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

• Difficulty Level : Medium
• Last Updated : 02 Feb, 2021

### Question 11. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = 6x2 + x â€“ 2, find the value of (Î±/Î²) +(Î²/Î±)

Solution:

Given that,

Î± and Î² are the zeroes of the quadratic polynomial f(x) = 6x2 + x â€“ 2.

therefore,

Sum of the zeroes = Î± + Î² = -1/6,

Product of the zeroes =Î± Ã— Î² = -1/3.

Now,

(Î±/Î²) +(Î²/Î±) = (Î±2 + Î²2) – 2Î±Î² / Î±Î²

Now substitute the values of the sum of zeroes and products of the zeroes and we will get,

= -25/12

Hence the value of (Î±/Î²) +(Î²/Î±) is -25/12.

### Question 12. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = 6x2 + x â€“ 2, find the value of Î±/Î² + 2(1/Î± + 1/Î²) + 3Î±Î²

Solution:

Given that,

Î± and Î² are the zeroes of the quadratic polynomial f(x) = 6x2 + x â€“ 2.

therefore,

Sum of the zeroes = Î± + Î² = 6/3

Product of the zeroes = Î± Ã— Î² = 4/3

Now,

Î±/Î² + 2(1/Î± + 1/Î²) + 3Î±Î² = [(Î±2 + Î²2) / Î±Î²] + 2(1/Î± + 1/Î²) + 3Î±Î²

[ ((Î± + Î²)2 – 2Î±Î²) / Î±Î²] + 2(1/Î± + 1/Î²) + 3Î±Î²

Now substitute the values of the sum of zeroes and products of the zeroes and we will get,

Î±/Î² + 2(1/Î± + 1/Î²) + 3Î±Î² = 8

Hence the value of Î±/Î² + 2(1/Î± + 1/Î²) + 3Î±Î² 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 Î± and Î².

Given that,

f(x) = x2 + px + 45

Now,

Sum of the zeroes = Î± + Î² = â€“ p

Product of the zeroes = Î± Ã— Î² = 45

therefore,

(Î± + Î²)2 – 4Î±Î² = (-p)2 – 4 x 45 = 144

(-p)2 = 144 + 180 = 324

p = âˆš324

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

### Question 14. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = x2 â€“ px + q, prove that [(Î±2 / Î²2) + (Î²2 / Î±2)] = [p4/q2] – [4p2/q] + 2

Solution:

Given that,

Î± and Î² are the roots of the quadratic polynomial.

f(x) = x2 â€“ px + q

Now,

Sum of the zeroes = p = Î± + Î²

Product of the zeroes = q = Î± Ã— Î²

therefore,

LHS = [(Î±2 / Î²2) + (Î²2 / Î±2)]

= [(Î±^4 + Î²4) / Î±2.Î²2]

= [((Î±+ Î²)^2 – 2Î±Î²)2 + 2(Î±Î²)2] / (Î±Î²)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 Î± and Î² are the zeroes of the quadratic polynomial f(x) = x2 â€“ p(x + 1) â€“ c, show that (Î± + 1)(Î² + 1) = 1 â€“ c.

Solution:

Given that,

Î± and Î² are the zeroes of the quadratic polynomial

f(x) = x2 â€“ p(x + 1)â€“ c

Now,

Sum of the zeroes = Î± + Î² = p

Product of the zeroes = Î± Ã— Î² = (- p â€“ c)

therefore,

(Î± + 1)(Î² + 1)

= Î±Î² + Î± + Î² + 1

= Î±Î² + (Î± + Î²) + 1

= (âˆ’ p â€“ c) + p + 1

= 1 â€“ c = RHS

therefore, LHS = RHS

Hence proved.

### Question 16. If Î± and Î² are the zeroes of the quadratic polynomial such that Î± + Î² = 24 and Î± â€“ Î² = 8, find a quadratic polynomial having Î± and Î² as its zeroes.

Solution:

Given that,

Î± + Î² = 24 ——(i)

Î± â€“ Î² = 8 ——(ii)

By solving the above two equations, we will get

2Î± = 32

Î± = 16

put the value of Î± in any of the equation.

Let we substitute it in (ii) and we will get,

Î² = 16 â€“ 8

Î² = 8

Now,

Sum of the zeroes of the new polynomial = Î± + Î² = 16 + 8 = 24

Product of the zeroes = Î±Î² = 16 Ã— 8 = 128

Then, The quadratic polynomial = x2â€“ (sum of the zeroes)x + (product of the zeroes) = x2 â€“ 24x + 128

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

### Question 17. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = x2 â€“ 1, find a quadratic polynomial whose zeroes are 2Î±/Î² and 2Î²/Î±.

Solution:

Given that,

f(x) = x2 â€“ 1

Sum of the zeroes = Î± + Î² = 0

Product of the zeroes = Î±Î² = â€“ 1

therefore,

Sum of the zeroes of the new polynomial

= [(2Î±2 + 2Î²2)] / Î±Î²

= [2(Î±2 + Î²2)] / Î±Î²

= [2((Î± + Î²)2 – 2Î±Î²)] / Î±Î² = 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

= (2Î±)(2Î²) / Î±Î² = 4

x2 â€“ (sum of the zeroes)x + (product of the zeroes)

= kx2 â€“ (âˆ’4)x + 4x2 â€“(âˆ’4)x + 4

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

### Question 18. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = x2 â€“ 3x â€“ 2, find a quadratic polynomial whose zeroes are 1/(2Î± + Î²) and 1/(2Î² + Î±).

Solution:

Given that,

f(x) = x2 â€“ 3x â€“ 2

Sum of the zeroes = Î± + Î² = 3

Product of the zeroes = Î±Î² = â€“ 2

therefore,

Sum of the zeroes of the new polynomial

= 1/(2Î± + Î²) + 1/(2Î² + Î±)

= (2Î± + Î² + 2Î² + Î±) / (2Î± + Î²)(2Î² + Î±)

= (3Î± + 3Î²) / (2(Î±2 + Î²2) + 5Î±Î²)

= (3 x 3) / 2[2(Î± + Î²)2 – 2Î±Î² + 5 x (-2)]

= 9 / 2[9-(-4)]-10 = 9/16

Product of zeroes = 1/(2Î± + Î²) x 1/(2Î² + Î±)

= 1 / (4Î±Î² + 2Î±2 + 2Î²2 + Î±Î²)

= 1 / [5Î±Î² + 2((Î± + Î²)2 – 2Î±Î²)]

= 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 Î± and Î² are the zeroes of the quadratic polynomial f(x) = x2 + px + q, form a polynomial whose zeroes are (Î± + Î²)2 and (Î± â€“ Î²)2.

Solution:

Given that,

f(x) = x2 + px + q

Sum of the zeroes = Î± + Î² = -p

Product of the zeroes = Î±Î² = q

therefore,

Sum of the zeroes of new polynomial = (Î± + Î²)2 + (Î± â€“ Î²)2

= (Î± + Î²)2 + Î±2 + Î²2 â€“ 2Î±Î²

= (Î± + Î²)2 + (Î± + Î²)2 â€“ 2Î±Î² â€“ 2Î±Î²

= (- p)2 + (- p)2 â€“ 2 Ã— q â€“ 2 Ã— q

= p2 + p2 â€“ 4q = p2 â€“ 4q

Product of the zeroes of new polynomial = (Î± + Î²)2 x (Î± â€“ Î²)2

= (- p)2((- p)2 – 4q)

= p2 (p2â€“4q)

x2 â€“ (sum of the zeroes)x + (product of the zeroes)

= x2 â€“ (2p2 â€“ 4q)x + p2(p2 â€“ 4q)

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

### Question 20. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = x2 â€“ 2x + 3, find a polynomial whose roots are:

(i) Î± + 2, Î² + 2

(ii) [Î±-1] / [Î±+1], [Î²-1] / [Î²+1]

Solution:

Given that,

f(x) = x2 â€“ 2x + 3

Sum of the zeroes = Î± + Î² = 2

Product of the zeroes = Î±Î² = 3

(i) Sum of the zeroes of new polynomial = (Î± + 2) + (Î² + 2)

= Î± + Î² + 4 = 2 + 4 = 6

Product of the zeroes of new polynomial = (Î± + 1)(Î² + 1)

= Î±Î² + 2Î± + 2Î² + 4

= Î±Î² + 2(Î± + Î²) + 4 = 3 + 2(2) + 4 = 11

x2 â€“ (sum of the zeroes)x + (product of the zeroes)

= x2 â€“ 6x +11

Hence, the required quadratic polynomial is f(x) = k(x2 â€“ 6x + 11).

(ii) Sum of the zeroes of new polynomial :

= [(Î±-1)/(Î±+1)] + [(Î²-1)/(Î²+1)]

= [(Î±-1)(Î²+1) + (Î²-1)(Î±+1)] / (Î±+1)(Î²+1)

= [Î±Î² + Î± – Î² – 1 + Î±Î² – Î± + Î² – 1)] / (Î±+1)(Î²+1)

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

Product of the zeroes of new polynomial :

= [(Î±-1)/(Î±+1)] + [(Î²-1)/(Î²+1)]

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

x2 â€“ (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 â€“ (2/3)x + (1/3))

### Question 21. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate:

(i) Î± â€“ Î²

(ii) 1/Î± – 1/Î²

(iii) 1/Î± + 1/Î² – 2Î±Î²

(iv) Î±2Î² + Î±Î²2

(v) Î±4 + Î²4

(vi) 1/(aÎ± + b) + 1/(aÎ² + b)

(vii) Î²/(aÎ± + b) + Î±/(aÎ² + b)

(viii) [(Î±2/Î²) + (Î²2/Î±)] + b[Î±/a + Î²/a]

Solution:

Given that,

f(x) = ax2 + bx + c

Sum of the zeroes of polynomial = Î± + Î² = -b/a

Product of zeroes of polynomial = Î±Î² = c/a

Since, Î± + Î² are the zeroes of the given polynomial therefore,

(i) Î± â€“ Î²

The two zeroes of the polynomials are :

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

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

= âˆš(b2-4ac) / a

(ii) 1/Î± – 1/Î²

= (Î²-1) / Î±Î² = -(Î±-Î²)/Î±Î² ——-(1)

From above question as we know that,

Î±-Î² = âˆš(b2-4ac) / a

and,

Î±Î² = c/a

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

= -[(âˆš(b2-4ac))/c]

(iii) (1/Î±) + (1/Î²) – 2Î±Î²

= (Î±+Î²)/Î±Î² – 2Î±Î² ———- (i)

Since,

Sum of the zeroes of polynomial = Î± + Î² = â€“ b/a

Product of zeroes of polynomial = Î±Î² = c/a

After putting it in (i), we will get

= (-b/a x a/c – 2c/a) = -[b/c + 2c/a]

(iv) Î±2Î² + Î±Î²2

= Î±Î²(Î± + Î²) ——–(i)

Since,

Sum of the zeroes of polynomial = Î± + Î² = â€“ b/ a

Product of zeroes of polynomial = Î±Î² = c/a

After putting it in (i), we will get

= c/a(-b/a) = -bc/a^2

(v) Î±4 + Î²4

= (Î±2 + Î²2)2 â€“ 2Î±2Î²2

= ((Î± + Î²)2 â€“ 2Î±Î²)2 â€“ (2Î±Î²)2 ———(i)

Since,

Sum of the zeroes of polynomial = Î± + Î² = â€“ b/a

Product of zeroes of polynomial = Î±Î² = 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/(aÎ± + b) + 1/(aÎ² + b)

= (aÎ² + b + aÎ± + b) / (aÎ± + b)(aÎ² + b)

= (a(Î± + Î²) + 2b) / (a2 x Î±Î² + abÎ± + abÎ² + b2)

Since,

Sum of the zeroes of polynomial = Î± + Î² = â€“ b/a

Product of zeroes of polynomial = Î±Î² = c/a

After putting it, we will get

= b / (ac – b2 + b2) = b/ac

(vii) Î²/(aÎ± + b) + Î±/(aÎ² + b)

= [Î²(aÎ² + b) + Î±(aÎ± + b)] / (aÎ² + b)(aÎ± + b)

= [aÎ±2 + bÎ²2 + bÎ± + bÎ²] / (a2 x (c/a) + ab(Î±+Î²) + b2)

Since,

Sum of the zeroes of polynomial = Î± + Î² = â€“ b/a

Product of zeroes of polynomial = Î±Î² = c/a

After putting it, we will get

= a[(Î±+Î²)2 – b(Î±+Î²)] / ac

= a[b2/a – 2c/a] – b2/a

= a[(b2 – 2c – b2)/a] / ac

= (b2 – 2c – b2) / ac = -2/a

(viii) [(Î±2/Î²) + (Î²2/Î±)] + b[Î±/a + Î²/a]

= a[(Î±2 + Î²2) / Î±Î²] + b[(Î±2+Î²2)/Î±Î²]

= a[(Î±+Î²)2 – 2Î±Î²] + b((Î±+Î²)2 – 2Î±Î²) / Î±Î²

Since,

Sum of the zeroes of polynomial= Î± + Î² = â€“ b/a

Product of zeroes of polynomial= Î±Î² = 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

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