# Class 10 RD Sharma Solutions- Chapter 2 Polynomials – Exercise 2.1 | Set 2

### Question 11. If α and β are the zeroes of the quadratic polynomial f(x) = 6x^{2} + x – 2, find the value of (α/β) +(β/α)

**Solution:**

Given that,

α and β are the zeroes of the quadratic polynomial f(x) = 6x

^{2}+ x – 2.therefore,

Sum of the zeroes = α + β = -1/6,

Product of the zeroes =α × β = -1/3.

Now,

(α/β) +(β/α) = (α

^{2}+ β^{2}) – 2αβ / αβNow substitute the values of the sum of zeroes and products of the zeroes and we will get,

= -25/12

Hence the value of (α/β) +(β/α) is -25/12.

### Question 12. If α and β are the zeroes of the quadratic polynomial f(x) = 6x^{2} + x – 2, find the value of α/β + 2(1/α + 1/β) + 3αβ

**Solution: **

Given that,

α and β are the zeroes of the quadratic polynomial f(x) = 6x

^{2}+ x – 2.therefore,

Sum of the zeroes = α + β = 6/3

Product of the zeroes = α × β = 4/3

Now,

α/β + 2(1/α + 1/β) + 3αβ = [(α

^{2}+ β^{2}) / αβ] + 2(1/α + 1/β) + 3αβ[ ((α + β)

^{2}– 2αβ) / αβ] + 2(1/α + 1/β) + 3αβNow substitute the values of the sum of zeroes and products of the zeroes and we will get,

α/β + 2(1/α + 1/β) + 3αβ = 8

Hence the value of α/β + 2(1/α + 1/β) + 3αβ is 8.

**Question 13. If the squared difference of the zeroes of the quadratic polynomial f(x) = x**^{2} + px + 45 is equal to 144, find the value of p.

^{2}+ px + 45 is equal to 144, find the value of p.

**Solution: **

Let as assume that the two zeroes of the polynomial are α and β.

Given that,

f(x) = x

^{2 }+ px + 45Now,

Sum of the zeroes = α + β = – p

Product of the zeroes = α × β = 45

therefore,

(α + β)

^{2}– 4αβ = (-p)^{2}– 4 x 45 = 144(-p)

^{2}= 144 + 180 = 324p = √324

Hence the value of p will be either 18 or -18.

### Question 14. If α and β are the zeroes of the quadratic polynomial f(x) = x^{2} – px + q, prove that [(α^{2} / β^{2}) + (β^{2} / α^{2})] = [p^{4}/q^{2}] – [4p^{2}/q] + 2

**Solution:**

Given that,

α and β are the roots of the quadratic polynomial.

f(x) = x

^{2}– px + qNow,

Sum of the zeroes = p = α + β

Product of the zeroes = q = α × β

therefore,

LHS = [(α

^{2}/ β^{2}) + (β^{2 }/ α^{2})]= [(α

^{^4}+ β^{4}) / α^{2}.β^{2}]= [((α+ β)

^{^2}– 2αβ)^{2}+ 2(αβ)^{2}] / (αβ)^{2}= [((p)

^{2}– 2q)^{2 }+ 2(q)^{2}] / (q)^{2}= [(p

^{4 }+ 4q^{2}– 4pq^{2}) – 2q^{2}] / q^{2}= (p

^{4 }+ 2q^{2 }– 4pq^{2}) / q^{2}= (p/q)^{2}– (4p^{2}/q) + 2LHS = RHS

Hence, proved.

**Q**uestion 15. If α and β are the zeroes of the quadratic polynomial f(x) = x^{2 }– p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.

**Solution: **

Given that,

α and β are the zeroes of the quadratic polynomial

f(x) = x

^{2}– p(x + 1)– cNow,

Sum of the zeroes = α + β = p

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

therefore,

(α + 1)(β + 1)

= αβ + α + β + 1

= αβ + (α + β) + 1

= (− p – c) + p + 1

= 1 – c = RHS

therefore, LHS = RHS

Hence proved.

### Question 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:**

Given that,

α + β = 24 ——(i)

α – β = 8 ——(ii)

By solving the above two equations, we will get

2α = 32

α = 16

put the value of α in any of the equation.

Let we substitute it in (ii) and 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 = x2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128

Hence, the required quadratic polynomial is f(x) = x^{2}+ 24x + 128

**Question 17**. If α and β are the zeroes of the quadratic polynomial f(x) = x^{2 }– 1, find a quadratic polynomial whose zeroes are 2α/β and 2β/α.

**Solution: **

Given that,

f(x) = x

^{2}– 1Sum of the zeroes = α + β = 0

Product of the zeroes = αβ = – 1

therefore,

Sum of the zeroes of the new polynomial

= [(2α

^{2}+ 2β^{2})] / αβ= [2(α

^{2}+ β^{2})] / αβ= [2((α + β)

^{2}– 2αβ)] / αβ = 4/(-1)After substituting the value of the sum and products of the zeroes we will get,

As given in the question,

Product of the zeroes

= (2α)(2β) / αβ = 4

Hence, the quadratic polynomial is

x

^{2}– (sum of the zeroes)x + (product of the zeroes)= kx

^{2}– (−4)x + 4x^{2}–(−4)x + 4

Hence, the required quadratic polynomial is f(x) = x^{2 }+ 4x + 4

**Question 18. If **α and β are the zeroes of the quadratic polynomial f(x) = x^{2} – 3x – 2, find a quadratic polynomial whose zeroes are 1/(2α + β) and 1/(2β + α).

**Solution:**

Given that,

f(x) = x

^{2}– 3x – 2Sum of the zeroes = α + β = 3

Product of the zeroes = αβ = – 2

therefore,

Sum of the zeroes of the new polynomial

= 1/(2α + β) + 1/(2β + α)

= (2α + β + 2β + α) / (2α + β)(2β + α)

= (3α + 3β) / (2(α

^{2}+ β^{2}) + 5αβ)= (3 x 3) / 2[2(α + β)

^{2}– 2αβ + 5 x (-2)]= 9 / 2[9-(-4)]-10 = 9/16

Product of zeroes = 1/(2α + β) x 1/(2β + α)

= 1 / (4αβ + 2α

^{2}+ 2β^{2}+ αβ)= 1 / [5αβ + 2((α + β)

^{2 }– 2αβ)]= 1 / [5 x (-2) + 2((3)

^{2}– 2 x (-2))] = 1/16therefore, the quadratic polynomial is,

x

^{2}– (sum of the zeroes)x + (product of the zeroes)= (x

^{2}+ (9/16)x +(1/16))

Hence, the required quadratic polynomial is (x^{2}+ (9/16)x +(1/16)).

**Question 19. I**f α and β are the zeroes of the quadratic polynomial f(x) = x^{2} + px + q, form a polynomial whose zeroes are (α + β)^{2} and (α – β)^{2.}

**Solution: **

Given that,

f(x) = x

^{2}+ px + qSum of the zeroes = α + β = -p

Product of the zeroes = αβ = q

therefore,

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

^{2}+ (α – β)^{2}= (α + β)

^{2}+ α^{2}+ β^{2}– 2αβ= (α + β)

^{2}+ (α + β)^{2 }– 2αβ – 2αβ= (- p)

^{2 }+ (- p)^{2}– 2 × q – 2 × q= p

^{2}+ p^{2}– 4q = p^{2}– 4qProduct of the zeroes of new polynomial = (α + β)

^{2 }x (α – β)^{2}= (- p)

^{2}((- p)^{2}– 4q)= p

^{2}(p^{2}–4q)therefore, the quadratic polynomial is,

x

^{2}– (sum of the zeroes)x + (product of the zeroes)= x

^{2}– (2p^{2}– 4q)x + p^{2}(p^{2}– 4q)

Hence, the required quadratic polynomial is f(x) = k(x^{2}– (2p^{2}–4q) x + p^{2}(p^{2}– 4q)).

**Question 20. I**f α and β** are the zeroes of the quadratic polynomial f(x) = x**^{2} – 2x + 3, find a polynomial whose roots are:

^{2}– 2x + 3, find a polynomial whose roots are:

**(i) α + 2, β + 2**

**(ii) [α-1] / [α+1], [β-1] / [β+1]**

**Solution: **

Given that,

f(x) = x

^{2}– 2x + 3Sum 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

therefore, quadratic polynomial is :

x

^{2}– (sum of the zeroes)x + (product of the zeroes)= x

^{2}– 6x +11Hence, the required quadratic polynomial is f(x) = k(x

^{2 }– 6x + 11).

(ii)Sum of the zeroes of new polynomial := [(α-1)/(α+1)] + [(β-1)/(β+1)]

= [(α-1)(β+1) + (β-1)(α+1)] / (α+1)(β+1)

= [αβ + α – β – 1 + αβ – α + β – 1)] / (α+1)(β+1)

= (3-1+3-1) / (3+1+2) = 2/3

Product of the zeroes of new polynomial :

= [(α-1)/(α+1)] + [(β-1)/(β+1)]

= 26 = 13(2/6) = 1/3

therefore, the quadratic polynomial is,

x

^{2}– (sum of the zeroes)x + (product of the zeroes)= x

^{2}– (2/3)x + (1/3)

Hence, the required quadratic polynomial is f(x) = k(x^{2}– (2/3)x + (1/3))

**Question 21. If **α and β** are the zeroes of the quadratic polynomial f(x) = ax**^{2} + bx + c, then evaluate:

^{2}+ bx + c, then evaluate:

**(i) α – β**

**(ii) 1/α – 1/β**

**(iii) 1/α + 1/β – 2αβ**

**(iv) α ^{2}β + αβ^{2}**

**(v) α ^{4} + β^{4}**

**(vi) 1/(aα + b) + 1/(aβ + b)**

**(vii) β/(aα + b) + α/(aβ + b)**

**(viii) [(α ^{2}/β) + (β^{2}/α)] + b[α/a + β/a]**

**Solution: **

Given that,

f(x) = ax

^{2}+ bx + cSum of the zeroes of polynomial = α + β = -b/a

Product of zeroes of polynomial = αβ = c/a

Since, α + β are the zeroes of the given polynomial therefore,

(i)α – βThe two zeroes of the polynomials are :

= [√-b+b

^{2}-4ac]/2a – ([-b+√(b^{2}-4ac)]/2a)= [-b+√(b

^{2}-4ac) + b+√(b^{2}-4ac)] / 2a=

√(b^{2}-4ac) / a

(ii)1/α – 1/β= (β-1) / αβ = -(α-β)/αβ ——-(1)

From above question as we know that,

α-β = √(b

^{2}-4ac) / aand,

αβ = c/a

Put the values in (i) and we will get,

=

-[(√(b^{2}-4ac))/c]

(iii)(1/α) + (1/β) – 2αβ= (α+β)/αβ – 2αβ ———- (i)

Since,

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

Product of zeroes of polynomial = αβ = c/a

After putting it in (i), we will get

= (-b/a x a/c – 2c/a) =

-[b/c + 2c/a]

(iv)α^{2}β + αβ^{2}= αβ(α + β) ——–(i)

Since,

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

Product of zeroes of polynomial = αβ = c/a

After putting it in (i), we will get

= c/a(-b/a) =

-bc/a^{^2}

(v)α^{4}+ β^{4}= (α

^{2}+ β^{2})^{2}– 2α^{2}β^{2}= ((α + β)

^{2}– 2αβ)^{2 }– (2αβ)^{2}———(i)Since,

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

Product of zeroes of polynomial = αβ = c/a

After substituting it in (i), we will get

= [(-b/a) -2(c/a)]

^{2}– [2(c/a)^{2}]= [(b

^{2}-2(ac)) / a^{2}]^{2}– [2(c/a)^{2}]=

[(b^{2}– 2ac)^{2}– 2a^{2}c^{2}] / a^{4}

(vi)1/(aα + b) + 1/(aβ + b)= (aβ + b + aα + b) / (aα + b)(aβ + b)

= (a(α + β) + 2b) / (a

^{2}x αβ + abα + abβ + b^{2})Since,

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

Product of zeroes of polynomial = αβ = c/a

After putting it, we will get

= b / (ac – b

^{2}+ b^{2}) =b/ac

(vii)β/(aα + b) + α/(aβ + b)= [β(aβ + b) + α(aα + b)] / (aβ + b)(aα + b)

= [aα

^{2}+ bβ^{2}+ bα + bβ] / (a^{2}x (c/a) + ab(α+β) + b^{2})Since,

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

Product of zeroes of polynomial = αβ = c/a

After putting it, we will get

= a[(α+β)

^{2 }– b(α+β)] / ac= a[b

^{2}/a – 2c/a] – b^{2}/a= a[(b

^{2}– 2c – b^{2})/a] / ac= (b

^{2}– 2c – b^{2}) / ac =-2/a

(viii)[(α^{2}/β) + (β^{2}/α)] + b[α/a + β/a]= a[(α

^{2}+ β^{2}) / αβ] + b[(α^{2}+β^{2})/αβ]= a[(α+β)

^{2}– 2αβ] + b((α+β)^{2}– 2αβ) / αβSince,

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

Product of zeroes of polynomial= αβ = c/a

After putting it, we will get

= a[(-ba)

^{2 }– 3x(c/a)] + b((-b/a)^{2 }– 2(c/a)) / (c/a)= [(-b

^{2}a^{2}/a^{2}c)+(3bca^{2}/a^{2})+(b/a)^{2}– (2bca^{2}/a^{2}c)] =b