# 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,

Hey! Looking for some great resources suitable for young ones? You've come to the right place. Check out our

self-paced courses designed for students of grades I-XII.Start with topics like

Python, HTML, ML, and learn to make some games and appsall with the help of our expertly designed content! So students worry no more, becauseGeeksforGeeks Schoolis now here!

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