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• RD Sharma Class 10 Solutions

# 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) = 6x2 + 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) = 6x2 + x – 2.

therefore,

Sum of the zeroes = α + β = -1/6,

Product of the zeroes =α × β = -1/3.

Now,

(α/β) +(β/α) = (α2 + β2) – 2αβ / αβ

Now 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) = 6x2 + 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) = 6x2 + 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 = [(α2 + β2) / αβ] + 2(1/α + 1/β) + 3αβ

[ ((α + β)2 – 2αβ) / αβ] + 2(1/α + 1/β) + 3αβ

Now 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) = x2 + 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) = x2 + px + 45

Now,

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 = 324

p = √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) = x2 – px + q, prove that [(α2 / β2) + (β2 / α2)] = [p4/q2] – [4p2/q] + 2

Solution:

Given that,

α and β are the roots of the quadratic polynomial.

f(x) = x2 – px + q

Now,

Sum of the zeroes = p = α + β

Product of the zeroes = q = α × β

therefore,

LHS = [(α2 / β2) + (β2 / α2)]

= [(α^4 + β4) / α22]

= [((α+ β)^2 – 2αβ)2 + 2(αβ)2] / (αβ)2

= [((p)2 – 2q)2 + 2(q)2] / (q)2

= [(p4 + 4q2 – 4pq2) – 2q2] / q2

= (p4 + 2q2 – 4pq2) / q2 = (p/q)2 – (4p2/q) + 2

LHS = RHS

Hence, proved.

### Question 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:

Given that,

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)

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) = x2 + 24x + 128

### Question 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:

Given that,

f(x) = x2 – 1

Sum of the zeroes = α + β = 0

Product of the zeroes = αβ = – 1

therefore,

Sum of the zeroes of the new polynomial

= [(2α2 + 2β2)] / αβ

= [2(α2 + β2)] / 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

x2 – (sum of the zeroes)x + (product of the zeroes)

= kx2 – (−4)x + 4x2 –(−4)x + 4

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

### Question 18. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 3x – 2, find a quadratic polynomial whose zeroes are 1/(2α + β) and 1/(2β + α).

Solution:

Given that,

f(x) = x2 – 3x – 2

Sum 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(α2 + β2) + 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 + 2α2 + 2β2 + αβ)

= 1 / [5αβ + 2((α + β)2 – 2αβ)]

= 1 / [5 x (-2) + 2((3)2 – 2 x (-2))] = 1/16

x2– (sum of the zeroes)x + (product of the zeroes)

= (x2 + (9/16)x +(1/16))

Hence, the required quadratic polynomial is (x2 + (9/16)x +(1/16)).

### Question 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:

Given that,

f(x) = x2 + px + q

Sum 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 + α2 + β2 – 2αβ

= (α + β)2 + (α + β)2 – 2αβ – 2αβ

= (- p)2 + (- p)2 – 2 × q – 2 × q

= p2 + p2 – 4q = p2 – 4q

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

= (- p)2((- p)2 – 4q)

= p2 (p2–4q)

x2 – (sum of the zeroes)x + (product of the zeroes)

= x2 – (2p2 – 4q)x + p2(p2 – 4q)

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

### Question 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) [α-1] / [α+1], [β-1] / [β+1]

Solution:

Given that,

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

x2 – (sum of the zeroes)x + (product of the zeroes)

= x2 – 6x +11

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

x2 – (sum of the zeroes)x + (product of the zeroes)

= x2 – (2/3)x + (1/3)

Hence, the required quadratic polynomial is f(x) = k(x2 – (2/3)x + (1/3))

### Question 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

(iii) 1/α + 1/β – 2αβ

(iv) α2β + αβ2

(v) α4 + β4

(vi) 1/(aα + b) + 1/(aβ + b)

(vii) β/(aα + b) + α/(aβ + b)

(viii) [(α2/β) + (β2/α)] + b[α/a + β/a]

Solution:

Given that,

f(x) = ax2 + bx + c

Sum 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 – β

The two zeroes of the polynomials are :

= [√-b+b2-4ac]/2a – ([-b+√(b2-4ac)]/2a)

= [-b+√(b2-4ac) + b+√(b2-4ac)] / 2a

= √(b2-4ac) / a

(ii) 1/α – 1/β

= (β-1) / αβ = -(α-β)/αβ ——-(1)

From above question as we know that,

α-β = √(b2-4ac) / a

and,

αβ = c/a

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

= -[(√(b2-4ac))/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) α2β + αβ2

= 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) α4 + β4

= (α2 + β2)2 – 2α2β2

= ((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]

= [(b2 -2(ac)) / a2]2 – [2(c/a)2]

= [(b2 – 2ac)2 – 2a2 c2] / a4

(vi) 1/(aα + b) + 1/(aβ + b)

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

= (a(α + β) + 2b) / (a2 x αβ + abα + abβ + b2)

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 – b2 + b2) = b/ac

(vii) β/(aα + b) + α/(aβ + b)

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

= [aα2 + bβ2 + bα + bβ] / (a2 x (c/a) + ab(α+β) + b2)

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[b2/a – 2c/a] – b2/a

= a[(b2 – 2c – b2)/a] / ac

= (b2 – 2c – b2) / ac = -2/a

(viii) [(α2/β) + (β2/α)] + b[α/a + β/a]

= a[(α2 + β2) / αβ] + b[(α22)/αβ]

= a[(α+β)2 – 2αβ] + b((α+β)2 – 2αβ) / αβ

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[(-ba)2 – 3x(c/a)] + b((-b/a)2 – 2(c/a)) / (c/a)

= [(-b2a2/a2c)+(3bca2/a2)+(b/a)2 – (2bca2/a2c)] = b