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

### Question 11. If Î± and Î² are the zeroes of the quadratic polynomial f(x) = 6x^{2} + 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) = 6x

^{2}+ 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}+ Î²^{2}) – 2Î±Î² / Î±Î²Now 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) = 6x^{2} + 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) = 6x

^{2}+ 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}+ Î²^{2}) / Î±Î²] + 2(1/Î± + 1/Î²) + 3Î±Î²[ ((Î± + Î²)

^{2}– 2Î±Î²) / Î±Î²] + 2(1/Î± + 1/Î²) + 3Î±Î²Now 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) = 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 Î± and Î².

Given that,

f(x) = x

^{2 }+ px + 45Now,

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 = 324p = 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) = x^{2} â€“ px + q, prove that [(Î±^{2} / Î²^{2}) + (Î²^{2} / Î±^{2})] = [p^{4}/q^{2}] – [4p^{2}/q] + 2

**Solution:**

Given that,

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

f(x) = x

^{2}â€“ px + qNow,

Sum of the zeroes = p = Î± + Î²

Product of the zeroes = q = Î± Ã— Î²

therefore,

LHS = [(Î±

^{2}/ Î²^{2}) + (Î²^{2 }/ Î±^{2})]= [(Î±

^{^4}+ Î²^{4}) / Î±^{2}.Î²^{2}]= [((Î±+ Î²)

^{^2}– 2Î±Î²)^{2}+ 2(Î±Î²)^{2}] / (Î±Î²)^{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 Î± and Î² are the zeroes of the quadratic polynomial f(x) = x^{2 }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) = x

^{2}â€“ p(x + 1)â€“ cNow,

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) = x^{2}+ 24x + 128

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

**Solution: **

Given that,

f(x) = x

^{2}â€“ 1Sum 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}+ 2Î²^{2})] / Î±Î²= [2(Î±

^{2}+ Î²^{2})] / 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

Hence, the quadratic polynomial is

x

^{2}â€“ (sum of the zeroes)x + (product of the zeroes)= kx

^{2}â€“ (âˆ’4)x + 4x^{2}â€“(âˆ’4)x + 4

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

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

**Solution:**

Given that,

f(x) = x

^{2}â€“ 3x â€“ 2Sum 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}+ Î²^{2}) + 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}+ 2Î²^{2}+ Î±Î²)= 1 / [5Î±Î² + 2((Î± + Î²)

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

**Solution: **

Given that,

f(x) = x

^{2}+ px + qSum 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}+ Î²^{2}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= p

^{2}+ p^{2}â€“ 4q = p^{2}â€“ 4qProduct of the zeroes of new polynomial = (Î± + Î²)

^{2 }x (Î± â€“ Î²)^{2}= (- p)

^{2}((- p)^{2}– 4q)= p

^{2}(p^{2}â€“4q)therefore, the quadratic polynomial is,

x

^{2}â€“ (sum of the zeroes)x + (product of the zeroes)= x

^{2}â€“ (2p^{2}â€“ 4q)x + p^{2}(p^{2}â€“ 4q)

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

**Question 20. I**f Î± and Î²** are the zeroes of the quadratic polynomial f(x) = x**^{2} â€“ 2x + 3, find a polynomial whose roots are:

^{2}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) = x

^{2}â€“ 2x + 3Sum 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

therefore, quadratic polynomial is :

x

^{2}â€“ (sum of the zeroes)x + (product of the zeroes)= x

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

^{2 }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

therefore, the quadratic polynomial is,

x

^{2}â€“ (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}â€“ (2/3)x + (1/3))

**Question 21. If **Î± and Î²** are the zeroes of the quadratic polynomial f(x) = ax**^{2} + bx + c, then evaluate:

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

**(i) Î± â€“ Î²**

**(ii) 1/Î± – 1/Î²**

**(iii) 1/Î± + 1/Î² – 2Î±Î²**

**(iv) Î± ^{2}Î² + Î±Î²^{2}**

**(v) Î± ^{4} + Î²^{4}**

**(vi) 1/(aÎ± + b) + 1/(aÎ² + b)**

**(vii) Î²/(aÎ± + b) + Î±/(aÎ² + b)**

**(viii) [(Î± ^{2}/Î²) + (Î²^{2}/Î±)] + b[Î±/a + Î²/a]**

**Solution: **

Given that,

f(x) = ax

^{2}+ bx + cSum 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€“ Î²The two zeroes of the polynomials are :

= [âˆš-b+b

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

^{2}-4ac) + b+âˆš(b^{2}-4ac)] / 2a=

âˆš(b^{2}-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(b

^{2}-4ac) / aand,

Î±Î² = c/a

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

=

-[(âˆš(b^{2}-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±^{2}Î² + Î±Î²^{2}= 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}+ Î²^{4}= (Î±

^{2}+ Î²^{2})^{2}â€“ 2Î±^{2}Î²^{2}= ((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}]= [(b

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

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

(vi)1/(aÎ± + b) + 1/(aÎ² + b)= (aÎ² + b + aÎ± + b) / (aÎ± + b)(aÎ² + b)

= (a(Î± + Î²) + 2b) / (a

^{2}x Î±Î² + abÎ± + abÎ² + b^{2})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 – b

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

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

= [aÎ±

^{2}+ bÎ²^{2}+ bÎ± + bÎ²] / (a^{2}x (c/a) + ab(Î±+Î²) + b^{2})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[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)[(Î±^{2}/Î²) + (Î²^{2}/Î±)] + b[Î±/a + Î²/a]= a[(Î±

^{2}+ Î²^{2}) / Î±Î²] + b[(Î±^{2}+Î²^{2})/Î±Î²]= a[(Î±+Î²)

^{2}– 2Î±Î²] + b((Î±+Î²)^{2}– 2Î±Î²) / Î±Î²Since,

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)= [(-b

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

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