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# Quadratic Equations(NCERT) Class 10 Notes | EduRev

## Class 10 : Quadratic Equations(NCERT) Class 10 Notes | EduRev

``` Page 1

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 2
? INTRODUCTION
When a polynomial f(x) is equated to zero, we get an equation which is known as a polynomial equation. If f(x)
is a linear polynomial then f(x) = 0 is called a linear equation. For example, 3x – 2 = 0, 4t +
3
5
= 0 etc.
are linear equations. If f(x) is a quadratic polynomial i.e., f(x) = ax
2
+ bx + c, a ? 0, then f(x) = 0 i.e., ax
2
+ bx + c = 0, a ? 0 is called a quadratic equation. Such equations arise in many real life situations. In this
chapter, we will learn about quadratic equations and various ways of finding their zeros or roots. In the end of
the chapter, we will also discuss some applications of quadratic equations in daily life situations.
? HISTORICAL FACTS
On clay tables dated between 1800 BC and 1600 BC, the ancient Babylonians left the earliest evidence of the
discovery of quadratic equations, and also gave early methods for solving them.
Indian mathematician Baudhayana who wrote a Sulba Sutra in ancient India circa 8th century BC first used quadratic
equations of the form : ax
2
= c and ax
2
+ bx = c and also gave methods for solving them.
Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used the method
of completing the square to solve quadratic equations with positive roots, but did not have a general formula.
Euclid, a Greek mathematician, produced a more abstract geometrical method around 300 BC.
The first mathematician to have found negative solutions with the general algebraic formula was Brahmagupta
(India, 7th century). He gave the first explicit (although still not completely general) solution of the quadratic
equation ax
2
+ bx = c as follows :
"To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient
of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice
the [coefficient of the] square is the value."
This is equivalent to :
2
4ac b – b
x
2a
?
?
Muhammad ibn Musa al-Kwarizmi (Persia, 9th century) developed a set of formulae that worked for positive solutions.
Bhaskara II (1114-1185), an Indian mathematician-astronomer, solved quadratic equations with more than one
unknown and is considered the originator of the equation.
Shridhara (India, 9th century) was one of the first mathematicians to give a general rule for solving a quadratic equation.
A polynomial equation of degree two is called a quadratic equation.
Ex. 2x
2
– 3x + 1 = 0, 4x – 3x
2
= 0 and 1 – x
2
= 0
General form of quadratic equations : ax
2
+ bx + c = 0, where a,b,c, are real numbers and a ? 0.
Moreover, it is general form of a quadratic equation in standard form.
Types of Quadratic Equations : A quadratic equation can be of the following types :
( i ) b = 0, c ? 0 i.e. of the type ax
2
+ c = 0 (Pure quadratic equation)
( i i ) b ? 0, c = 0 i.e. of the type ax
2
+ bx = 0
( i i i ) b = 0, c = 0 i.e. of the type ax
2
= 0
( i v ) b ? 0, c ? 0 i.e. of the type ax
2
+ bx + c = 0 (Mixed or complete quadratic equation)
Roots of quadratic equation : x = ? is said to be root of the quadratic equation ax
2
+ bx + c = 0, a
? 0 iff x = ? satisfies the quadratic equation i.e. in other words the value of a ?
2
+ b ? + c is zero.
Page 2

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 2
? INTRODUCTION
When a polynomial f(x) is equated to zero, we get an equation which is known as a polynomial equation. If f(x)
is a linear polynomial then f(x) = 0 is called a linear equation. For example, 3x – 2 = 0, 4t +
3
5
= 0 etc.
are linear equations. If f(x) is a quadratic polynomial i.e., f(x) = ax
2
+ bx + c, a ? 0, then f(x) = 0 i.e., ax
2
+ bx + c = 0, a ? 0 is called a quadratic equation. Such equations arise in many real life situations. In this
chapter, we will learn about quadratic equations and various ways of finding their zeros or roots. In the end of
the chapter, we will also discuss some applications of quadratic equations in daily life situations.
? HISTORICAL FACTS
On clay tables dated between 1800 BC and 1600 BC, the ancient Babylonians left the earliest evidence of the
discovery of quadratic equations, and also gave early methods for solving them.
Indian mathematician Baudhayana who wrote a Sulba Sutra in ancient India circa 8th century BC first used quadratic
equations of the form : ax
2
= c and ax
2
+ bx = c and also gave methods for solving them.
Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used the method
of completing the square to solve quadratic equations with positive roots, but did not have a general formula.
Euclid, a Greek mathematician, produced a more abstract geometrical method around 300 BC.
The first mathematician to have found negative solutions with the general algebraic formula was Brahmagupta
(India, 7th century). He gave the first explicit (although still not completely general) solution of the quadratic
equation ax
2
+ bx = c as follows :
"To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient
of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice
the [coefficient of the] square is the value."
This is equivalent to :
2
4ac b – b
x
2a
?
?
Muhammad ibn Musa al-Kwarizmi (Persia, 9th century) developed a set of formulae that worked for positive solutions.
Bhaskara II (1114-1185), an Indian mathematician-astronomer, solved quadratic equations with more than one
unknown and is considered the originator of the equation.
Shridhara (India, 9th century) was one of the first mathematicians to give a general rule for solving a quadratic equation.
A polynomial equation of degree two is called a quadratic equation.
Ex. 2x
2
– 3x + 1 = 0, 4x – 3x
2
= 0 and 1 – x
2
= 0
General form of quadratic equations : ax
2
+ bx + c = 0, where a,b,c, are real numbers and a ? 0.
Moreover, it is general form of a quadratic equation in standard form.
Types of Quadratic Equations : A quadratic equation can be of the following types :
( i ) b = 0, c ? 0 i.e. of the type ax
2
+ c = 0 (Pure quadratic equation)
( i i ) b ? 0, c = 0 i.e. of the type ax
2
+ bx = 0
( i i i ) b = 0, c = 0 i.e. of the type ax
2
= 0
( i v ) b ? 0, c ? 0 i.e. of the type ax
2
+ bx + c = 0 (Mixed or complete quadratic equation)
Roots of quadratic equation : x = ? is said to be root of the quadratic equation ax
2
+ bx + c = 0, a
? 0 iff x = ? satisfies the quadratic equation i.e. in other words the value of a ?
2
+ b ? + c is zero.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 1 2 2
Solving a quadratic equation : The determination of all the roots of a quadratic equation is called solving
Ex.1 Check whether the following are quadratic equations :
(i) (x + 1)
2
= 2(x – 3)    (ii) (x – 2) (x + 1) = (x – 1) (x + 3)     (iii) (x – 3) (2x + 1) = x (x + 5)
Sol. ( i ) Here, the given equation is (x + 1)
2
= 2(x – 3)
? x
2
+ 2x + 1 = 2x – 6 ? x
2
+ 2x – 2x + 1 + 6 = 0
? x
2
+ 7 = 0 ? x
2
+ 0.x + 7 = 0, which is of the form ax
2
+ bx + c = 0
Hence, (x + 1)
2
= 2(x – 3) is a quadratic equation.
( i i ) Here, the given equation is (x – 2) (x + 1) = (x – 1) (x + 3)
? x
2
+ x – 2x – 2 = x
2
+ 3x – x – 3 ? x
2
– x
2
– x – 2x – 2 + 3 = 0 ? – 3x + 1 = 0,
Which is not of the form ax
2
+ bx + c = 0
Hence, (x – 2) (x + 1) = (x – 1) (x + 3) is not a quadratic equation
( i i i ) Here, the given equation is (x – 3) (2x + 1) = x (x + 5)
? 2x
2
+ x – 6x – 3 = x
2
+ 5x ? 2x
2
– x
2
– 5x – 5x – 3 = 0 ? x
2
– 10x – 3 = 0,
Which is of the form ax
2
+ bx + c = 0.
Hence, (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
Ex.2 In each of the following, determine whether the given values are the solution of the given equation or not :
(i)
2
2 5
–
x x
+ 2 = 0; x = 5, x =
1
2
(ii) a
2
x
2
– 3abx + 2b
2
= 0 ; x =
a b
, x
b a
?
Sol. ( i ) Putting x = 5 and x =
1
2
in the given equation.
2
2 5
– 2
(5 ) 5
?
and
2
2 5
– 2
1
1
2
2
?
? ?
? ?
? ?
? ?
? ?
? ?
?
2
25
– 1 + 2 and
2 5
– 2
1 1
4 2
?
?
2
25
+ 1 and 8 – 10 + 2 ?
27
25
and 0
i.e., x = 5 does not satisfy but x =
1
2
satisfies the given equation.
Hence, x = 5 is not a solution but x =
1
2
is a solution of
2
2 5
– 2 0
x x
? ? .
( i i ) Putting x =
a
b
and x =
b
a
in the given equation.
2
2 2
a a
a – 3ab 2b
b b
? ? ? ?
?
? ? ? ?
? ? ? ?
and
2
2 2
b b
a – 3ab 2b
a a
? ? ? ?
?
? ? ? ?
? ? ? ?
?
4
2
a
b
?2b
2
– 3a
2
and 0
i.e., x =
a
b
does not satisfy but x =
b
a
satisfies the given equation.
Hence x =
b
a
is a solution but x =
a
b
is not a solution of a
2
x
2
– 3abx + 2b
2
= 0.
Page 3

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 2
? INTRODUCTION
When a polynomial f(x) is equated to zero, we get an equation which is known as a polynomial equation. If f(x)
is a linear polynomial then f(x) = 0 is called a linear equation. For example, 3x – 2 = 0, 4t +
3
5
= 0 etc.
are linear equations. If f(x) is a quadratic polynomial i.e., f(x) = ax
2
+ bx + c, a ? 0, then f(x) = 0 i.e., ax
2
+ bx + c = 0, a ? 0 is called a quadratic equation. Such equations arise in many real life situations. In this
chapter, we will learn about quadratic equations and various ways of finding their zeros or roots. In the end of
the chapter, we will also discuss some applications of quadratic equations in daily life situations.
? HISTORICAL FACTS
On clay tables dated between 1800 BC and 1600 BC, the ancient Babylonians left the earliest evidence of the
discovery of quadratic equations, and also gave early methods for solving them.
Indian mathematician Baudhayana who wrote a Sulba Sutra in ancient India circa 8th century BC first used quadratic
equations of the form : ax
2
= c and ax
2
+ bx = c and also gave methods for solving them.
Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used the method
of completing the square to solve quadratic equations with positive roots, but did not have a general formula.
Euclid, a Greek mathematician, produced a more abstract geometrical method around 300 BC.
The first mathematician to have found negative solutions with the general algebraic formula was Brahmagupta
(India, 7th century). He gave the first explicit (although still not completely general) solution of the quadratic
equation ax
2
+ bx = c as follows :
"To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient
of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice
the [coefficient of the] square is the value."
This is equivalent to :
2
4ac b – b
x
2a
?
?
Muhammad ibn Musa al-Kwarizmi (Persia, 9th century) developed a set of formulae that worked for positive solutions.
Bhaskara II (1114-1185), an Indian mathematician-astronomer, solved quadratic equations with more than one
unknown and is considered the originator of the equation.
Shridhara (India, 9th century) was one of the first mathematicians to give a general rule for solving a quadratic equation.
A polynomial equation of degree two is called a quadratic equation.
Ex. 2x
2
– 3x + 1 = 0, 4x – 3x
2
= 0 and 1 – x
2
= 0
General form of quadratic equations : ax
2
+ bx + c = 0, where a,b,c, are real numbers and a ? 0.
Moreover, it is general form of a quadratic equation in standard form.
Types of Quadratic Equations : A quadratic equation can be of the following types :
( i ) b = 0, c ? 0 i.e. of the type ax
2
+ c = 0 (Pure quadratic equation)
( i i ) b ? 0, c = 0 i.e. of the type ax
2
+ bx = 0
( i i i ) b = 0, c = 0 i.e. of the type ax
2
= 0
( i v ) b ? 0, c ? 0 i.e. of the type ax
2
+ bx + c = 0 (Mixed or complete quadratic equation)
Roots of quadratic equation : x = ? is said to be root of the quadratic equation ax
2
+ bx + c = 0, a
? 0 iff x = ? satisfies the quadratic equation i.e. in other words the value of a ?
2
+ b ? + c is zero.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 1 2 2
Solving a quadratic equation : The determination of all the roots of a quadratic equation is called solving
Ex.1 Check whether the following are quadratic equations :
(i) (x + 1)
2
= 2(x – 3)    (ii) (x – 2) (x + 1) = (x – 1) (x + 3)     (iii) (x – 3) (2x + 1) = x (x + 5)
Sol. ( i ) Here, the given equation is (x + 1)
2
= 2(x – 3)
? x
2
+ 2x + 1 = 2x – 6 ? x
2
+ 2x – 2x + 1 + 6 = 0
? x
2
+ 7 = 0 ? x
2
+ 0.x + 7 = 0, which is of the form ax
2
+ bx + c = 0
Hence, (x + 1)
2
= 2(x – 3) is a quadratic equation.
( i i ) Here, the given equation is (x – 2) (x + 1) = (x – 1) (x + 3)
? x
2
+ x – 2x – 2 = x
2
+ 3x – x – 3 ? x
2
– x
2
– x – 2x – 2 + 3 = 0 ? – 3x + 1 = 0,
Which is not of the form ax
2
+ bx + c = 0
Hence, (x – 2) (x + 1) = (x – 1) (x + 3) is not a quadratic equation
( i i i ) Here, the given equation is (x – 3) (2x + 1) = x (x + 5)
? 2x
2
+ x – 6x – 3 = x
2
+ 5x ? 2x
2
– x
2
– 5x – 5x – 3 = 0 ? x
2
– 10x – 3 = 0,
Which is of the form ax
2
+ bx + c = 0.
Hence, (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
Ex.2 In each of the following, determine whether the given values are the solution of the given equation or not :
(i)
2
2 5
–
x x
+ 2 = 0; x = 5, x =
1
2
(ii) a
2
x
2
– 3abx + 2b
2
= 0 ; x =
a b
, x
b a
?
Sol. ( i ) Putting x = 5 and x =
1
2
in the given equation.
2
2 5
– 2
(5 ) 5
?
and
2
2 5
– 2
1
1
2
2
?
? ?
? ?
? ?
? ?
? ?
? ?
?
2
25
– 1 + 2 and
2 5
– 2
1 1
4 2
?
?
2
25
+ 1 and 8 – 10 + 2 ?
27
25
and 0
i.e., x = 5 does not satisfy but x =
1
2
satisfies the given equation.
Hence, x = 5 is not a solution but x =
1
2
is a solution of
2
2 5
– 2 0
x x
? ? .
( i i ) Putting x =
a
b
and x =
b
a
in the given equation.
2
2 2
a a
a – 3ab 2b
b b
? ? ? ?
?
? ? ? ?
? ? ? ?
and
2
2 2
b b
a – 3ab 2b
a a
? ? ? ?
?
? ? ? ?
? ? ? ?
?
4
2
a
b
?2b
2
– 3a
2
and 0
i.e., x =
a
b
does not satisfy but x =
b
a
satisfies the given equation.
Hence x =
b
a
is a solution but x =
a
b
is not a solution of a
2
x
2
– 3abx + 2b
2
= 0.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 4
Ex.3 Find the values of p and q for which x =
3
4
and x = – 2 are the roots of the equation px
2
+ qx – 6 = 0.
Sol. Since x =
3
4
and x = – 2 are the roots of the equation px
2
+ qx – 6 = 0.
? p
2
3
4
? ?
? ?
? ?
+ q
3
4
? ?
? ?
? ?
– 6 = 0 and p(–2)
2
+ q(–2) – 6 = 0
? p ×
9
16
+ q ×
3
4
– 6 = 0 and 4p – 2q – 6 = 0
?
9p 12q 96
16
? ?
= 0 and 4p – 2q – 6 = 0
? 9p + 12q – 96 = 0 and 4p – 2q – 6 = 0
? 3p + 4q – 32 = 0 . . . ( i )
and 2p – q – 3 = 0 . . . ( i i )
Multiplying (2) by 4, we get 8p – 4q – 12 = 0 . . . ( i i i )
Adding (1) and (3), we get p = 4
Putting the value of p in equation (2), we get
23 × 4 – q – 3 = 0 ? q = 5
Hence, p = 4, q = 5.
? METHODS OF SOLVING QUADRATIC EQUATIONS
Solution by factorisation method
Algorithm :
Step-I : Factorize the constant term of the given quadratic equation.
Step-II : Express the coefficient of middle term as the sum or difference of the factors obtained in step-I.Clearly, the
product of these two factors will be equal to the product of the coefficient of x
2
and constant term.
Step-III: Split the middle term in two parts obtained in step-II
Step-IV : Factorize the quadratic equation obtained in step-III by grouping method.
Ex.4 Solve the following quadratic equation by factorization method x
2
– 2ax + a
2
– b
2
= 0
Sol. Factors of the constant term a
2
– b
2
are (a – b) & (a + b) also coefficient of the middle term = – 2a = – [(a – b) + (a + b)]
? x
2
– 2ax + a
2
– b
2
= 0
? x
2
– {(a – b) + (a + b)} x + (a + b) (a – b) = 0
? x
2
– (a – b) x – (a + b) x + (a – b) (a + b) = 0
? x [x – (a – b)] – (a + b) [x – (a – b)] = 0
? [x – (a – b)] [x – (a + b)] = 0
x – (a – b) = 0 or x – (a + b) = 0
x = a – b, x = a + b
Ex.5 Solve the quadratic equation 5x
2
= – 16x – 12 by factorisation method.
Sol. 5x
2
= – 16x – 12
5x
2
+ 16x + 12 = 0
5x
2
+ 10x + 6x + 12 = 0
5x(x + 2) + 6 (x + 2) = 0
(x + 2) (5x + 6) = 0
x + 2 = 0 ? x = – 2
Page 4

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 2
? INTRODUCTION
When a polynomial f(x) is equated to zero, we get an equation which is known as a polynomial equation. If f(x)
is a linear polynomial then f(x) = 0 is called a linear equation. For example, 3x – 2 = 0, 4t +
3
5
= 0 etc.
are linear equations. If f(x) is a quadratic polynomial i.e., f(x) = ax
2
+ bx + c, a ? 0, then f(x) = 0 i.e., ax
2
+ bx + c = 0, a ? 0 is called a quadratic equation. Such equations arise in many real life situations. In this
chapter, we will learn about quadratic equations and various ways of finding their zeros or roots. In the end of
the chapter, we will also discuss some applications of quadratic equations in daily life situations.
? HISTORICAL FACTS
On clay tables dated between 1800 BC and 1600 BC, the ancient Babylonians left the earliest evidence of the
discovery of quadratic equations, and also gave early methods for solving them.
Indian mathematician Baudhayana who wrote a Sulba Sutra in ancient India circa 8th century BC first used quadratic
equations of the form : ax
2
= c and ax
2
+ bx = c and also gave methods for solving them.
Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used the method
of completing the square to solve quadratic equations with positive roots, but did not have a general formula.
Euclid, a Greek mathematician, produced a more abstract geometrical method around 300 BC.
The first mathematician to have found negative solutions with the general algebraic formula was Brahmagupta
(India, 7th century). He gave the first explicit (although still not completely general) solution of the quadratic
equation ax
2
+ bx = c as follows :
"To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient
of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice
the [coefficient of the] square is the value."
This is equivalent to :
2
4ac b – b
x
2a
?
?
Muhammad ibn Musa al-Kwarizmi (Persia, 9th century) developed a set of formulae that worked for positive solutions.
Bhaskara II (1114-1185), an Indian mathematician-astronomer, solved quadratic equations with more than one
unknown and is considered the originator of the equation.
Shridhara (India, 9th century) was one of the first mathematicians to give a general rule for solving a quadratic equation.
A polynomial equation of degree two is called a quadratic equation.
Ex. 2x
2
– 3x + 1 = 0, 4x – 3x
2
= 0 and 1 – x
2
= 0
General form of quadratic equations : ax
2
+ bx + c = 0, where a,b,c, are real numbers and a ? 0.
Moreover, it is general form of a quadratic equation in standard form.
Types of Quadratic Equations : A quadratic equation can be of the following types :
( i ) b = 0, c ? 0 i.e. of the type ax
2
+ c = 0 (Pure quadratic equation)
( i i ) b ? 0, c = 0 i.e. of the type ax
2
+ bx = 0
( i i i ) b = 0, c = 0 i.e. of the type ax
2
= 0
( i v ) b ? 0, c ? 0 i.e. of the type ax
2
+ bx + c = 0 (Mixed or complete quadratic equation)
Roots of quadratic equation : x = ? is said to be root of the quadratic equation ax
2
+ bx + c = 0, a
? 0 iff x = ? satisfies the quadratic equation i.e. in other words the value of a ?
2
+ b ? + c is zero.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 1 2 2
Solving a quadratic equation : The determination of all the roots of a quadratic equation is called solving
Ex.1 Check whether the following are quadratic equations :
(i) (x + 1)
2
= 2(x – 3)    (ii) (x – 2) (x + 1) = (x – 1) (x + 3)     (iii) (x – 3) (2x + 1) = x (x + 5)
Sol. ( i ) Here, the given equation is (x + 1)
2
= 2(x – 3)
? x
2
+ 2x + 1 = 2x – 6 ? x
2
+ 2x – 2x + 1 + 6 = 0
? x
2
+ 7 = 0 ? x
2
+ 0.x + 7 = 0, which is of the form ax
2
+ bx + c = 0
Hence, (x + 1)
2
= 2(x – 3) is a quadratic equation.
( i i ) Here, the given equation is (x – 2) (x + 1) = (x – 1) (x + 3)
? x
2
+ x – 2x – 2 = x
2
+ 3x – x – 3 ? x
2
– x
2
– x – 2x – 2 + 3 = 0 ? – 3x + 1 = 0,
Which is not of the form ax
2
+ bx + c = 0
Hence, (x – 2) (x + 1) = (x – 1) (x + 3) is not a quadratic equation
( i i i ) Here, the given equation is (x – 3) (2x + 1) = x (x + 5)
? 2x
2
+ x – 6x – 3 = x
2
+ 5x ? 2x
2
– x
2
– 5x – 5x – 3 = 0 ? x
2
– 10x – 3 = 0,
Which is of the form ax
2
+ bx + c = 0.
Hence, (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
Ex.2 In each of the following, determine whether the given values are the solution of the given equation or not :
(i)
2
2 5
–
x x
+ 2 = 0; x = 5, x =
1
2
(ii) a
2
x
2
– 3abx + 2b
2
= 0 ; x =
a b
, x
b a
?
Sol. ( i ) Putting x = 5 and x =
1
2
in the given equation.
2
2 5
– 2
(5 ) 5
?
and
2
2 5
– 2
1
1
2
2
?
? ?
? ?
? ?
? ?
? ?
? ?
?
2
25
– 1 + 2 and
2 5
– 2
1 1
4 2
?
?
2
25
+ 1 and 8 – 10 + 2 ?
27
25
and 0
i.e., x = 5 does not satisfy but x =
1
2
satisfies the given equation.
Hence, x = 5 is not a solution but x =
1
2
is a solution of
2
2 5
– 2 0
x x
? ? .
( i i ) Putting x =
a
b
and x =
b
a
in the given equation.
2
2 2
a a
a – 3ab 2b
b b
? ? ? ?
?
? ? ? ?
? ? ? ?
and
2
2 2
b b
a – 3ab 2b
a a
? ? ? ?
?
? ? ? ?
? ? ? ?
?
4
2
a
b
?2b
2
– 3a
2
and 0
i.e., x =
a
b
does not satisfy but x =
b
a
satisfies the given equation.
Hence x =
b
a
is a solution but x =
a
b
is not a solution of a
2
x
2
– 3abx + 2b
2
= 0.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 4
Ex.3 Find the values of p and q for which x =
3
4
and x = – 2 are the roots of the equation px
2
+ qx – 6 = 0.
Sol. Since x =
3
4
and x = – 2 are the roots of the equation px
2
+ qx – 6 = 0.
? p
2
3
4
? ?
? ?
? ?
+ q
3
4
? ?
? ?
? ?
– 6 = 0 and p(–2)
2
+ q(–2) – 6 = 0
? p ×
9
16
+ q ×
3
4
– 6 = 0 and 4p – 2q – 6 = 0
?
9p 12q 96
16
? ?
= 0 and 4p – 2q – 6 = 0
? 9p + 12q – 96 = 0 and 4p – 2q – 6 = 0
? 3p + 4q – 32 = 0 . . . ( i )
and 2p – q – 3 = 0 . . . ( i i )
Multiplying (2) by 4, we get 8p – 4q – 12 = 0 . . . ( i i i )
Adding (1) and (3), we get p = 4
Putting the value of p in equation (2), we get
23 × 4 – q – 3 = 0 ? q = 5
Hence, p = 4, q = 5.
? METHODS OF SOLVING QUADRATIC EQUATIONS
Solution by factorisation method
Algorithm :
Step-I : Factorize the constant term of the given quadratic equation.
Step-II : Express the coefficient of middle term as the sum or difference of the factors obtained in step-I.Clearly, the
product of these two factors will be equal to the product of the coefficient of x
2
and constant term.
Step-III: Split the middle term in two parts obtained in step-II
Step-IV : Factorize the quadratic equation obtained in step-III by grouping method.
Ex.4 Solve the following quadratic equation by factorization method x
2
– 2ax + a
2
– b
2
= 0
Sol. Factors of the constant term a
2
– b
2
are (a – b) & (a + b) also coefficient of the middle term = – 2a = – [(a – b) + (a + b)]
? x
2
– 2ax + a
2
– b
2
= 0
? x
2
– {(a – b) + (a + b)} x + (a + b) (a – b) = 0
? x
2
– (a – b) x – (a + b) x + (a – b) (a + b) = 0
? x [x – (a – b)] – (a + b) [x – (a – b)] = 0
? [x – (a – b)] [x – (a + b)] = 0
x – (a – b) = 0 or x – (a + b) = 0
x = a – b, x = a + b
Ex.5 Solve the quadratic equation 5x
2
= – 16x – 12 by factorisation method.
Sol. 5x
2
= – 16x – 12
5x
2
+ 16x + 12 = 0
5x
2
+ 10x + 6x + 12 = 0
5x(x + 2) + 6 (x + 2) = 0
(x + 2) (5x + 6) = 0
x + 2 = 0 ? x = – 2
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 1 2 4
5x + 6 = 0 ? x =
–6
5
Solution by completing the square
Algorithm :
Step-I : Obtain the quadratic equation. Let the quadratic equation be ax
2
+ bx + c = 0, a ? 0.
Step-II : Make the coefficient of x
2
unity by dividing throughout by it, if it is not unity that is obtain x
2
+
b
a
x +
c
a
= 0
Step-III: Shift the constant term
c
a
on R.H.S. to get x
2
+
b
a
x = –
c
a
Step-IV : Add square of half of the coefficient of x. i.e.
2
b
2a
? ?
? ?
? ?
on both sides to obtain.
x
2
+ 2
b
2a
? ?
? ?
? ?
x +
2
b
2a
? ?
? ?
? ?
=
2
b
2a
? ?
? ?
? ?
–
c
a
Step-V: Write L.H.S. as the perfect square and simplify R.H.S. to get
2
b
x
2a
? ?
?
? ?
? ?
=
2
2
b – 4ac
4a
Step-VI : Take square root of both sides to get
b
x
2a
? = ±
2
2
b – 4ac
4a
Step-VII : Obtain the values of x by shifting the constant term
b
2a
on R.H.S. i.e. x = –
b
2a
±
2
2
b – 4ac
4a
Ex.6 Solve : 9x
2
– 15x + 6 = 0
Sol. Here, 9x
2
– 15x + 6 = 0
? x
2
–
15
9
x +
6
9
= 0 [Dividing throughout by 9]
? x
2
–
5
3
x +
2
3
= 0
? x
2
–
5
3
x = –
2
3
[Shifting the constant term on RHS)
? x
2
– 2
5
6
? ?
? ?
? ?
x +
2
5
6
? ?
? ?
? ?
=
2
5
6
? ?
? ?
? ?
–
2
3
[Adding square of half of coefficient of x on both sides]
?
2
5
x –
6
? ?
? ?
? ?
=
25
36
–
2
3
?
2
5
x –
6
? ?
? ?
? ?
=
25 – 24
36
?? ?
2
5
x –
6
? ?
? ?
? ?
=
1
36
?
5
x –
6
= ±
1
6
[Taking square root of both sides]
?
5
x
6
? ±
1
6
?
5
x
6
? +
1
6
= 1 or,
5
x
6
? –
1
6
=
4
6
=
2
3
? x = 1 or, x =
2
3
Ex.7 Solve the equation x
2
– (
3
+ 1) x +
3
= 0  by the method of completing the square.
Sol. We have,
x
2
– (
3
+ 1) x +
3
= 0
? x
2
– (
3
+ 1) x = –
3
Page 5

97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 2
? INTRODUCTION
When a polynomial f(x) is equated to zero, we get an equation which is known as a polynomial equation. If f(x)
is a linear polynomial then f(x) = 0 is called a linear equation. For example, 3x – 2 = 0, 4t +
3
5
= 0 etc.
are linear equations. If f(x) is a quadratic polynomial i.e., f(x) = ax
2
+ bx + c, a ? 0, then f(x) = 0 i.e., ax
2
+ bx + c = 0, a ? 0 is called a quadratic equation. Such equations arise in many real life situations. In this
chapter, we will learn about quadratic equations and various ways of finding their zeros or roots. In the end of
the chapter, we will also discuss some applications of quadratic equations in daily life situations.
? HISTORICAL FACTS
On clay tables dated between 1800 BC and 1600 BC, the ancient Babylonians left the earliest evidence of the
discovery of quadratic equations, and also gave early methods for solving them.
Indian mathematician Baudhayana who wrote a Sulba Sutra in ancient India circa 8th century BC first used quadratic
equations of the form : ax
2
= c and ax
2
+ bx = c and also gave methods for solving them.
Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used the method
of completing the square to solve quadratic equations with positive roots, but did not have a general formula.
Euclid, a Greek mathematician, produced a more abstract geometrical method around 300 BC.
The first mathematician to have found negative solutions with the general algebraic formula was Brahmagupta
(India, 7th century). He gave the first explicit (although still not completely general) solution of the quadratic
equation ax
2
+ bx = c as follows :
"To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient
of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice
the [coefficient of the] square is the value."
This is equivalent to :
2
4ac b – b
x
2a
?
?
Muhammad ibn Musa al-Kwarizmi (Persia, 9th century) developed a set of formulae that worked for positive solutions.
Bhaskara II (1114-1185), an Indian mathematician-astronomer, solved quadratic equations with more than one
unknown and is considered the originator of the equation.
Shridhara (India, 9th century) was one of the first mathematicians to give a general rule for solving a quadratic equation.
A polynomial equation of degree two is called a quadratic equation.
Ex. 2x
2
– 3x + 1 = 0, 4x – 3x
2
= 0 and 1 – x
2
= 0
General form of quadratic equations : ax
2
+ bx + c = 0, where a,b,c, are real numbers and a ? 0.
Moreover, it is general form of a quadratic equation in standard form.
Types of Quadratic Equations : A quadratic equation can be of the following types :
( i ) b = 0, c ? 0 i.e. of the type ax
2
+ c = 0 (Pure quadratic equation)
( i i ) b ? 0, c = 0 i.e. of the type ax
2
+ bx = 0
( i i i ) b = 0, c = 0 i.e. of the type ax
2
= 0
( i v ) b ? 0, c ? 0 i.e. of the type ax
2
+ bx + c = 0 (Mixed or complete quadratic equation)
Roots of quadratic equation : x = ? is said to be root of the quadratic equation ax
2
+ bx + c = 0, a
? 0 iff x = ? satisfies the quadratic equation i.e. in other words the value of a ?
2
+ b ? + c is zero.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 1 2 2
Solving a quadratic equation : The determination of all the roots of a quadratic equation is called solving
Ex.1 Check whether the following are quadratic equations :
(i) (x + 1)
2
= 2(x – 3)    (ii) (x – 2) (x + 1) = (x – 1) (x + 3)     (iii) (x – 3) (2x + 1) = x (x + 5)
Sol. ( i ) Here, the given equation is (x + 1)
2
= 2(x – 3)
? x
2
+ 2x + 1 = 2x – 6 ? x
2
+ 2x – 2x + 1 + 6 = 0
? x
2
+ 7 = 0 ? x
2
+ 0.x + 7 = 0, which is of the form ax
2
+ bx + c = 0
Hence, (x + 1)
2
= 2(x – 3) is a quadratic equation.
( i i ) Here, the given equation is (x – 2) (x + 1) = (x – 1) (x + 3)
? x
2
+ x – 2x – 2 = x
2
+ 3x – x – 3 ? x
2
– x
2
– x – 2x – 2 + 3 = 0 ? – 3x + 1 = 0,
Which is not of the form ax
2
+ bx + c = 0
Hence, (x – 2) (x + 1) = (x – 1) (x + 3) is not a quadratic equation
( i i i ) Here, the given equation is (x – 3) (2x + 1) = x (x + 5)
? 2x
2
+ x – 6x – 3 = x
2
+ 5x ? 2x
2
– x
2
– 5x – 5x – 3 = 0 ? x
2
– 10x – 3 = 0,
Which is of the form ax
2
+ bx + c = 0.
Hence, (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
Ex.2 In each of the following, determine whether the given values are the solution of the given equation or not :
(i)
2
2 5
–
x x
+ 2 = 0; x = 5, x =
1
2
(ii) a
2
x
2
– 3abx + 2b
2
= 0 ; x =
a b
, x
b a
?
Sol. ( i ) Putting x = 5 and x =
1
2
in the given equation.
2
2 5
– 2
(5 ) 5
?
and
2
2 5
– 2
1
1
2
2
?
? ?
? ?
? ?
? ?
? ?
? ?
?
2
25
– 1 + 2 and
2 5
– 2
1 1
4 2
?
?
2
25
+ 1 and 8 – 10 + 2 ?
27
25
and 0
i.e., x = 5 does not satisfy but x =
1
2
satisfies the given equation.
Hence, x = 5 is not a solution but x =
1
2
is a solution of
2
2 5
– 2 0
x x
? ? .
( i i ) Putting x =
a
b
and x =
b
a
in the given equation.
2
2 2
a a
a – 3ab 2b
b b
? ? ? ?
?
? ? ? ?
? ? ? ?
and
2
2 2
b b
a – 3ab 2b
a a
? ? ? ?
?
? ? ? ?
? ? ? ?
?
4
2
a
b
?2b
2
– 3a
2
and 0
i.e., x =
a
b
does not satisfy but x =
b
a
satisfies the given equation.
Hence x =
b
a
is a solution but x =
a
b
is not a solution of a
2
x
2
– 3abx + 2b
2
= 0.
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 4
Ex.3 Find the values of p and q for which x =
3
4
and x = – 2 are the roots of the equation px
2
+ qx – 6 = 0.
Sol. Since x =
3
4
and x = – 2 are the roots of the equation px
2
+ qx – 6 = 0.
? p
2
3
4
? ?
? ?
? ?
+ q
3
4
? ?
? ?
? ?
– 6 = 0 and p(–2)
2
+ q(–2) – 6 = 0
? p ×
9
16
+ q ×
3
4
– 6 = 0 and 4p – 2q – 6 = 0
?
9p 12q 96
16
? ?
= 0 and 4p – 2q – 6 = 0
? 9p + 12q – 96 = 0 and 4p – 2q – 6 = 0
? 3p + 4q – 32 = 0 . . . ( i )
and 2p – q – 3 = 0 . . . ( i i )
Multiplying (2) by 4, we get 8p – 4q – 12 = 0 . . . ( i i i )
Adding (1) and (3), we get p = 4
Putting the value of p in equation (2), we get
23 × 4 – q – 3 = 0 ? q = 5
Hence, p = 4, q = 5.
? METHODS OF SOLVING QUADRATIC EQUATIONS
Solution by factorisation method
Algorithm :
Step-I : Factorize the constant term of the given quadratic equation.
Step-II : Express the coefficient of middle term as the sum or difference of the factors obtained in step-I.Clearly, the
product of these two factors will be equal to the product of the coefficient of x
2
and constant term.
Step-III: Split the middle term in two parts obtained in step-II
Step-IV : Factorize the quadratic equation obtained in step-III by grouping method.
Ex.4 Solve the following quadratic equation by factorization method x
2
– 2ax + a
2
– b
2
= 0
Sol. Factors of the constant term a
2
– b
2
are (a – b) & (a + b) also coefficient of the middle term = – 2a = – [(a – b) + (a + b)]
? x
2
– 2ax + a
2
– b
2
= 0
? x
2
– {(a – b) + (a + b)} x + (a + b) (a – b) = 0
? x
2
– (a – b) x – (a + b) x + (a – b) (a + b) = 0
? x [x – (a – b)] – (a + b) [x – (a – b)] = 0
? [x – (a – b)] [x – (a + b)] = 0
x – (a – b) = 0 or x – (a + b) = 0
x = a – b, x = a + b
Ex.5 Solve the quadratic equation 5x
2
= – 16x – 12 by factorisation method.
Sol. 5x
2
= – 16x – 12
5x
2
+ 16x + 12 = 0
5x
2
+ 10x + 6x + 12 = 0
5x(x + 2) + 6 (x + 2) = 0
(x + 2) (5x + 6) = 0
x + 2 = 0 ? x = – 2
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009 1 2 4
5x + 6 = 0 ? x =
–6
5
Solution by completing the square
Algorithm :
Step-I : Obtain the quadratic equation. Let the quadratic equation be ax
2
+ bx + c = 0, a ? 0.
Step-II : Make the coefficient of x
2
unity by dividing throughout by it, if it is not unity that is obtain x
2
+
b
a
x +
c
a
= 0
Step-III: Shift the constant term
c
a
on R.H.S. to get x
2
+
b
a
x = –
c
a
Step-IV : Add square of half of the coefficient of x. i.e.
2
b
2a
? ?
? ?
? ?
on both sides to obtain.
x
2
+ 2
b
2a
? ?
? ?
? ?
x +
2
b
2a
? ?
? ?
? ?
=
2
b
2a
? ?
? ?
? ?
–
c
a
Step-V: Write L.H.S. as the perfect square and simplify R.H.S. to get
2
b
x
2a
? ?
?
? ?
? ?
=
2
2
b – 4ac
4a
Step-VI : Take square root of both sides to get
b
x
2a
? = ±
2
2
b – 4ac
4a
Step-VII : Obtain the values of x by shifting the constant term
b
2a
on R.H.S. i.e. x = –
b
2a
±
2
2
b – 4ac
4a
Ex.6 Solve : 9x
2
– 15x + 6 = 0
Sol. Here, 9x
2
– 15x + 6 = 0
? x
2
–
15
9
x +
6
9
= 0 [Dividing throughout by 9]
? x
2
–
5
3
x +
2
3
= 0
? x
2
–
5
3
x = –
2
3
[Shifting the constant term on RHS)
? x
2
– 2
5
6
? ?
? ?
? ?
x +
2
5
6
? ?
? ?
? ?
=
2
5
6
? ?
? ?
? ?
–
2
3
[Adding square of half of coefficient of x on both sides]
?
2
5
x –
6
? ?
? ?
? ?
=
25
36
–
2
3
?
2
5
x –
6
? ?
? ?
? ?
=
25 – 24
36
?? ?
2
5
x –
6
? ?
? ?
? ?
=
1
36
?
5
x –
6
= ±
1
6
[Taking square root of both sides]
?
5
x
6
? ±
1
6
?
5
x
6
? +
1
6
= 1 or,
5
x
6
? –
1
6
=
4
6
=
2
3
? x = 1 or, x =
2
3
Ex.7 Solve the equation x
2
– (
3
+ 1) x +
3
= 0  by the method of completing the square.
Sol. We have,
x
2
– (
3
+ 1) x +
3
= 0
? x
2
– (
3
+ 1) x = –
3
97/1, 3F, Adhchini, Sri Aurobindo Marg, Near NCERT, New Delhi |  011-32044009
1 2 6
? x
2
– 2
3 1
2
? ?
?
? ?
? ?
? ?
x +
2
3 1
2
? ?
?
? ?
? ?
? ?
= – 3 +
2
3 1
2
? ?
?
? ?
? ?
? ?
?
2
3 1
x –
2
? ?
?
? ?
? ?
? ?
=
2
– 4 ( 3 1)
3
4
?
?
?
2
3 1
x –
2
? ?
?
? ?
? ?
? ?
=
2
–
3 1
2
? ?
? ?
? ?
? ?
?
3 1
x –
2
?
= ±
–
3 1
2
?
3 1
x
2
?
?
±
–
3 1
2
? x = 3 , 1
Hence, the roots are 3 and 1.
Solution by Quadratic Formula "Sreedharacharya's  Rule"
Consider quadratic equation ax
2
+ bx + c = 0, a ? 0 then x =
2
– b b – 4ac
2a
?
? The roots of x are
x =
2
– b b – 4ac
2a
?
and
2
– b – b – 4ac
2a
? x =
– b D
2a
?
or, x =
– b – D
2a
, where D = b
2
– 4ac
Thus, if D = b
2
– 4ac ? 0, then the quadratic equation ax
2
+ bx + c = 0 has real roots ? and ? given
by
? =
– b D
2a
?
and ? =
– b – D
2a
Discriminant : If ax
2
+ bx + c = 0, a ? 0 (a,b,c ? R) is a quadratic equation, then the expression b
2
–
4ac is known as its discriminant and is generally denoted by D or ?.
Ex.8 Solve the quadratic equation x
2
– 6x + 4 = 0 by using quadratic formula (Sreedharacharya's Rule).
Sol. On comparing the given equation x
2
– 6x + 4 = 0 with the standard quadratic equation ax
2
+ bx + c = 0,
we get a = 1, b = – 6, c = 4
Hence the required roots are
x =
2
–(– 6) (–6) – 4( 1)(4)
2( 1)
?
=
6 36 – 16
2
?
=
6 20
2
?
=
6 4 5
2
? ?
=
2(3 5 )
2
?
= 3 ± 5
Thus the roots of the equation are 3 + 5 & 3 – 5 .
```
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