RD Sharma Maths Solutions Class 10

RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two
Variables Ex 3.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.10
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

Question 1.
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs. 3, and a game of Hoopla costs Rs. 4. If she spend Rs. 20 in the fair, represent this situation algebraically and graphically.
Solution:
Let number of rides on the wheel = x
and number of play of Hoopla = y
According to the given conditions x = 2y ⇒ x – 2y = 0 ….(i)
and cost of ride on wheel at the rate of Rs. 3 = 3x
and cost on Hoopla = 4y
and total cost = Rs. 20
3x + 4y = 20 ….(ii)
Now we shall solve these linear equations graphically as under
We take three points of each line and join them to get a line in each case the point of intersection will be the solution
From equation (i)
x = 2y

X	4	0	6
y	2	0	3

y = 2, then x = 2 x 2 = 4
y = 0, then x = 2 x 0 = 0
y = 3, then x = 2 x 3 = 6
Now, we plot these points on the graphs and join them to get a line
Similarly in equation (ii)
3x + 4y = 20 ⇒ 3x = 20 – 4y

Now we plot these points and get another line by joining them
These two lines intersect eachother at the point (4, 2)
Its solution is (4, 2)
Which is a unique Hence x = 4, y = 2

Question 2.
Aftab tells his daughter, “Seven years ago, I w as seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Is not this interesting ? Represent this situation algebraically and graphically.
Solution:
Seven years ago
Let age of Aftab’s daughter = x years
and age of Aftab = y years
and 3 years later
Age of daughter = x + 10 years
and age of Aftab = y + 10 years
According to the conditions,
y = 7x ⇒ 7x – y = 0 ……….(i)
y + 10 = 3 (x + 10)
=> y + 10 = 3x + 30
3x – y = 10 – 30 = -20
3x – y = -20 ….(ii)
Equations are
7x – y = 0
3 x – y = -20
Now we shall solve these linear equations graphically as under
7x – y = 0 ⇒ y = 7x

X	0	1	-1
y	0	7	-7

If x = 0, y = 7 x 0 = 0
If x = 1, y = 7 x 1=7
If x = -1, y = 7 x (-1) = -7
Now plot these points on the graph and join
then
3x – y = -20
y = 3x + 20

X	-1	-2	-3
y	17	14	11

If x = -1, y = 3 x (-1) + 20 = -3 + 20= 17
If x = -2, y = 3 (-2) + 20 = -6 + 20 = 14
If x = -3, y = 3 (-3) + 20 = -9 + 20= 11
Now plot the points on the graph and join them we see that lines well meet at a point on producing at (5, 35).

Question 3.
The path of a train A is given by the equation 3x + 4y – 12 = 0 and the path of another train B is given by the equation 6x + 8y – 48 = 0. Represent this situation graphically.
Solution:
Path of A train is 3x + 4y – 12 = 0
and path of B train is 6x + 8y – 48 = 0
Graphically, we shall represent these on the graph as given under 3x + 4y- 12 = 0

Question 4.
Gloria is walking along the path joining (-2, 3) and (2, -2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.
Solution:
Plot the points (-2, 3) and (2, -2) and join them to get a line
and also plot the points (0, 5), (4, 0) and joint them to get another line as shown on the graph
We see that these two lines are parallel to each other

Question 5.
On comparing the ratios , and without drawing them, find out whether the lines representing following pairs of linear equations intersect at a point, are parallel or coincide :
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
(ii) 9x + 3y +12 = 0
18x + 6y + 24 = 0
(iii) 6x – 3y +10 = 0
2x – y + 9 = 0
Solution:

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Solution:

Question 7.
The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4kg of apples and 2kg of grapes is Rs. 300. Represent the situation algebraically and geo-metrically.
Solution:
Let cost of 1kg of apples = Rs. x
and cost of 1kg of grapes = Rs. y
Now according to the condition, the system of equation will be
2x + y = 160
4x + 2y = 300
Now 2x + y = 160
y = 160 – 2x

X	20	40	60
y	120	80	40

If x = 20, then y = 160 – 2 x 20 = 160 – 40 = 120
If x = 40, then y = 160 – 2 x 40 = 160 – 80 = 80
If x = 60, then y = 160 – 2 x 60 = 160 – 120 = 40
Now plot the points and join them and 4x + 2y = 300
=> 2x + y = 150
=> y = 150 – 2x

X	40	50	60
y	70	50	30

If x = 40, then y = 150 – 2 x 40 = 150 – 80 = 70
If x = 50, then y = 150 – 2 x 50 = 150 – 100 = 50
If x = 60, then y = 150 – 2 x 60 = 150 – 120 = 30
Now plot the points and join them We see that these two lines are parallel

 

Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3
• RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS
• RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

Answer each of the following questions in one word or one sentence or as per the exact requirement of the questions :
Question 1.
Define a polynomial with real co-efficients.
Solution:

Question 2.
Define degree of a polynomial.
Solution:
The exponent of the highest degree term in a polynomial is known as its degree. A polynomial of degree O is called a constant polynomial.

Question 3.
Write the standard form of a linear polynomial with real co-efficients.
Solution:
ax + b is the standard form of a linear polynomial with real co-efficients and a ≠ 0

Question 4.
Write the standard form of a quadratic polynomial with real co-efficients.
Solution:
ax² + bx + c is a standard form of quadratic polynomial with real co-efficients and a ≠ 0.

Question 5.
Write the standard form of a cubic polynomial with real co-efficients.
Solution:
ax³ + bx² + cx + d is a standard form of cubic polynomial with real co-efficients and a ≠ 0.

Question 6.
Define value of a polynomial at a point.
Solution:
If f(x) is a polynomial and a is any real number then the real number obtained by replacing x by α in f(x) is called the value of f(x) at x = α and is denoted by f(α).

Question 7.
Define zero of a polynomial.
Solution:
A real number a is a zero of a polynomial f(x) if f(α) = 0.

Question 8.
The sum and product of the zeros of a quadratic polynomial are – \(\frac { 1 }{ 2 }\) and -3 respectively. What is the quadratic polynomial ?
Solution:

Question 9.
Write the family of quadratic polynomials having – \(\frac { 1 }{ 4 }\) and 1 as its zeros.
Solution:

Question 10.
If the product of zeros of the quadratic polynomial f(x) = x² – 4x + k is 3, find the value of k.
Solution:
We know that a quadratic polynomial x² – (sum of zeros) x + product of zeros
In the given polynomial f(x) = x² – 4x + k is the product of zeros which is equal to 3
k = 3

Question 11.
If the sum of the zeros of a quadratic polynomial f(x) = kx² – 3x + 5 is 1, write the value of k.
Solution:
f (x) = kx² – 3x + 5
Here a = k, b = -3, c = 5

Question 12.
In the figure, the graph of a polynomial p (x) is given. Find the zeros of the polynomial.

Solution:
The graph of the given polynomial meets the x-axis at -1 and -3
Zero will be -1 and -3
Zero of polynomial is 3

Question 13.
The graph of a polynomial y = f(x) is given below. Find the number of real zeros of f (x).

Solution:
The curve touches x-axis at one point and also intersects at one point So number of zeros will be 3, two equal and one distinct

Question 14.
The graph of the polynomial f(x) = ax² + bx + c is as shown below (in the figure) write the signs of ‘a’ and b² – 4ac.

Solution:
The shape of parabola is up word a > 0
and b² – 4ac >0 i.e., both are positive.

Question 15.
The graph of the polynomial f(x) = ax² + bx + c is as shown in the figure write the value of b² – 4ac and the number of real zeros of f(x).

Solution:
The curve parabola touches the x-axis at one point
It has two equal zeros
b² – 4ac = 0

Question 16.
In Q. No. 14, write the sign of c
Solution:
The mouth of parabola is upward and intersect y-axis above x-axis
c > 0

Question 17.
In Q. No. 15, write the sign of c.
Solution:
The mouth of parabola is downward and intersects y-axis below x-axis
c < 0

Question 18.
The graph of a polynomial f (x) is as shown in the figure. Write the number of real zeros of f (x).

Solution:
The curves touches the x-axis at two distinct point
It has a pair of two equal zeros i.e., it has 4 real zeros

Question 19.
If x = 1, is a zero of the polynomial f(x) = x³ – 2x² + 4x + k, write the value of k.
Solution:

Question 20.
State division algorithm for polynomials.
Solution:
If f(x) is a polynomial and g (x) is a non zero polynomial, there exist two polynomials q (x) and r (x) such that
f(x) = g (x) x q (x) + r (x)
where r (x) = 0 or degree r (x) < degree g (x)
This is called division algorithm

Question 21.
Give an example of polynomials f(x), g (x), q (x) and r (x) satisfying f(x) = g (x) . q (x) + r (x), where degree r (x) = 0.
Solution:
f (x) = x³ + x² + x + 4
g (x) = x + 1
q (x) = x² + 1
r (x) = 3
is an example of f (x) = g (x) x q (x) + r (x)
where degree of r (x) is zero.

Question 22.
Write a quadratic polynomial, sum of whose zeros is 2√3 and their product is 2.
Solution:
Sum of zeros = 2 √3
and product of zeros = 2
Quadratic polynomial will be f (x) = x2 – (sum of zeros) x + product of zeros
= x² – 2 √3 x + 2

Question 23.
If fourth degree polynomial is divided by a quadratic polynomial, write the degree of the remainder.
Solution:
Degree of the given polynomial = 4
and degree of divisor = 2
Degree of quotient will be 4 – 2 = 2
and degree of remainder will be less than 2 In other words equal to or less than one degree

Question 24.
If f(x) = x³ + x² – ax + b is divisible by x² – x, write the value of a and b.
Solution:

Question 25.
If a – b, a and a + b are zeros of the polynomial f(x) = 2x³ – 6x² + 5x – 7, write the value of a.
Solution:

Question 26.
Write the coefficients of the polynomial p (z) = z⁵ – 2z² + 4.
Solution:
p (z) = z⁵ + oz⁴ + oz³ – 2z² + oz + 4
Coefficient of z⁵ = 1
Coefficient of z⁴ = 0
Coefficient of z³ = 0
Coefficient of z² = – 2
Coefficient of z = 0
Constant = 4

Question 27.
Write the zeros of the polynomial x² – x – 6. (C.B.S.E. 2008)
Solution:

Question 28.
If (x + a) is a factor of 2x² + 2ax + 5x + 10, find a. (C.B.S.E. 2008)
Solution:
x + a is a factor of

Question 29.
For what value of k, -4 is a zero of the polynomial x² – x – (2k + 2) ? (CBSE 2009)
Solution:

Question 30.
If 1 is a zero of the polynomial p (x) = ax² – 3 (a – 1) x – 1, then find the value of a.
Solution:

Question 31.
If α, β are the zeros of a polynomial such that α + β = -6 and α β = -4, then write the polynomial. [CBSE 2010]
Solution:

Question 32.
If α, β are the zeros of the polynomial 2y² + 7y + 5, write the value of α + β + αβ. [CBSE 2010]
Solution:

Question 33.
For what value of k, is 3 a zero of the polynomial 2x² + x + k ? [CBSE 2010]
Solution:
3 is a zero of f(x) = 2x² + x + k
It will satisfy the polynomial
f(x) = 0 ⇒ f(3) = 0
Now 2x² + x + k = 0
=> 2 (3)² + 3 + k = 0
=> 18 + 3 + k = 0
=> 21 + k = 0
=> k = -21

Question 34.
For what value of k, is -3 a zero of the polynomial x² + 11x + k ? [CBSE 2010]
Solution:
-3 is a zero of polynomial f(x) = x² + 11x + k
It will satisfy the polynomial
f (x) = 0 => f(-3) = 0
Now x² + 11x + k = 0
=> (-3)²+ 11 x (-3) + k = 0
⇒ 9 – 33 + k = 0
⇒ -24 + k = 0
⇒ k = 24

Question 35.
For what value of k, is -2 a zero of the polynomial 3x² + 4x + 2k ? [CBSE 2010]
Solution:
-2 is a zero of the polynomial
f(x) = 3x² + 4x + 2k
f(-2) = 0
=> 3 (-2)² + 4 (-2) + 2k = 0
=> 12 – 8 + 2k = 0
=> 4 + 2k = 0
=> 2k = -4
=> k = -2

Question 36.
If a quadratic polynomial f(x) is factorizable into linear distinct factors, then what is the total number of real and distinct zeros of f (x) ?
Solution:
In a quadratic polynomial f(x) its degree is 2 and it can be factorised in to two distinct linear factors.
f(x) has two distinct zeros

Question 37.
If a quadratic polynomiaI f(x) is a square of a linear polynomial, then its two zeros are coincident. (True / False)
Solution:
In a quadratic polynomial f(x), it is the square of a linear polynomial It has two zeros which are equal i.e. coincident
It is true

Question 38.
If a quadratic polynomial f(x) is not factorizable into linear factors, then it has no real zero. (True / False)
Solution:
A quadratic polynomial f(x) is not factorised into linear factors It has no real zeros It is true

Question 39.
If f(x) is a polynomial such that f(a) f(b) < 0, then what is the number of zeros lying between a and b ?
Solution:
f(x) is a polynomial such that f(a) f(b) < 0
At least one of its zeros will be between a and b

Question 40.
If graph of quadratic polynomial ax² + bx + c cuts positive direction of y-axis, then what is the sign of c ?
Solution:
The graph of quadratic polynomial ax² + bx + c cuts positive direction of y-axis Then sign of constant term c will be also positive.

Question 41.
If the graph of quadratic polynomial ax² + bx + c cuts negative direction of y-axis, then what is the sigh of c ?
Solution:
The graph of quadratic polynomial ax² + bx + c cuts negative side of y-axis
Then sign of constant term c will be negative

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3

Other Exercises

• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3
• RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS
• RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

Question 1.
Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following :

Solution:

Question 2.
Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm.

Solution:

Question 3.
Obtain all zeros of the polynomial f(x) = 2x⁴ + x³ – 14x² – 19x – 6, if two of its zeros are -2 and -1.
Solution:

Question 4.
Obtain all zeros of f(x) = x³ + 13x² + 32x + 20, if one of its zeros is -2.
Solution:
f(x) = x³ + 13x² + 32x + 20
One zero = -2 or x = -2
x + 2 is a factor of f (x)
Now dividing f(x) by x + 2, we get

x + 1 = 0 => x = -1
and x + 10 = 0
=> x = -10
-1 and -10
Hence zeros are -10, -1, -2

Question 5.
Obtain all zeros of the polynomial f(x) = x⁴ – 3x³ – x² + 9x – 6, if two of its zeros are – √3 and √3
Solution:

Question 6.
Find all zeros of the polynomial f(x) = 2x⁴ – 2x³ – 7x² + 3x + 6, if its two zeros are \(\surd \frac { 3 }{ 2 }\) and – \(\surd \frac { 3 }{ 2 }\)
Solution:

Question 7.
Find all the zeros of the polynomial x⁴ + x³ – 34x² -4x+ 120, if two of its zeros are 2 and -2. [CBSE 2008]
Solution:

Either x + 6 = 0, then x = -6
or x – 5 = 0, then x = 5
Hence other two zeros are -6, 5
and all zeros are 2, -2, -6, 5

Question 8.
Find all zeros of the polynomial 2x⁴ + 7x³ – 19x² – 14x + 30, if two of its zeros are √2 and -√2
Solution:

Question 9.
Find all the zeros of the polynomial 2x³ + x² – 6x – 3, if two of its zeros are – √3 and √3. (CBSE 2009)
Solution:
Let f(x) = 2x³ + x² – 6x – 3
and two zeros of f(x) are – √3 and √3

Question 10.
Find all the zeros of the polynomial x³ + 3x² – 2x – 6, if two of its zeros are – √2 and √2 (CBSE2009)
Solution:

Question 11.
What must be added to the polynomial f(x) = x⁴ + 2x³ – 2x² + x – 1 so that the resulting polynomial is exactly divisible by x² + 2x – 3 ?
Solution:

Question 12.
What must be subtracted from the polynomial f(x) = x⁴ + 2x³ – 13x² – 12x + 21 so that the resulting polynomial is exactly divisible by x³ – 4x + 3 ?
Solution:

The resulting polynomial is exactly divisible by x² – 4x + 3
Remainder = 0
=> 2x -3 – k = 0
=> k = 2x – 3
(2x – 3) must be subtracted

Question 13.
Given that √2 is a zero of the cubic polynomial 16x³ + √2x² – 10x – 4√2 , find its other two zeroes. [NCERT Exemplar]
Solution:

Question 14.
Given that x – √5 is a factor of the cubic polynomial x³ – 3 √5 x² + 13x – 3 √5 , find all the zeroes of the polynomial. [NCERT Exemplar|
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3
• RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS
• RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

Question 1.
Verify that 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 ; \(\frac { 1 }{ 2 }\) , 1, -2
(ii) g(x) = x³ – 4x² + 5x – 2 ; 2, 1, 1
Solution:

Question 2.
Find a cubic polynomial with the sum, sum of product of its zeros taken two at a time and product of its zeros as 3, -1 and -3 respectively.
Solution:

Question 3.
If the zeros of the polynomial f(x) = 2x³ – 15x² + 37x – 30 are in AP. Find them.
Solution:

Question 4.
Find the condition that the zeros of the polynomial f(x) = x³ + 3px² + 3qx + r may be in AP.
Solution:

Question 5.
If the zeros of the polynomial f(x) = ax³ + 3bx² + 3cx + d are in A.P., prove that 2b³ – 3abc + a²d = 0.
Solution:

Question 6.
If the zeros of the polynomial f(x) = x³ – 12x² + 39x + k are in AP, find the value of k.
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.2
• RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.3
• RD Sharma Class 10 Solutions Chapter 2 Polynomials VSAQS
• RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

Question 1.
Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their co-efficients :

Solution:
(i) f(x) = x² – 2x – 8

Question 2.
For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization.

Solution:
(i) Given that, sum of zeroes (S) = – \(\frac { 8 }{ 3 }\)
and product of zeroes (P) = \(\frac { 4 }{ 3 }\)
Required quadratic expression,

Question 3.
If α and β are the zeros of the quadratic polynomial f(x) = x² – 5x + 4, find the value of \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } -2\alpha \beta\).
Solution:

Question 4.
If α and β are the zeros of the quadratic polynomial p(y) = 5y² – 7y + 1, find the value of \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta }\)
Solution:

Question 5.
If α and β are the zeros of the quadratic polynomial f(x) = x² – x – 4, find the value of \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } -\alpha \beta\)
Solution:

Question 6.
If α and β are the zeros of the quadratic polynomial f(x) = x² + x – 2, find the value of \(\frac { 1 }{ \alpha } -\frac { 1 }{ \beta }\)
Solution:

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

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

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

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

Question 11.
If α and β are the zeros of the quadratic polynomial f (x) = 6x⁴ + x – 2, find the value of \(\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha }\)
Solution:

Question 12.
If α and β are the zeros of the quadratic polynomial p(s) = 3s² – 6s + 4, find the value of \(\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha } +2\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } \right) +3\alpha \beta\)
Solution:

Question 13.
If the squared difference of the zeros of the quadratic polynomial f(x) = x² + px + 45 is equal to 144, find the value of p
Solution:

Question 14.
If α and β are the zeros of the quadratic polynomial f(x) = x² – px + q, prove that:

Solution:

Question 15.
If α and β are the zeros of the quadratic polynomial f(x) = x² – p(x + 1) – c, show that (α + 1) (β + 1) = 1 – c.
Solution:

Question 16.
If α and β are the zeros of the quadratic polynomial such that α + β = 24 and α – β = 8, find a quadratic polynomial having α and β as its zeros.
Solution:

Question 17.
If α and β are the zeros of the quadratic polynomial f(x) = x² – 1, find a quadratic polynomial whose zeros are \(\frac { 2\alpha }{ \beta }\) and \(\frac { 2\beta }{ \alpha }\)
Solution:

Question 18.
If α and β are the zeros of the quadratic polynomial f(x) = x² – 3x – 2, find a quadratic polynomial whose zeros are \(\frac { 1 }{ 2\alpha +\beta }\) and \(\frac { 1 }{ 2\beta +\alpha }\)
Solution:

Question 19.
If α and β are the zeroes of the polynomial f(x) = x² + px + q, form a polynomial whose zeros are (α + β)² and (α – β)².
Solution:

Question 20.
If α and β are the zeros of the quadratic polynomial f(x) = x² – 2x + 3, find a polynomial whose roots are :
(i) α + 2, β + 2
(ii) \(\frac { \alpha -1 }{ \alpha +1 } ,\frac { \beta -1 }{ \beta +1 }\)
Solution:

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


Solution:

 

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials Ex 2.1 are helpful to complete your math homework.

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