = 5

Product of the roots =  αβ  = 4

So,

Q.4: If α and β are the zeroes of the quadratic polynomial p(y) = 5y² – 7y + 1, find the value of 1/α + 1/β.

Solution: Since, α and β are the zeroes of the quadratic polynomial.

p(y) = 5y² – 7y + 1

Sum of the zeroes =  α + β  = 7

Product of the roots =  αβ  = 1

So,

Q.5: If α and β are the zeroes of the quadratic polynomial f(x) = x² – x – 4, find the value of 1/α + 1/β – αβ.

Solution: Since, α and β are the zeroes of the quadratic polynomial.

We have,

f(x) = x² – x – 4

Sum of zeroes =  α + β  = 1

Product of the zeroes =  αβ  = -4

So,

Q.6: If α and β are the zeroes of the quadratic polynomial f(x) = x² + x – 2, find the value of 1/α – 1/β.

Solution: Since, α and β are the zeroes of the quadratic polynomial.

We have,

f(x) = x² + x – 2

Sum of zeroes =  α + β  = 1

Product of the zeroes =  αβ  = -2

So,

Q.7: If one of the zero of the quadratic polynomial f(x) = 4x² – 8kx – 9 is negative of the other, then find the value of k.

Solution: Let, the two zeroes of the polynomial f(x) = 4x² – 8kx – 9 be α and −α.

Product of the zeroes =  α x −α  = -9

Sum of the zeroes =  α + (−α)  = -8k = 0                              Since,α – α = 0

⇒ 8k = 0

⇒ k = 0

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

Solution: Let the two zeroes of the polynomial f(t) = kt² + 2t + 3k be α and β.

Sum of the zeroes =  α + β  = 2

Product of the zeroes =  α × β  = 3k

Now,

−2/k = 3k/k

⇒ 3k = −2

⇒ k = −2/3

So, k = 0 and  ⇒ k = −2/3

Q.9: If α and β are the zeroes of the quadratic polynomial p(x) = 4x² – 5x – 1, find the value of α²β + αβ².

Solution: Since, α and β are the zeroes of the quadratic polynomial p(x) = 4x² – 5x – 1

So, Sum of the zeroes =  α + β  =  5/4

Product of the zeroes =  α × β  =   −1/4

Now,

α²β + αβ²  =  αβ(α + β)

=  5/4(−1/4)

=  −5/16

Q.10: If α and β are the zeroes of the quadratic polynomial f(t) = t² – 4t + 3, find the value of α⁴β³ + α³β⁴.

Solution: Since, α and β are the zeroes of the quadratic polynomial f(t) = t² – 4t + 3

So, Sum of the zeroes =  α + β  = 4

Product of the zeroes =  α × β  =   3

Now,

α⁴β³ + α³β⁴  =  α³β³(α + β)

=  (3)³(4)  = 108

Q.11: If α and β are the zeroes of the quadratic polynomial f(x) = 6x² + x – 2, find the value of α/β + β/α.

Solution: Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x² + x – 2.

Sum of the zeroes =   α + β  =  −1/6

Product of the zeroes =  α × β  =   −1/3

Now,

By substitution the values of the sum of zeroes and products of the zeroes, we will get

=  −25/12

Q.12: If α and β are the zeroes of the quadratic polynomial f(x) = 6x² + x – 2, find the value of 

Solution: Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x² + x – 2.

Sum of the zeroes =  α + β  =  6/3

Product of the zeroes =  α × β  =   4/3

Now,

By substituting the values of sum and product of the zeroes, we will get

Q.13: If the squared difference of the zeroes of the quadratic polynomial f(x) = x² + px + 45 is equal to 144, find the value of p.

Solution: Let the two zeroes of the polynomial be α and β.

We have,

f(x) = x² + px + 45

Now,

Sum of the zeroes =  α + β  = -p

Product of the zeroes =  α × β  =   45

So,

(α + β)² – 4αβ = 144

(p)² – 4 × 45 = 144

(p)² = 144 + 180

(p)² = 324

p = 

p = ±18

Thus, in the given equation, p will be either 18 or -18.

Q.14: If α and β are the zeroes of the quadratic polynomial f(x) = x² – px + q,  prove that 

Solution: Since, α and β are the roots of the quadratic polynomial given in the question.

f(x) = x² – px + q

Now,

Sum of the zeroes = p =  α + β

Product of the zeroes = q =  α × β

LHS = RHS

Hence, proved.

Q.15: If α and β are the zeroes of the quadratic polynomial f(x) = x² – p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.

Solution: Since, α and β are the zeroes of the quadratic polynomial

f(x) = x² – p(x + 1) – c

Now,

Sum of the zeroes =  α + β  = p

Product of the zeroes =  α × β  = (- p – c)

So,

(α + 1)(β + 1)

=  αβ + α + β + 1

=  αβ + (α + β) + 1

=  (−p – c) + p + 1

= 1 – c = RHS

So, LHS = RHS

Hence, proved.

Q.16: If α and β are the zeroes of the quadratic polynomial such that α + β = 24 and α – β = 8, find a quadratic polynomial having α and β as its zeroes.

Solution: We have,

α + β = 24 …………E-1

α – β = 8 ………….E-2

By solving the above two equations accordingly, we will get

2α = 32

α = 16

Substitute the value of α, in any of the equation. Let we substitute it in E-2, we will get

β = 16 – 8

β = 8

Now,

Sum of the zeroes of the new polynomial =  α + β  = 16 + 8 = 24

Product of the zeroes =  αβ  =  16 × 8  = 128

Then, the quadratic polynomial is-

K x² – (sumofthezeroes)x + (productofthezeroes)  =  x² – 24x + 128

Hence, the required quadratic polynomial is f(x) = x² + 24x + 128

Q.17: If α and β are the zeroes of the quadratic polynomial f(x) = x² – 1, find a quadratic polynomial whose zeroes are 2α/β and 2β/α.

Solution: We have,

f(x) = x² – 1

Sum of the zeroes =  α + β  = 0

Product of the zeroes =  αβ  = -1

From the question,

Sum of the zeroes of the new polynomial =  2α/β and 2β/α

 { By substituting the value of the sum and products of the zeroes }

As given in the question,

Product of the zeroes =

Hence, the quadratic polynomial is

x² – (sumofthezeroes)x + (productofthezeroes)

= kx² – (−4)x + 4  =  x² + 4x + 4

Hence, the required quadratic polynomial is f(x) = x² + 4x + 4

Q.18: If α and β are the zeroes of the quadratic polynomial f(x) = x² – 3x – 2, find a quadratic polynomial whose zeroes are 

Solution: We have,

f(x) = x² – 3x – 2

Sum of the zeroes =  α + β  = 3

Product of the zeroes =  αβ  = -2

From the question,

Sum of the zeroes of the new polynomial    

Product of the zeroes =  

So, the quadratic polynomial is,

x² – (sumofthezeroes)x + (productofthezeroes)

Hence, the required quadratic polynomial is k 
 

Q.19: If α and β are the zeroes of the quadratic polynomial f(x) = x² + px + q, form a polynomial whose zeroes are (α + β)² and (α – β)².

Solution: We have,

f(x) = x² + px + q

Sum of the zeroes =  α + β  = -p

Product of the zeroes =  αβ  = q

From the question,

Sum of the zeroes of new polynomial =  (α + β)² + (α – β)²

                                                                        =  (α + β)² + α² + β² – 2αβ

=  (α + β)² + (α + β)² – 2αβ – 2αβ

=  (−p)² + (−p)² – 2×q – 2×q

=  p² + p² – 4q

=  2p² – 4q

Product of the zeroes of new polynomial =  (α + β)²(α – β)²

=  (−p)²((−p)² – 4q)

=  p²(p² – 4q)

So, the quadratic polynomial is ,

x² – (sumofthezeroes)x + (productofthezeroes)

=  x² – (2p² – 4q)x + p²(p² – 4q)

Hence, the required quadratic polynomial is f(x) = k(x² – (2p² – 4q)x + p²(p² – 4q)).

Q.20: If α and β are the zeroes of the quadratic polynomial   f(x) = x² – 2x + 3, find a polynomial whose roots are:

(i) α + 2,β + 2             

(ii) 

Solution: We have,

f(x) = x² – 2x + 3

Sum of the zeroes =  α + β  = 2

Product of the zeroes =  αβ  = 3

(i) Sum of the zeroes of new polynomial =   (α + 2) + (β + 2)

=    α + β + 4

=    2 + 4 = 6

Product of the zeroes of new polynomial =  (α + 1)(β + 1)

=  αβ + 2α + 2β + 4

=  αβ + 2(α + β) + 4  =  3 + 2(2) + 4  = 11

So, quadratic polynomial is:

x² – (sumofthezeroes)x + (productofthezeroes)

=  x² – 6x + 11

Hence, the required quadratic polynomial is f(x) = k(x² – 6x + 11)

(ii) Sum of the zeroes of new polynomial =

Product of the zeroes of new polynomial =

=  2/6 = 1/3

So, the quadratic polynomial is,

x² – (sumofthezeroes)x + (productofthezeroes)

=  

Thus, the required quadratic polynomial is f(x) =

Q.21: If α and β are the zeroes of the quadratic polynomial  f(x) = ax² + bx + c, then evaluate:

(i) α – β

(ii) 1/α – 1/β

(ii) 1/α + 1/β – 2αβ

(iv) α²β + αβ²

(v) α⁴ + β⁴

Solution:

f(x) = ax² + bx + c

Here,

Sum of the zeroes of polynomial =   α + β  =  −b/a

Product of zeroes of polynomial =   αβ  =  c/a

Since, α + β are the roots (or) zeroes of the given polynomial, so

(i) α – β

The two zeroes of the polynomials are-

(ii) 1/α – 1/β

        ….. E.1

From previous question we know that,

Also,

αβ  =  c/a

Putting the values in E.1, we will get

(iii) 1/α + 1/β – 2αβ

Since,

Sum of the zeroes of polynomial =   α + β  =  −b/a

Product of zeroes of polynomial =   αβ  =  c/a

After substituting it in E-1, we will get

(iv) α²β + αβ²

=  αβ(α + β) …….. E-1.

Since,

Sum of the zeroes of polynomial =   α + β  =  −b/a

Product of zeroes of polynomial =   αβ  =  c/a

After substituting it in E-1, we will get

(v) α⁴ + β⁴

=  (α² + β²)² – 2α²β²

=  ((α + β)² – 2αβ)² – (2αβ)² ……. E- 1

Since,

Sum of the zeroes of polynomial =   α + β  =  −b/a

Product of zeroes of polynomial =   αβ  =  c/a

After substituting it in E-1, we will get

Since,

Sum of the zeroes of polynomial =   α + β  =  −b/a

Product of zeroes of polynomial =   αβ  =  c/a

After substituting it , we will get

Since,

Sum of the zeroes of polynomial =   α + β  =  −b/a

Product of zeroes of polynomial =   αβ  =  c/a

After substituting it , we will get

Since,

Sum of the zeroes of polynomial =   α + β  =  −b/a

Product of zeroes of polynomial =   αβ  =  c/a

After substituting it , we will get