NCERT Textbook - Quadratic Equation Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Created by: Indu Gupta

Class 10 : NCERT Textbook - Quadratic Equation Class 10 Notes | EduRev

 Page 1


70 MATHEMA TICS
4
4.1 Introduction
In Chapter 2, you have studied different types of polynomials. One type was the
quadratic polynomial of the form ax
2
 + bx + c, a ? 0. When we equate this polynomial
to zero, we get a quadratic equation. Quadratic equations come up when we deal with
many real-life situations. For instance, suppose a
charity trust decides to build a prayer hall having
a carpet area of 300 square metres with its length
one metre more than twice its breadth. What
should be the length and breadth of the hall?
Suppose the breadth of the hall is x metres. Then,
its length should be (2x + 1) metres. W e can depict
this information pictorially as shown in Fig. 4.1.
Now, area of the hall = (2x + 1). x m
2
 = (2x
2
 + x) m
2
So, 2x
2
 + x = 300 (Given)
Therefore, 2x
2
 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x
2
 + x – 300 = 0 which is a
quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum
and a given positive product, and this problem is equivalent to solving a quadratic
equation of the form x
2
 – px + q = 0. Greek mathematician Euclid developed a
geometrical approach for finding out lengths which, in our present day terminology,
are solutions of quadratic equations. Solving of quadratic equations, in general form, is
often credited to ancient Indian mathematicians. In fact, Brahmagupta (A.D.598–665)
gave an explicit formula to solve a quadratic equation of the form ax
2
 + bx = c. Later,
QUADRA TIC EQUA TIONS
Fig. 4.1
Page 2


70 MATHEMA TICS
4
4.1 Introduction
In Chapter 2, you have studied different types of polynomials. One type was the
quadratic polynomial of the form ax
2
 + bx + c, a ? 0. When we equate this polynomial
to zero, we get a quadratic equation. Quadratic equations come up when we deal with
many real-life situations. For instance, suppose a
charity trust decides to build a prayer hall having
a carpet area of 300 square metres with its length
one metre more than twice its breadth. What
should be the length and breadth of the hall?
Suppose the breadth of the hall is x metres. Then,
its length should be (2x + 1) metres. W e can depict
this information pictorially as shown in Fig. 4.1.
Now, area of the hall = (2x + 1). x m
2
 = (2x
2
 + x) m
2
So, 2x
2
 + x = 300 (Given)
Therefore, 2x
2
 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x
2
 + x – 300 = 0 which is a
quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum
and a given positive product, and this problem is equivalent to solving a quadratic
equation of the form x
2
 – px + q = 0. Greek mathematician Euclid developed a
geometrical approach for finding out lengths which, in our present day terminology,
are solutions of quadratic equations. Solving of quadratic equations, in general form, is
often credited to ancient Indian mathematicians. In fact, Brahmagupta (A.D.598–665)
gave an explicit formula to solve a quadratic equation of the form ax
2
 + bx = c. Later,
QUADRA TIC EQUA TIONS
Fig. 4.1
QUADRA TIC EQUA TIONS 71
Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula,
(as quoted by Bhaskara II) for solving a quadratic equation by the method of completing
the square. An  Arab mathematician Al-Khwarizmi (about A.D. 800) also studied
quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book
‘Liber embadorum’ published in Europe in A.D. 1145 gave complete solutions of
different quadratic equations.
In this chapter, you will study quadratic equations, and various ways of finding
their roots. You will also see some applications of quadratic equations in daily life
situations.
4.2 Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax
2
 + bx + c = 0, where
a, b, c are real numbers, a ? 0. For example, 2x
2
 + x – 300 = 0 is a quadratic equation.
Similarly, 2x
2
 – 3x + 1 = 0, 4x – 3x
2
 + 2 = 0 and 1 – x
2
 + 300 = 0 are also quadratic
equations.
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree
2, is a quadratic equation. But when we write the terms of p(x) in descending order of
their degrees, then we get the standard form of the equation. That is, ax
2
 + bx + c = 0,
a ? 0 is called the standard form of a quadratic equation.
Quadratic equations arise in several situations in the world around us and in
different fields of mathematics. Let us consider a few examples.
Example 1 : Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and
the product of the number of marbles they now have is 124. We would like to find
out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of
production of each toy (in rupees) was found to be 55 minus the number of toys
produced in a day. On a particular day, the total cost of production was
Rs 750. We would like to find out the number of toys produced on that day.
Solution :
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
= 40 – x
Page 3


70 MATHEMA TICS
4
4.1 Introduction
In Chapter 2, you have studied different types of polynomials. One type was the
quadratic polynomial of the form ax
2
 + bx + c, a ? 0. When we equate this polynomial
to zero, we get a quadratic equation. Quadratic equations come up when we deal with
many real-life situations. For instance, suppose a
charity trust decides to build a prayer hall having
a carpet area of 300 square metres with its length
one metre more than twice its breadth. What
should be the length and breadth of the hall?
Suppose the breadth of the hall is x metres. Then,
its length should be (2x + 1) metres. W e can depict
this information pictorially as shown in Fig. 4.1.
Now, area of the hall = (2x + 1). x m
2
 = (2x
2
 + x) m
2
So, 2x
2
 + x = 300 (Given)
Therefore, 2x
2
 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x
2
 + x – 300 = 0 which is a
quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum
and a given positive product, and this problem is equivalent to solving a quadratic
equation of the form x
2
 – px + q = 0. Greek mathematician Euclid developed a
geometrical approach for finding out lengths which, in our present day terminology,
are solutions of quadratic equations. Solving of quadratic equations, in general form, is
often credited to ancient Indian mathematicians. In fact, Brahmagupta (A.D.598–665)
gave an explicit formula to solve a quadratic equation of the form ax
2
 + bx = c. Later,
QUADRA TIC EQUA TIONS
Fig. 4.1
QUADRA TIC EQUA TIONS 71
Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula,
(as quoted by Bhaskara II) for solving a quadratic equation by the method of completing
the square. An  Arab mathematician Al-Khwarizmi (about A.D. 800) also studied
quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book
‘Liber embadorum’ published in Europe in A.D. 1145 gave complete solutions of
different quadratic equations.
In this chapter, you will study quadratic equations, and various ways of finding
their roots. You will also see some applications of quadratic equations in daily life
situations.
4.2 Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax
2
 + bx + c = 0, where
a, b, c are real numbers, a ? 0. For example, 2x
2
 + x – 300 = 0 is a quadratic equation.
Similarly, 2x
2
 – 3x + 1 = 0, 4x – 3x
2
 + 2 = 0 and 1 – x
2
 + 300 = 0 are also quadratic
equations.
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree
2, is a quadratic equation. But when we write the terms of p(x) in descending order of
their degrees, then we get the standard form of the equation. That is, ax
2
 + bx + c = 0,
a ? 0 is called the standard form of a quadratic equation.
Quadratic equations arise in several situations in the world around us and in
different fields of mathematics. Let us consider a few examples.
Example 1 : Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and
the product of the number of marbles they now have is 124. We would like to find
out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of
production of each toy (in rupees) was found to be 55 minus the number of toys
produced in a day. On a particular day, the total cost of production was
Rs 750. We would like to find out the number of toys produced on that day.
Solution :
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
= 40 – x
72 MATHEMA TICS
Therefore, their product = (x – 5) (40 – x)
=40x – x
2
 – 200 + 5x
=– x
2
 + 45x – 200
So, – x
2
 + 45x – 200 = 124 (Given that product = 124)
i.e., – x
2
 + 45x – 324 = 0
i.e., x
2
 – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x
2
 – 45x + 324 = 0
which is the required representation of the problem mathematically.
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
Therefore, x (55 – x) = 750
i.e., 55x – x
2
 = 750
i.e., – x
2
 + 55x – 750 = 0
i.e., x
2
 – 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
x
2
 – 55x + 750 = 0
which is the required representation of the problem mathematically.
Example 2 : Check whether the following are quadratic equations:
(i) (x – 2)
2
 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2)
(iii) x (2x + 3) = x
2
 + 1 (iv) (x + 2)
3
 = x
3
 – 4
Solution :
(i) LHS = (x – 2)
2
 + 1 = x
2
 – 4x + 4 + 1 = x
2
 – 4x + 5
Therefore, (x – 2)
2 
+
 
1 = 2x – 3 can be rewritten as
x
2
 – 4x + 5 = 2x – 3
i.e., x
2
 – 6x + 8 = 0
It is of the form ax
2
 + bx + c = 0.
Therefore, the given equation is a quadratic equation.
Page 4


70 MATHEMA TICS
4
4.1 Introduction
In Chapter 2, you have studied different types of polynomials. One type was the
quadratic polynomial of the form ax
2
 + bx + c, a ? 0. When we equate this polynomial
to zero, we get a quadratic equation. Quadratic equations come up when we deal with
many real-life situations. For instance, suppose a
charity trust decides to build a prayer hall having
a carpet area of 300 square metres with its length
one metre more than twice its breadth. What
should be the length and breadth of the hall?
Suppose the breadth of the hall is x metres. Then,
its length should be (2x + 1) metres. W e can depict
this information pictorially as shown in Fig. 4.1.
Now, area of the hall = (2x + 1). x m
2
 = (2x
2
 + x) m
2
So, 2x
2
 + x = 300 (Given)
Therefore, 2x
2
 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x
2
 + x – 300 = 0 which is a
quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum
and a given positive product, and this problem is equivalent to solving a quadratic
equation of the form x
2
 – px + q = 0. Greek mathematician Euclid developed a
geometrical approach for finding out lengths which, in our present day terminology,
are solutions of quadratic equations. Solving of quadratic equations, in general form, is
often credited to ancient Indian mathematicians. In fact, Brahmagupta (A.D.598–665)
gave an explicit formula to solve a quadratic equation of the form ax
2
 + bx = c. Later,
QUADRA TIC EQUA TIONS
Fig. 4.1
QUADRA TIC EQUA TIONS 71
Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula,
(as quoted by Bhaskara II) for solving a quadratic equation by the method of completing
the square. An  Arab mathematician Al-Khwarizmi (about A.D. 800) also studied
quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book
‘Liber embadorum’ published in Europe in A.D. 1145 gave complete solutions of
different quadratic equations.
In this chapter, you will study quadratic equations, and various ways of finding
their roots. You will also see some applications of quadratic equations in daily life
situations.
4.2 Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax
2
 + bx + c = 0, where
a, b, c are real numbers, a ? 0. For example, 2x
2
 + x – 300 = 0 is a quadratic equation.
Similarly, 2x
2
 – 3x + 1 = 0, 4x – 3x
2
 + 2 = 0 and 1 – x
2
 + 300 = 0 are also quadratic
equations.
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree
2, is a quadratic equation. But when we write the terms of p(x) in descending order of
their degrees, then we get the standard form of the equation. That is, ax
2
 + bx + c = 0,
a ? 0 is called the standard form of a quadratic equation.
Quadratic equations arise in several situations in the world around us and in
different fields of mathematics. Let us consider a few examples.
Example 1 : Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and
the product of the number of marbles they now have is 124. We would like to find
out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of
production of each toy (in rupees) was found to be 55 minus the number of toys
produced in a day. On a particular day, the total cost of production was
Rs 750. We would like to find out the number of toys produced on that day.
Solution :
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
= 40 – x
72 MATHEMA TICS
Therefore, their product = (x – 5) (40 – x)
=40x – x
2
 – 200 + 5x
=– x
2
 + 45x – 200
So, – x
2
 + 45x – 200 = 124 (Given that product = 124)
i.e., – x
2
 + 45x – 324 = 0
i.e., x
2
 – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x
2
 – 45x + 324 = 0
which is the required representation of the problem mathematically.
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
Therefore, x (55 – x) = 750
i.e., 55x – x
2
 = 750
i.e., – x
2
 + 55x – 750 = 0
i.e., x
2
 – 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
x
2
 – 55x + 750 = 0
which is the required representation of the problem mathematically.
Example 2 : Check whether the following are quadratic equations:
(i) (x – 2)
2
 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2)
(iii) x (2x + 3) = x
2
 + 1 (iv) (x + 2)
3
 = x
3
 – 4
Solution :
(i) LHS = (x – 2)
2
 + 1 = x
2
 – 4x + 4 + 1 = x
2
 – 4x + 5
Therefore, (x – 2)
2 
+
 
1 = 2x – 3 can be rewritten as
x
2
 – 4x + 5 = 2x – 3
i.e., x
2
 – 6x + 8 = 0
It is of the form ax
2
 + bx + c = 0.
Therefore, the given equation is a quadratic equation.
QUADRA TIC EQUA TIONS 73
(ii) Since x(x + 1) + 8 = x
2
 + x + 8 and (x + 2)(x – 2) = x
2
 – 4
Therefore, x
2
 + x + 8 = x
2
 – 4
i.e., x + 12 = 0
It is not of the form ax
2
 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(iii) Here, LHS = x (2x + 3) = 2x
2
 + 3x
So, x (2x + 3) = x
2
 + 1 can be rewritten as
2x
2
 + 3x = x
2
 + 1
Therefore, we get x
2
 + 3x – 1 = 0
It is of the form ax
2
 + bx + c = 0.
So, the given equation is a quadratic equation.
(iv) Here, LHS = (x + 2)
3
 = x
3
 + 6x
2
 + 12x + 8
Therefore, (x + 2)
3
 = x
3
 – 4 can be rewritten as
x
3
 + 6x
2
 + 12x + 8 = x
3
 – 4
i.e., 6x
2
 + 12x + 12 = 0 or, x
2
 + 2x + 2 = 0
It is of the form ax
2
 + bx + c = 0.
So, the given equation is a quadratic equation.
Remark : Be careful! In (ii) above, the given equation appears to be a quadratic
equation, but it is not a quadratic equation.
In (iv) above, the given equation appears to be a cubic equation (an equation of
degree 3) and not a quadratic equation. But it turns out to be a quadratic equation. As
you can see, often we need to simplify the given equation before deciding whether it
is quadratic or not.
EXERCISE 4.1
1. Check whether the following are quadratic equations :
(i) (x + 1)
2
 = 2(x – 3) (ii) x
2
 – 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x
2
 + 3x + 1 = (x – 2)
2
(vii) (x + 2)
3
 = 2x (x
2
 – 1) (viii) x
3
 – 4x
2
 – x + 1 = (x – 2)
3
2. Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m
2
. The length of the plot (in metres) is one
more than twice its breadth. We need to find the length and breadth of the plot.
Page 5


70 MATHEMA TICS
4
4.1 Introduction
In Chapter 2, you have studied different types of polynomials. One type was the
quadratic polynomial of the form ax
2
 + bx + c, a ? 0. When we equate this polynomial
to zero, we get a quadratic equation. Quadratic equations come up when we deal with
many real-life situations. For instance, suppose a
charity trust decides to build a prayer hall having
a carpet area of 300 square metres with its length
one metre more than twice its breadth. What
should be the length and breadth of the hall?
Suppose the breadth of the hall is x metres. Then,
its length should be (2x + 1) metres. W e can depict
this information pictorially as shown in Fig. 4.1.
Now, area of the hall = (2x + 1). x m
2
 = (2x
2
 + x) m
2
So, 2x
2
 + x = 300 (Given)
Therefore, 2x
2
 + x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x
2
 + x – 300 = 0 which is a
quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum
and a given positive product, and this problem is equivalent to solving a quadratic
equation of the form x
2
 – px + q = 0. Greek mathematician Euclid developed a
geometrical approach for finding out lengths which, in our present day terminology,
are solutions of quadratic equations. Solving of quadratic equations, in general form, is
often credited to ancient Indian mathematicians. In fact, Brahmagupta (A.D.598–665)
gave an explicit formula to solve a quadratic equation of the form ax
2
 + bx = c. Later,
QUADRA TIC EQUA TIONS
Fig. 4.1
QUADRA TIC EQUA TIONS 71
Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula,
(as quoted by Bhaskara II) for solving a quadratic equation by the method of completing
the square. An  Arab mathematician Al-Khwarizmi (about A.D. 800) also studied
quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book
‘Liber embadorum’ published in Europe in A.D. 1145 gave complete solutions of
different quadratic equations.
In this chapter, you will study quadratic equations, and various ways of finding
their roots. You will also see some applications of quadratic equations in daily life
situations.
4.2 Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax
2
 + bx + c = 0, where
a, b, c are real numbers, a ? 0. For example, 2x
2
 + x – 300 = 0 is a quadratic equation.
Similarly, 2x
2
 – 3x + 1 = 0, 4x – 3x
2
 + 2 = 0 and 1 – x
2
 + 300 = 0 are also quadratic
equations.
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree
2, is a quadratic equation. But when we write the terms of p(x) in descending order of
their degrees, then we get the standard form of the equation. That is, ax
2
 + bx + c = 0,
a ? 0 is called the standard form of a quadratic equation.
Quadratic equations arise in several situations in the world around us and in
different fields of mathematics. Let us consider a few examples.
Example 1 : Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and
the product of the number of marbles they now have is 124. We would like to find
out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of
production of each toy (in rupees) was found to be 55 minus the number of toys
produced in a day. On a particular day, the total cost of production was
Rs 750. We would like to find out the number of toys produced on that day.
Solution :
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
= 40 – x
72 MATHEMA TICS
Therefore, their product = (x – 5) (40 – x)
=40x – x
2
 – 200 + 5x
=– x
2
 + 45x – 200
So, – x
2
 + 45x – 200 = 124 (Given that product = 124)
i.e., – x
2
 + 45x – 324 = 0
i.e., x
2
 – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x
2
 – 45x + 324 = 0
which is the required representation of the problem mathematically.
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
Therefore, x (55 – x) = 750
i.e., 55x – x
2
 = 750
i.e., – x
2
 + 55x – 750 = 0
i.e., x
2
 – 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
x
2
 – 55x + 750 = 0
which is the required representation of the problem mathematically.
Example 2 : Check whether the following are quadratic equations:
(i) (x – 2)
2
 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2)
(iii) x (2x + 3) = x
2
 + 1 (iv) (x + 2)
3
 = x
3
 – 4
Solution :
(i) LHS = (x – 2)
2
 + 1 = x
2
 – 4x + 4 + 1 = x
2
 – 4x + 5
Therefore, (x – 2)
2 
+
 
1 = 2x – 3 can be rewritten as
x
2
 – 4x + 5 = 2x – 3
i.e., x
2
 – 6x + 8 = 0
It is of the form ax
2
 + bx + c = 0.
Therefore, the given equation is a quadratic equation.
QUADRA TIC EQUA TIONS 73
(ii) Since x(x + 1) + 8 = x
2
 + x + 8 and (x + 2)(x – 2) = x
2
 – 4
Therefore, x
2
 + x + 8 = x
2
 – 4
i.e., x + 12 = 0
It is not of the form ax
2
 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(iii) Here, LHS = x (2x + 3) = 2x
2
 + 3x
So, x (2x + 3) = x
2
 + 1 can be rewritten as
2x
2
 + 3x = x
2
 + 1
Therefore, we get x
2
 + 3x – 1 = 0
It is of the form ax
2
 + bx + c = 0.
So, the given equation is a quadratic equation.
(iv) Here, LHS = (x + 2)
3
 = x
3
 + 6x
2
 + 12x + 8
Therefore, (x + 2)
3
 = x
3
 – 4 can be rewritten as
x
3
 + 6x
2
 + 12x + 8 = x
3
 – 4
i.e., 6x
2
 + 12x + 12 = 0 or, x
2
 + 2x + 2 = 0
It is of the form ax
2
 + bx + c = 0.
So, the given equation is a quadratic equation.
Remark : Be careful! In (ii) above, the given equation appears to be a quadratic
equation, but it is not a quadratic equation.
In (iv) above, the given equation appears to be a cubic equation (an equation of
degree 3) and not a quadratic equation. But it turns out to be a quadratic equation. As
you can see, often we need to simplify the given equation before deciding whether it
is quadratic or not.
EXERCISE 4.1
1. Check whether the following are quadratic equations :
(i) (x + 1)
2
 = 2(x – 3) (ii) x
2
 – 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x
2
 + 3x + 1 = (x – 2)
2
(vii) (x + 2)
3
 = 2x (x
2
 – 1) (viii) x
3
 – 4x
2
 – x + 1 = (x – 2)
3
2. Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m
2
. The length of the plot (in metres) is one
more than twice its breadth. We need to find the length and breadth of the plot.
74 MATHEMA TICS
(ii) The product of two consecutive positive integers is 306. We need to find the
integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years)
3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been
8 km/h less, then it would have taken 3 hours more to cover the same distance. W e
need to find the speed of the train.
4.3 Solution of a Quadratic Equation by Factorisation
Consider the quadratic equation 2x
2
 – 3x + 1 = 0. If we replace x by 1 on the
LHS of this equation, we get (2 × 1
2
) – (3 × 1) + 1 = 0 = RHS of the equation.
We say that 1 is a root of the quadratic equation 2x
2
 – 3x + 1 = 0. This also means that
1 is a zero of the quadratic polynomial 2x
2
 – 3x + 1.
In general, a real number a is called a root of the quadratic equation
ax
2
 + bx + c = 0, a ? 0 if a a
2
 + ba + c = 0. We also say that x = a a a a a is a solution of
the quadratic equation, or that a a a a a satisfies the quadratic equation. Note that the
zeroes of the quadratic polynomial ax
2
 + bx + c and the roots of the quadratic
equation ax
2
 + bx + c = 0 are the same.
You have observed, in Chapter 2, that a quadratic polynomial can have at most
two zeroes. So, any quadratic equation can have atmost two roots.
You have learnt in Class IX, how to factorise quadratic polynomials by splitting
their middle terms. We shall use this knowledge for finding the roots of a quadratic
equation. Let us see how.
Example 3 : Find the roots of the equation 2x
2
 – 5x + 3 = 0, by factorisation.
Solution : Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) =
6x
2
 = (2x
2
) × 3].
So, 2x
2
 – 5x + 3 = 2x
2
 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1)
Now, 2x
2
 – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.
So, the values of x for which 2x
2
 – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0,
i.e., either 2x – 3 = 0 or x – 1 = 0.
Now, 2x – 3 = 0 gives 
3
2
x =
 and x – 1 = 0 gives x = 1.
So, 
3
2
x =
 and x = 1 are the solutions of the equation.
In other words, 1 and 
3
2
 are the roots of the equation 2x
2
 – 5x + 3 = 0.
Verify that these are the roots of the given equation.
Read More

Complete Syllabus of Class 10

Dynamic Test

Content Category

Related Searches

Exam

,

Objective type Questions

,

video lectures

,

shortcuts and tricks

,

Semester Notes

,

NCERT Textbook - Quadratic Equation Class 10 Notes | EduRev

,

NCERT Textbook - Quadratic Equation Class 10 Notes | EduRev

,

ppt

,

Free

,

Viva Questions

,

past year papers

,

pdf

,

Extra Questions

,

NCERT Textbook - Quadratic Equation Class 10 Notes | EduRev

,

Summary

,

study material

,

practice quizzes

,

Sample Paper

,

Previous Year Questions with Solutions

,

Important questions

,

mock tests for examination

,

MCQs

;