RD Sharma Solutions: Polynomials (Exercise 2.1)

RD Sharma Solutions: Polynomials (Exercise 2.1) Notes | Study Mathematics (Maths) Class 10 - Class 10

Document Description: RD Sharma Solutions: Polynomials (Exercise 2.1) for Class 10 2022 is part of Mathematics (Maths) Class 10 preparation. The notes and questions for RD Sharma Solutions: Polynomials (Exercise 2.1) have been prepared according to the Class 10 exam syllabus. Information about RD Sharma Solutions: Polynomials (Exercise 2.1) covers topics like and RD Sharma Solutions: Polynomials (Exercise 2.1) Example, for Class 10 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Polynomials (Exercise 2.1).

Introduction of RD Sharma Solutions: Polynomials (Exercise 2.1) in English is available as part of our Mathematics (Maths) Class 10 for Class 10 & RD Sharma Solutions: Polynomials (Exercise 2.1) in Hindi for Mathematics (Maths) Class 10 course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: RD Sharma Solutions: Polynomials (Exercise 2.1) Notes | Study Mathematics (Maths) Class 10 - Class 10
 1 Crore+ students have signed up on EduRev. Have you?

Question: 1
Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) f(x) = x2 – 2x – 8
(ii) g(s) = 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) h(t) = t2 – 15
(v) p(x) = x2 + 2√2 x – 6
(vi) q(x) = √3 x2 + 10x + 7√3
(vii) f(x) = x2 - (√3 + 1)x + √3
(viii) g(x) = a(x2 + 1) – x(a2 + 1)

Solution:
(i) f(x) = x2 – 2x – 8
We have,
f(x) = x2 – 2x – 8
=  x2 – 4x + 2x – 8
=  x (x – 4) + 2(x – 4)
=  (x + 2)(x – 4)
Zeroes of the polynomials are – 2 and 4.
Now,

Hence, the relationship is verified.

(ii)  g(s) = 4s2 – 4s + 1
We have,
g(s) = 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1

= 2s(2s – 1)− 1(2s – 1)
= (2s – 1)(2s – 1)
Zeroes of the polynomials are 1/2 and 1/2.

Hence, the relationship is verified.

(iii) 6s2 − 3 − 7x
= 6s2 − 7x − 3 = (3x + 11) (2x – 3)
Zeros of the polynomials are 3/2 and (-1)/3

Hence, the relationship is verified.

(iv) h(t) = t2 – 15
We have,

Zeroes of the polynomials are - √15 and √15
Sum of the zeroes = 0 - √15 + √15 = 0
0 = 0

Hence, the relationship verified.

(v) p(x) = x2 + 2√2 x – 6
We have,
p(x) = x2 + 22 - 6
= x2 + 3√2x + 3√2x - 6
= x(x + 3√2) - √2(x + 3√2)
= (x + 3√2)(x - √2)
Zeroes of the polynomials are 3√2 and –3√2 Sum of the zeroes

-  6 = – 6
Hence, the relationship is verified.

(vi) q(x) = √3 x2 + 10x + 7√3

q(x) = √3x2 + 10x + 7√3

= 3x2 + 7x + 3x + 7√3

= 3x(x+√3)7(x+√3)

= (x + √3)(7 + √3x)
Zeros of the polynomials are -√3 and -7/√3

Product of the polynomials are - √3, 7/√3
7 = 7
Hence, the relationship is verified.

Zeros of the polynomials are 1 and √3

Hence, the relationship is verified

(viii) g(x) = a[(x2 + 1)–  x(a2 + 1)]2
= ax2 + a − a2x − x
= ax2 − [(a2x + 1)] + a
= ax2 − a2x – x + a
= ax(x − a) − 1(x – a) = (x – a)(ax – 1)
Zeros of the polynomials are 1/a and 1 Sum of the zeros

Product of zeros = a/a

Hence, the relationship is verified.

Question: 2
If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of
Solution:
We have,
α and β are the roots of the quadratic polynomial.
f(x) = x2 – 5x + 4
Sum of the roots = α + β = 5
Product of the roots = αβ = 4
So,

Question: 3
If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of
Solution:

Since, α and β are the zeroes of the quadratic polynomial.
p(y) = x2 – 5x + 4
Sum of the zeroes = α + β = 5
Product of the roots = αβ = 4
So,

Question: 4
If α and β are the zeroes of the quadratic polynomial p (y) = 5y2 – 7y + 1, find the value of
Solution:

Since, α and β are the zeroes of the quadratic polynomial.
p(y) = 5y2 – 7y + 1
Sum of the zeroes = α + β = 7
Product of the roots = αβ = 1
So,

Question: 5
If α and β are the zeroes of the quadratic polynomial f(x) = x2 – x – 4, find the value of
Solution:

Since, α and β are the zeroes of the quadratic polynomial.
We have,
f(x) = x2 – x – 4
Sum of zeroes = α + β = 1
Product of the zeroes = αβ = - 4
So,

Question: 6
If α and β are the zeroes of the quadratic polynomial f(x) = x2 + x – 2, find the value of

Solution:

Since, α and β are the zeroes of the quadratic polynomial.
We have,
f(x) = x2 + x – 2
Sum of zeroes = α + β = 1
Product of the zeroes = αβ = – 2
So,

Question: 7
If one of the zero of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is negative of the other, then find the value of k.
Solution:

Let, the two zeroes of the polynomial f(x) = 4x2 – 8kx – 9 be α and − α.
Product of the zeroes = α × − α = – 9
Sum of the zeroes = α + (− α) = – 8k = 0
Since, α – α = 0
⇒ 8k = 0 ⇒ k = 0

Question: 8
If the sum of the zeroes of the quadratic polynomial f(t) = kt2 + 2t + 3k is equal to their product, then find the value of k.
Solution:

Let the two zeroes of the polynomial f(t) = kt2 + 2t + 3k be α and β.
Sum of the zeroes = α + β = 2
Product of the zeroes = α × β = 3k
Now,

Question: 9
If α and β are the zeroes of the quadratic polynomial p(x) = 4x2 – 5x – 1, find the value of α2β + αβ2.

Solution:

Since, α and β are the zeroes of the quadratic polynomial p(x) = 4x2 – 5x – 1
So, Sum of the zeroes α + β = 5/4
Product of the zeroes  α × β = – ¼
Now,
α2β + αβ2 = αβ (α + β)

Question: 10
If α and β are the zeroes of the quadratic polynomial
f(t) = t2 – 4t + 3, find the value of α4β3 + α3β4.
Solution:

Since, α and β are the zeroes of the quadratic polynomial f(t) = t2 – 4t + 3
So, Sum of the zeroes = α + β = 4
Product of the zeroes = α × β = 3
Now,
α4β3 + α3β4 = α3β3(α + β)
= (3)3(4) = 108

Question: 11
If α and β are the zeroes of the quadratic polynomial
f(x) = 6x2 + x – 2, find the value of
Solution:

Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
Sum of the zeroes = α + β = -⅙
Product of the zeroes =α × β = -⅓
Now,

By substitution the values of the sum of zeroes and products of the zeroes, we will get
= - 25/12

Question: 12
If α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of
Solution:

Since, α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
Sum of the zeroes = α + β = 6/3
Product of the zeroes = α × β = 4/3
Now,

By substituting the values of sum and product of the zeroes, we will get

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 the two zeroes of the polynomial be αand β.
We have,
f(x) = x2 + px + 45
Now,
Sum of the zeroes =  α + β = – p
Product of the zeroes =  α × β = 45
So,

Thus, in the given equation, 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

Solution:

Since, α and β are the roots of the quadratic polynomial given in the question.
f(x) = x2 – px + q
Now,
Sum of the zeroes = p = α + β
Product of the zeroes = q = α × β

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:

Since, 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)
So,
(α + 1)(β + 1)
= αβ + α + β + 1
= αβ + (α + β) + 1
= (− p – c) + p + 1
= 1 – c = RHS
So, 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:

We have,
α + β = 24         …… E-1
α – β = 8              …. E-2
By solving the above two equations accordingly, we will get
2α = 32 α = 16
Substitute the value of α, in any of the equation. Let we substitute it in E-2, 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 is-K
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
Solution:

We have,
f(x) = x2 – 1
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = – 1
From the question,

Sum of the zeroes of the new polynomial

{By substituting the value of the sum and products of the zeroes}
As given in the question,
Product of the zeroes

Hence, the quadratic polynomial is
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

Solution:
We have,
f(x) = x2 – 3x – 2
Sum of the zeroes =  α + β = 3
Product of the zeroes = αβ = – 2
From the question,
Sum of the zeroes of the new polynomial

So, the quadratic polynomial is,
x2- (sum of the zeroes)x + (product of the zeroes)
x2 - (sum of the zeroes)x + (product of the zeroes)
Hence, the required quadratic polynomial is k

Hence, the required quadratic polynomial is k

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:
We have,
f(x) = x2 + px + q
Sum of the zeroes = α + β = -p
Product of the zeroes = αβ = q
From the question,
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 (α – β)2
= (- p)2((- p)2 - 4q)
= p2 (p2–4q)
So, the quadratic polynomial is,
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)
Solution:
We have,
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
So, quadratic polynomial is:
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)
f(x) = k(x2 – 6x + 11)
(ii) Sum of the zeroes of new polynomial

Product of the zeroes of new polynomial

So, the quadratic polynomial is,
x2 – (sum of the zeroes)x + (product of the zeroes)

Thus, the required quadratic polynomial is f(x) = k(x2 – 23x + 13)

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

(iv) α2β + αβ2
(v) α4 + β4

Solution:

f(x) = ax2 + bx + c
Here,
Sum of the zeroes of polynomial = α + β = -b/a
Product of zeroes of polynomial = αβ = c/a
Since, α + β are the roots (or) zeroes of the given polynomial, so
(i) α – β
The two zeroes of the polynomials are -

From the previous question, we know that,

Also,
αβ = c/a
Putting the values in E.1, we will get

Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After substituting it in E-1, we will get

(iv) α2β + αβ2
= αβ(α + β) …….. E- 1.
Since,
Sum of the zeroes of polynomial = α + β = – b/ a
Product of zeroes of polynomial = αβ = c/a
After substituting it in E-1, we will get

(v)  α4 + β4
= (α2 + β2)2 – 2α2β2
= ((α + β)2 – 2αβ)2 – (2αβ)2 ……. E- 1
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After substituting it in E-1, we will get

Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After substituting it, we will get
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After substituting it, we will get

Since,
Sum of the zeroes of polynomial= α + β = – b/a
Product of zeroes of polynomial= αβ = c/a
After substituting it, we will get

The document RD Sharma Solutions: Polynomials (Exercise 2.1) Notes | Study Mathematics (Maths) Class 10 - Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

Mathematics (Maths) Class 10

53 videos|395 docs|138 tests
 Use Code STAYHOME200 and get INR 200 additional OFF

Mathematics (Maths) Class 10

53 videos|395 docs|138 tests

Download free EduRev App

Track your progress, build streaks, highlight & save important lessons and more!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;