RD Sharma Solutions: Polynomials (Exercise 2.1)

# Polynomials (Exercise 2.1) RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download

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
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

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

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)
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
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

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 Polynomials (Exercise 2.1) RD Sharma Solutions | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Polynomials (Exercise 2.1) RD Sharma Solutions - Mathematics (Maths) Class 10

 1. What are polynomials?
Ans. Polynomials are algebraic expressions that consist of variables, coefficients, and exponents, combined using addition, subtraction, multiplication, and non-negative integer exponents. They are widely used in various branches of mathematics and have many applications in real-life problems.
 2. How do you determine the degree of a polynomial?
Ans. The degree of a polynomial is determined by the highest power of the variable present in the polynomial. For example, if the polynomial is 3x^2 + 2x + 1, the degree is 2 because the highest power of x is 2. The degree helps in understanding the behavior and properties of the polynomial.
 3. Can a polynomial have more than one variable?
Ans. Yes, a polynomial can have more than one variable. Polynomials with more than one variable are called multivariate polynomials. For example, 3x^2y + 2xy + 1 is a polynomial with two variables, x and y. The degree of a multivariate polynomial is determined by the sum of the exponents of all variables in any term with the highest degree.
 4. What are the different types of polynomials based on the number of terms?
Ans. Polynomials can be classified based on the number of terms they have. Some common types of polynomials include: - Monomial: A polynomial with only one term, such as 5x^2. - Binomial: A polynomial with two terms, such as 3x + 2. - Trinomial: A polynomial with three terms, such as 4x^2 + 2x + 1. - Polynomial: A polynomial with more than three terms.
 5. How can polynomials be used to solve real-life problems?
Ans. Polynomials can be used to model and solve various real-life problems. They are often used in physics, engineering, economics, and many other fields. For example, polynomials can be used to represent and analyze the growth of populations, the trajectory of projectiles, the behavior of electrical circuits, and the optimization of resources. By converting real-life problems into polynomial equations, we can apply mathematical techniques to find solutions and make predictions.

## Mathematics (Maths) Class 10

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