Chapter 2 - Polynomials, RD Sharma Solutions - (Part-3) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 2.35

Ques.21. If α and β are the zeroes of the quadratic polynomial f(x) = ax² + bx + c, then evaluate :


Ans. (i) Given α and β are the zeros of the quadratic polynomial f(x) = ax²  + bx + c
α + β = -Coefficient of x/Coefficient of x²
= -b/a
αβ = Constant term/Coefficient of x²
= c/a
We have, (α - β)

Substituting α + β = (-b)/a and αβ = c/a then we get, 


Hence, the value of α - β is. 
(ii) Given α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c
α + β = -Coefficient of x/Coefficient of x²
= (-b)/a
αβ = Constant term/Coefficient of x²
= c/a
We have,

Substituting α + β = (-b)/a and αβ = c/a then we get,


By taking least common factor we get,

Hence the value of  is . 
(iii) Given α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c 
α + β = -Coefficient of x/Coefficient of x²
= (-b)/a
αβ = Constant term/Coefficient of x²
= c/a
We have, 
By cross multiplication we get,

By substituting α + β = -b/a and αβ = c/a we get,
 

Hence the value of  is. 
(iv) Given α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c 
α + β = -Coefficient of x/Coefficient of x²
= (-b)/a
αβ = Constant term/Coefficient of x²
= c/a
We have, α²β + αβ²
By taking common factor αβ we get, = 2β(α + β)
By substituting α + β = -b/a and αβ = c/a we get, 

Hence the value of α²β + αβ² is (-cb)/a². 
(v) Given α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c 
α + β = -Coefficient of x/Coefficient of x²
= (-b)/a
αβ = Constant term/Coefficient of x²
= c/a
We have,

By substituting α + β = -b/a and αβ = c/a we get,

By taking least common factor we get


Hence the value of α⁴ + β⁴ is
(vi) Since α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c
α + β = -Coefficient of x/Coefficient of x²
= (-b)/a
αβ = Constant term/Coefficient of x²
= c/a
We have,


By substituting α + β = -b/a and αβ = c/a we get,


Hence, the value of  is . 
(vii) Since α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c
α + β = -Coefficient of x/Coefficient of x²
= (-b)/a
αβ = Constant term/Coefficient of x²
= c/a
We have, 






By substituting α + β = -b/a and αβ = c/a we get,










Hence, the value of  is . 
(viii) Since α and β are the zeros of the quadratic polynomial f(x) = ax² + bx + c
α + β = -Coefficient of x/Coefficient of x²
= (-b)/a
αβ = Constant term/Coefficient of x²
= c/a
We have,




By substituting α + β = -b/a and αβ = c/a we get,









Hence, the value of  is . 

Page No 2.43

Ques.1. Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case:
(i) f(x) = 2x³+x² − 5x + 2; 1/2, 1, −2
(ii) g(x) = x³ − 4x² + 5x − 2; 2, 1, 1
Ans. (i) We have,
f(x) = 2x³ + 2x² − 5x + 2

f(1/2) = 0
f(1) = 2(1)³ + (1)² - 5(1) + 2
f(1) = 2 + 1 - 5 + 2
f(1) = 0
f(-2) = 2(-2)³ + (-2)² - 5(-2) + 2
f(-2) = -16 + 4 + 10 + 2
f(-2) = 0
So, 1/2, 1 and 2 are the zeros of polynomial p(x)
Let α = 1/2, β = 1 and γ = -2. Then 
α + β + γ = 1/2 + 1 - 2

= (-1)/2
From f(x) = 2x³ + 2x² - 5x + 2
α + β + γ = (-Coefficient of x²)/Coefficient of x³
α + β + γ = -(1/2)

= 1/2 - 2 - 2/2
Taking least common factor we get,

From f(x) = 2x³ + 2x² - 5x + 2 
αβ + βγ + γα = Coefficient of x/Coefficient of x³
αβ + βγ + γα = (-5)/2

From f(x) = 2x³ + 2x² - 5x + 2
αβγ = (-Constant term)/Coefficient of x²
αβγ =
αβγ = -1
Hence, it is verified that the numbers given along side of the cubic polynomials are their zeros and also verified the relationship between the zeros and coefficients
(ii) We have, 
g(x) = x³ - 4x² + 5x - 2
g(2) = (2)³ - 4(2)² + 5(2) - 2
g(2) = 8 - 16 + 10 - 2
g(2) = 0
g(1) = (1)³ - 4(1)² + 5(1) - 2
= 1 - 4 + 5 - 2
= 0
g(1) = (1)³ - 4(1)² + 5(1) - 2
= 1 - 4 + 5 - 2
= 0
So 2, 1 and 1 are the zeros of the polynomial g(x)
Let  α = 2, β = 1 and λ = 1. Then, 
α + β + γ = 2 + 1 + 1
α + β + γ = 4
From g(x) = x³ - 4x² + 5x - 2
α + β + γ = Coefficient of x²/Coefficient of x³
α + β + γ = 4/1
α + β + γ = 4
αβ + βγ + γα = 2(1) + 1(1) + 1(2)
αβ + βγ + γα = 2 + 1 + 2
αβ + βγ + γα = 5
From g(x) = x³ - 4x² + 5x - 2
αβ + βγ + γα = Coefficient of x/Coefficient of x³
αβ + βγ + γα = 5/1
αβ + βγ + γα = 5
αβγ = 2 x 1 x 1
αβγ = 2
From g(x) = x³ - 4x² + 5x - 2
αβγ = (-Constant term)/Coefficient of x²
αβγ = -((-2)/1)
αβγ = 2
Hence, it is verified that the numbers given along side of the cubic polynomials are their zeros and also verified the relationship between the zeros and coefficients.

Ques.2. Find the cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and product of its zeros as 3, −1 and − 3 respectively.
Ans. If α, β and γ are the zeros of a cubic polynomial f(x), then
where k is any non-zero real number.  
Here,
α + β + γ = 3,
αβ + βγ + γα = -1
αβγ = -3
Therefore
f(x) = k{x³ - (3)x² + (-1)x - (-3)}
f(x) = k{x³ - 3x² - 1x + 3}
Hence, cubic polynomial is f(x) = k{x³ - 3x² - 1x + 3}, where k is any non-zero real number.

Ques.3. If the zeros of the polynomial f(x) = 2x³ − 15x² + 37x − 30 are in A.P., find them.
Ans. Let α = a - d, β = a and γ = a + d be the zeros of the polynomial
f(x) = 2x³ - 15x² + 37x - 30 
Therefore
α + β + γ = Coefficient of x²/Coefficient of x³
= -((-15)/2)
= 15/2
αβγ = (-Constant term)/Coefficient of x²
= -((-30)/2)
= 15
Sum of the zeros = Coefficient of x²/Coefficient of x³



Product of the zeros = (-Constant term)/Coefficient of x²
αβγ = 15
(a - d) + a + (a + d) = 15
a(a² - d²) = 15
Substituting a = 5/2 we get



Therefore, substituting a = 5/2 and d = 1/2 in α = a - d, β = a and γ = a + d
α = a - d

α = 4/2
α = 2
β = a
β = 5/2
γ = a + d

γ = 6/2
γ = 3
Hence, the zeros of the polynomial are 2, 5/2, 3.

Ques.4. Find the condition that the zeros of the polynomial f(x) = x³ + 3px² + 3qx + r may be in A.P.
Ans. Let a-d, a and a+ d be the zeros of the polynomials f(x).Then, 
Sum of the zeros = Coefficient of x²/Coefficient of x³

3a = -3p


a = -p
Since a is a zero of the polynomial f(x). Therefore, 

a³ + 3pa² + 3qa + r = 0
Substituting a = -p we get, 
(-p)³ + 3p(-p)² + 3q(-p) + r = 0
-p³ + 3p³ - 3pq + r = 0
2p³ - 3pq + r = 0
Hence, the condition for the given polynomial is 2p³ - 3pq + r = 0.

Ques.5. If the zeros of the polynomial f(x) = ax³ + 3bx² + 3cx + d are in A.P., prove that 2b³ − 3abc + a²d = 0.
Ans. Let a - d, a and a + d be the zeros of the polynomial f(x). Then, 
Sum of the zeros = Coefficient of x²/Coefficient of x³


a = (-b)/a
Since a is a zero of the polynomial f(x).
Therefore,
f(x) = ax³ + 3bx² + 3cx + d
f(a) = 0
f(a) = aa³ + 3ba² + 3ca + d
aa³ + 3ba² + 3ca + d = 0



2b³ + 3abc +a²d = 0 x a²
2b³ - 3abc + a²d = 0
Hence, it is proved that 2b³ - 3abc + a²d = 0.

Ques.6. If the zeros of the polynomial f(x) = x³ − 12x² + 39x + k are in A.P., find the value of k.
Ans. Let a - d, a  and a + d be the zeros of the polynomial f(x).
Then,
Sum of the zeros = Coefficient of x²/Coefficient of x<sup>3

3a = 12
a = 12/3
a = 4
Since a is a zero of the polynomial f(x)


-28 = k
Hence, the value of k is -28.</sup>