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# Quadratic Equations JEE Notes | EduRev

## JEE : Quadratic Equations JEE Notes | EduRev

``` Page 1

30
E
JEE- Mathe ma ti cs
1 . INTRODUCTION :
The algebraic expression of the form ax
2
+ bx + c,
a 0 ?
is called a quadratic expression, because the highest t
order term in it is of second degree. Quadratic equation means, ax
2
+ bx + c = 0. In general whenever one says
zeroes of the expression ax
2
+bx + c, it implies roots of the equation ax
2
+ bx + c = 0, unless specified
otherwise.
A quadratic equation has exactly two roots which may be real (equal or unequal) or imaginary.
2 . SOLUTION OF QUADR ATIC EQUATION & RELATION BETWEEN ROOTS & CO-EFFICIENTS :
( a ) The general form of quadratic equation is ax
2
+ bx + c = 0, a ? 0.
The roots can be found in following manner :
2
b c
a x x
a a
? ?
? ?
? ?
? ?
= 0 ?
2
2
2
b c b
x 0
2a a 4a
? ?
? ? ? ?
? ?
? ?
2
2
2
b b c
x
2a a 4a
? ?
? ? ?
? ?
? ?
?
2
b b – 4ac
x
2a
? ?
?
This expression can be directly used to find the two roots of a quadratic equation.
( b ) The expression b
2
– 4 ac ? D is called the discriminant of the quadratic equation.
( c ) If & ? ? are the roots of the quadratic equation ax
2
+ bx + c = 0 , then :
(i) ? ? ? = – b/a (ii)  c / a ? ? ? (iii)  D /| a| ? ? ? ?
( d ) A  quadratic equation whose roots are & ? ? is ( x – ? ) ( x – ? ) = 0 i.e.
x
2
– ( ? ? ? ) x + ?? = 0 i.e. x
2
– (sum of roots) x + product of roots = 0.
Illustration 1 : If ? ? ? ? are the roots of a quadratic equation x
2
– 3x + 5 = 0, then the equation whose roots are
( ?
2
– 3 ? + 7) and ( ?
2
– 3 ? + 7) is -
(A) x
2
+ 4x + 1 = 0 (B) x
2
– 4x + 4 = 0 (C) x
2
– 4x – 1 = 0 (D) x
2
+ 2x + 3 = 0
Solution : Since ? ? ? ? are the roots of equation x
2
– 3x + 5 = 0
So ?
2
– 3 ? + 5 = 0
?
2
– 3 ? + 5 = 0
? ?
2
– 3 ?  = – 5
?
2
– 3 ?  = – 5
Putting in ( ?
2
– 3 ? + 7) & ( ?
2
– 3 ? + 7) .........(i)
– 5 + 7, – 5 + 7
? 2 and 2 are the roots.
? The required equation is
x
2
– 4x + 4 = 0. Ans. (B)
Illustration 2 : If ? and ? are the roots of ax
2
+ bx + c = 0, find the value of (a ? + b)
–2
+ (a ? + b)
–2
.
Solution : We know that ? + ? ?=
b
a
?   & ? ? =
c
a
(a ? + b)
–2
+ (a ? + b)
–2
= 2 2
1 1
(a b) (a b)
?
? ? ? ?
=
2 2 2 2 2 2
2 2 2
a b 2ab a b 2ab
(a ba ba b )
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
=
2 2 2 2
2 2 2
a ( ) 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
( ?
2
+ ?
2
can always be written as ( ? + ?)
2
– 2 ? ?)
JEEMAIN.GURU
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30
E
JEE- Mathe ma ti cs
1 . INTRODUCTION :
The algebraic expression of the form ax
2
+ bx + c,
a 0 ?
is called a quadratic expression, because the highest t
order term in it is of second degree. Quadratic equation means, ax
2
+ bx + c = 0. In general whenever one says
zeroes of the expression ax
2
+bx + c, it implies roots of the equation ax
2
+ bx + c = 0, unless specified
otherwise.
A quadratic equation has exactly two roots which may be real (equal or unequal) or imaginary.
2 . SOLUTION OF QUADR ATIC EQUATION & RELATION BETWEEN ROOTS & CO-EFFICIENTS :
( a ) The general form of quadratic equation is ax
2
+ bx + c = 0, a ? 0.
The roots can be found in following manner :
2
b c
a x x
a a
? ?
? ?
? ?
? ?
= 0 ?
2
2
2
b c b
x 0
2a a 4a
? ?
? ? ? ?
? ?
? ?
2
2
2
b b c
x
2a a 4a
? ?
? ? ?
? ?
? ?
?
2
b b – 4ac
x
2a
? ?
?
This expression can be directly used to find the two roots of a quadratic equation.
( b ) The expression b
2
– 4 ac ? D is called the discriminant of the quadratic equation.
( c ) If & ? ? are the roots of the quadratic equation ax
2
+ bx + c = 0 , then :
(i) ? ? ? = – b/a (ii)  c / a ? ? ? (iii)  D /| a| ? ? ? ?
( d ) A  quadratic equation whose roots are & ? ? is ( x – ? ) ( x – ? ) = 0 i.e.
x
2
– ( ? ? ? ) x + ?? = 0 i.e. x
2
– (sum of roots) x + product of roots = 0.
Illustration 1 : If ? ? ? ? are the roots of a quadratic equation x
2
– 3x + 5 = 0, then the equation whose roots are
( ?
2
– 3 ? + 7) and ( ?
2
– 3 ? + 7) is -
(A) x
2
+ 4x + 1 = 0 (B) x
2
– 4x + 4 = 0 (C) x
2
– 4x – 1 = 0 (D) x
2
+ 2x + 3 = 0
Solution : Since ? ? ? ? are the roots of equation x
2
– 3x + 5 = 0
So ?
2
– 3 ? + 5 = 0
?
2
– 3 ? + 5 = 0
? ?
2
– 3 ?  = – 5
?
2
– 3 ?  = – 5
Putting in ( ?
2
– 3 ? + 7) & ( ?
2
– 3 ? + 7) .........(i)
– 5 + 7, – 5 + 7
? 2 and 2 are the roots.
? The required equation is
x
2
– 4x + 4 = 0. Ans. (B)
Illustration 2 : If ? and ? are the roots of ax
2
+ bx + c = 0, find the value of (a ? + b)
–2
+ (a ? + b)
–2
.
Solution : We know that ? + ? ?=
b
a
?   & ? ? =
c
a
(a ? + b)
–2
+ (a ? + b)
–2
= 2 2
1 1
(a b) (a b)
?
? ? ? ?
=
2 2 2 2 2 2
2 2 2
a b 2ab a b 2ab
(a ba ba b )
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
=
2 2 2 2
2 2 2
a ( ) 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
( ?
2
+ ?
2
can always be written as ( ? + ?)
2
– 2 ? ?)
JEEMAIN.GURU
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JEE- Mathe ma ti cs
=
2 2 2
2 2 2
a ( ) 2 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ?
=
2
2 2
2
2
2 2
b 2ac b
a 2ab 2b
a a
c b
a ab b
a a
? ? ? ? ?
? ? ?
? ? ? ?
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
=
2
2 2
b 2ac
a c
?
Alternatively :
Take b = – ( ? ?+ ?) a
(a ? + b)
–2
+ (a ? + b)
–2
=
2 2 2
1 1 1
a ( ) ( )
? ?
?
? ?
? ? ? ? ? ? ? ? ? ?
? ?
=
2 2
2 2 2
1
a
? ? ? ? ?
? ?
? ?
? ?
=
2 2
2 2 2 2
2
2
1 b 2ac b 2ac
a c a c
a .
a
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
Do yourself - 1 :
( i ) Find  the roots of following equations :
(a) x
2
+ 3x + 2 = 0 (b) x
2
– 8x + 16 = 0 (c) x
2
– 2x – 1 = 0
(i i) Find the roots of the equation a(x
2
+ 1) – (a
2
+ 1)x = 0, where a ? 0.
(iii) Solve :
2
6 x x
2
x 2
x 4
?
? ?
?
?
( i v) If the roots of 4x
2
+ 5k = (5k + 1)x differ by unity, then find the values of k.
3 . NATURE  OF  ROOTS :
( a ) Consider the quadratic equation ax
2
+ bx + c = 0 where a, b , c ? R & a ? 0 then ;
b D
x
2a
? ?
?
(i) D > 0 ? roots are real & distinct (unequal).
(ii) D = 0 ? roots are real & coincident (equal)
(iii) D < 0 ? roots are imaginary.
(iv) If p + i q  is  one  root  of  a  quadratic equation, then the other root must be the conjugate
p – i q & vice versa.  (p, q ? R  &  i = 1 ? ) .
( b ) Consider the quadratic equation ax
2
+ bx + c = 0  where a, b, c ? Q & a ? 0 then ;
(i) If D is a perfect square, then roots are rational.
(ii) If ? = p + q is one root in this case, ( where p is rational & q is a surd) then other root will
be p – q .
Illustration 3 : If the coefficient of the quadratic equation are rational & the coefficient of x
2
is 1, then find the
equation one of whose roots is tan
8
?
.
Solution : We know that  tan 2 1
8
?
? ?
Irrational roots always occur in conjugational pairs.
Hence if one root is (–1 +
2
), the other root will be (–1 –
2
). Equation is
(x –(–1+
2
)) (x–(–1–
2
)) =0 ? x
2
+ 2x –1 = 0
JEEMAIN.GURU
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30
E
JEE- Mathe ma ti cs
1 . INTRODUCTION :
The algebraic expression of the form ax
2
+ bx + c,
a 0 ?
is called a quadratic expression, because the highest t
order term in it is of second degree. Quadratic equation means, ax
2
+ bx + c = 0. In general whenever one says
zeroes of the expression ax
2
+bx + c, it implies roots of the equation ax
2
+ bx + c = 0, unless specified
otherwise.
A quadratic equation has exactly two roots which may be real (equal or unequal) or imaginary.
2 . SOLUTION OF QUADR ATIC EQUATION & RELATION BETWEEN ROOTS & CO-EFFICIENTS :
( a ) The general form of quadratic equation is ax
2
+ bx + c = 0, a ? 0.
The roots can be found in following manner :
2
b c
a x x
a a
? ?
? ?
? ?
? ?
= 0 ?
2
2
2
b c b
x 0
2a a 4a
? ?
? ? ? ?
? ?
? ?
2
2
2
b b c
x
2a a 4a
? ?
? ? ?
? ?
? ?
?
2
b b – 4ac
x
2a
? ?
?
This expression can be directly used to find the two roots of a quadratic equation.
( b ) The expression b
2
– 4 ac ? D is called the discriminant of the quadratic equation.
( c ) If & ? ? are the roots of the quadratic equation ax
2
+ bx + c = 0 , then :
(i) ? ? ? = – b/a (ii)  c / a ? ? ? (iii)  D /| a| ? ? ? ?
( d ) A  quadratic equation whose roots are & ? ? is ( x – ? ) ( x – ? ) = 0 i.e.
x
2
– ( ? ? ? ) x + ?? = 0 i.e. x
2
– (sum of roots) x + product of roots = 0.
Illustration 1 : If ? ? ? ? are the roots of a quadratic equation x
2
– 3x + 5 = 0, then the equation whose roots are
( ?
2
– 3 ? + 7) and ( ?
2
– 3 ? + 7) is -
(A) x
2
+ 4x + 1 = 0 (B) x
2
– 4x + 4 = 0 (C) x
2
– 4x – 1 = 0 (D) x
2
+ 2x + 3 = 0
Solution : Since ? ? ? ? are the roots of equation x
2
– 3x + 5 = 0
So ?
2
– 3 ? + 5 = 0
?
2
– 3 ? + 5 = 0
? ?
2
– 3 ?  = – 5
?
2
– 3 ?  = – 5
Putting in ( ?
2
– 3 ? + 7) & ( ?
2
– 3 ? + 7) .........(i)
– 5 + 7, – 5 + 7
? 2 and 2 are the roots.
? The required equation is
x
2
– 4x + 4 = 0. Ans. (B)
Illustration 2 : If ? and ? are the roots of ax
2
+ bx + c = 0, find the value of (a ? + b)
–2
+ (a ? + b)
–2
.
Solution : We know that ? + ? ?=
b
a
?   & ? ? =
c
a
(a ? + b)
–2
+ (a ? + b)
–2
= 2 2
1 1
(a b) (a b)
?
? ? ? ?
=
2 2 2 2 2 2
2 2 2
a b 2ab a b 2ab
(a ba ba b )
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
=
2 2 2 2
2 2 2
a ( ) 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
( ?
2
+ ?
2
can always be written as ( ? + ?)
2
– 2 ? ?)
JEEMAIN.GURU
E
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JEE- Mathe ma ti cs
=
2 2 2
2 2 2
a ( ) 2 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ?
=
2
2 2
2
2
2 2
b 2ac b
a 2ab 2b
a a
c b
a ab b
a a
? ? ? ? ?
? ? ?
? ? ? ?
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
=
2
2 2
b 2ac
a c
?
Alternatively :
Take b = – ( ? ?+ ?) a
(a ? + b)
–2
+ (a ? + b)
–2
=
2 2 2
1 1 1
a ( ) ( )
? ?
?
? ?
? ? ? ? ? ? ? ? ? ?
? ?
=
2 2
2 2 2
1
a
? ? ? ? ?
? ?
? ?
? ?
=
2 2
2 2 2 2
2
2
1 b 2ac b 2ac
a c a c
a .
a
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
Do yourself - 1 :
( i ) Find  the roots of following equations :
(a) x
2
+ 3x + 2 = 0 (b) x
2
– 8x + 16 = 0 (c) x
2
– 2x – 1 = 0
(i i) Find the roots of the equation a(x
2
+ 1) – (a
2
+ 1)x = 0, where a ? 0.
(iii) Solve :
2
6 x x
2
x 2
x 4
?
? ?
?
?
( i v) If the roots of 4x
2
+ 5k = (5k + 1)x differ by unity, then find the values of k.
3 . NATURE  OF  ROOTS :
( a ) Consider the quadratic equation ax
2
+ bx + c = 0 where a, b , c ? R & a ? 0 then ;
b D
x
2a
? ?
?
(i) D > 0 ? roots are real & distinct (unequal).
(ii) D = 0 ? roots are real & coincident (equal)
(iii) D < 0 ? roots are imaginary.
(iv) If p + i q  is  one  root  of  a  quadratic equation, then the other root must be the conjugate
p – i q & vice versa.  (p, q ? R  &  i = 1 ? ) .
( b ) Consider the quadratic equation ax
2
+ bx + c = 0  where a, b, c ? Q & a ? 0 then ;
(i) If D is a perfect square, then roots are rational.
(ii) If ? = p + q is one root in this case, ( where p is rational & q is a surd) then other root will
be p – q .
Illustration 3 : If the coefficient of the quadratic equation are rational & the coefficient of x
2
is 1, then find the
equation one of whose roots is tan
8
?
.
Solution : We know that  tan 2 1
8
?
? ?
Irrational roots always occur in conjugational pairs.
Hence if one root is (–1 +
2
), the other root will be (–1 –
2
). Equation is
(x –(–1+
2
)) (x–(–1–
2
)) =0 ? x
2
+ 2x –1 = 0
JEEMAIN.GURU
32
E
JEE- Mathe ma ti cs
Illustration 4 : Find all the integral values of a for which the quadratic equation (x – a)(x – 10) + 1 = 0 has integral
roots.
Solution : Here the equation is x
2
– (a + 10)x + 10a + 1 = 0. Since integral roots will always be rational it
means D should be a perfect square.
From (i) D = a
2
– 20a + 96.
? D = (a – 10)
2
– 4 ? 4 = (a – 10)
2
– D
If D is a perfect square it means we want difference of two perfect square as 4 which is possible
only when (a –10)
2
= 4 and D = 0.
? (a – 10)

= ± 2 ? a = 12, 8 Ans.
Do yourself - 2 :
( i ) If 2 3 ? is a root of the equation x
2
+ bx + c = 0, where  b, c ? Q, find b, c.
(i i) For the following equations, find the nature of the roots (real & distinct, real & coincident or imaginary).
(a) x
2
– 6x + 10 = 0
(b)
2
x (7 3)x 6(1 3) 0 ? ? ? ? ?
(c) 4x
2
+ 28x + 49 = 0
(iii) If ?, m are real and ? ? m, then show that the roots of ( ? – m)x
2
– 5( ? + m)x – 2( ? – m) = 0 are real and
unequal.
4 . ROOTS UNDER PARTICULAR CASES :
2
+ bx + c = 0 has real roots and
( a ) If b = 0 ? roots are equal in magnitude but opposite in sign
( b ) If c = 0 ? one root is zero other is – b/a
( c ) If a = c ? roots are reciprocal to each other
( d ) If
a 0 c 0
a 0 c 0
? ? ?
?
? ?
?
? roots are of opposite signs
( e ) If
a 0, b 0,c 0
a 0, b 0,c 0
? ? ? ?
?
? ? ?
?
? both roots are negative.
( f ) If
a 0, b 0,c 0
a 0, b 0,c 0
? ? ? ?
?
? ? ?
?
? both roots are positive.
( g ) If sign of a = sign of b ? sign of c  ? Greater root in magnitude is negative.
( h ) If sign of b = sign of c ? sign of a  ? Greater root in magnitude is positive.
( i ) If a + b + c = 0 ? one root is 1 and second root is c/a or (–b–a)/a.
Illustration 5 : If equation
2
x bx k 1
ax c k 1
? ?
?
? ?
has roots equal in magnitude & opposite in sign, then the value of k
is -
(A)
a b
a b
?
?
(B)
a b
a b
?
?
(C)
a
1
b
? (D)
a
1
b
?
Solution : Let the roots are ? ? ?? ?? .
given equation is
(x
2
– bx)(k + 1) = (k – 1)(ax – c) {Considering, x ? c/a & k ? –1}
? x
2
(k + 1) – bx(k + 1) = ax (k – 1) – c(k – 1)
? x
2
(k + 1) – bx(k + 1) – ax (k – 1) + c(k – 1)  = 0
Now sum of roots = 0 ( ? ? ? ? ?? ? = 0)
? b(k + 1) + a(k – 1) = 0 ? k =
a b
a b
?
?
Ans. (B)
JEEMAIN.GURU
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30
E
JEE- Mathe ma ti cs
1 . INTRODUCTION :
The algebraic expression of the form ax
2
+ bx + c,
a 0 ?
is called a quadratic expression, because the highest t
order term in it is of second degree. Quadratic equation means, ax
2
+ bx + c = 0. In general whenever one says
zeroes of the expression ax
2
+bx + c, it implies roots of the equation ax
2
+ bx + c = 0, unless specified
otherwise.
A quadratic equation has exactly two roots which may be real (equal or unequal) or imaginary.
2 . SOLUTION OF QUADR ATIC EQUATION & RELATION BETWEEN ROOTS & CO-EFFICIENTS :
( a ) The general form of quadratic equation is ax
2
+ bx + c = 0, a ? 0.
The roots can be found in following manner :
2
b c
a x x
a a
? ?
? ?
? ?
? ?
= 0 ?
2
2
2
b c b
x 0
2a a 4a
? ?
? ? ? ?
? ?
? ?
2
2
2
b b c
x
2a a 4a
? ?
? ? ?
? ?
? ?
?
2
b b – 4ac
x
2a
? ?
?
This expression can be directly used to find the two roots of a quadratic equation.
( b ) The expression b
2
– 4 ac ? D is called the discriminant of the quadratic equation.
( c ) If & ? ? are the roots of the quadratic equation ax
2
+ bx + c = 0 , then :
(i) ? ? ? = – b/a (ii)  c / a ? ? ? (iii)  D /| a| ? ? ? ?
( d ) A  quadratic equation whose roots are & ? ? is ( x – ? ) ( x – ? ) = 0 i.e.
x
2
– ( ? ? ? ) x + ?? = 0 i.e. x
2
– (sum of roots) x + product of roots = 0.
Illustration 1 : If ? ? ? ? are the roots of a quadratic equation x
2
– 3x + 5 = 0, then the equation whose roots are
( ?
2
– 3 ? + 7) and ( ?
2
– 3 ? + 7) is -
(A) x
2
+ 4x + 1 = 0 (B) x
2
– 4x + 4 = 0 (C) x
2
– 4x – 1 = 0 (D) x
2
+ 2x + 3 = 0
Solution : Since ? ? ? ? are the roots of equation x
2
– 3x + 5 = 0
So ?
2
– 3 ? + 5 = 0
?
2
– 3 ? + 5 = 0
? ?
2
– 3 ?  = – 5
?
2
– 3 ?  = – 5
Putting in ( ?
2
– 3 ? + 7) & ( ?
2
– 3 ? + 7) .........(i)
– 5 + 7, – 5 + 7
? 2 and 2 are the roots.
? The required equation is
x
2
– 4x + 4 = 0. Ans. (B)
Illustration 2 : If ? and ? are the roots of ax
2
+ bx + c = 0, find the value of (a ? + b)
–2
+ (a ? + b)
–2
.
Solution : We know that ? + ? ?=
b
a
?   & ? ? =
c
a
(a ? + b)
–2
+ (a ? + b)
–2
= 2 2
1 1
(a b) (a b)
?
? ? ? ?
=
2 2 2 2 2 2
2 2 2
a b 2ab a b 2ab
(a ba ba b )
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
=
2 2 2 2
2 2 2
a ( ) 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
( ?
2
+ ?
2
can always be written as ( ? + ?)
2
– 2 ? ?)
JEEMAIN.GURU
E
31
JEE- Mathe ma ti cs
=
2 2 2
2 2 2
a ( ) 2 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ?
=
2
2 2
2
2
2 2
b 2ac b
a 2ab 2b
a a
c b
a ab b
a a
? ? ? ? ?
? ? ?
? ? ? ?
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
=
2
2 2
b 2ac
a c
?
Alternatively :
Take b = – ( ? ?+ ?) a
(a ? + b)
–2
+ (a ? + b)
–2
=
2 2 2
1 1 1
a ( ) ( )
? ?
?
? ?
? ? ? ? ? ? ? ? ? ?
? ?
=
2 2
2 2 2
1
a
? ? ? ? ?
? ?
? ?
? ?
=
2 2
2 2 2 2
2
2
1 b 2ac b 2ac
a c a c
a .
a
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
Do yourself - 1 :
( i ) Find  the roots of following equations :
(a) x
2
+ 3x + 2 = 0 (b) x
2
– 8x + 16 = 0 (c) x
2
– 2x – 1 = 0
(i i) Find the roots of the equation a(x
2
+ 1) – (a
2
+ 1)x = 0, where a ? 0.
(iii) Solve :
2
6 x x
2
x 2
x 4
?
? ?
?
?
( i v) If the roots of 4x
2
+ 5k = (5k + 1)x differ by unity, then find the values of k.
3 . NATURE  OF  ROOTS :
( a ) Consider the quadratic equation ax
2
+ bx + c = 0 where a, b , c ? R & a ? 0 then ;
b D
x
2a
? ?
?
(i) D > 0 ? roots are real & distinct (unequal).
(ii) D = 0 ? roots are real & coincident (equal)
(iii) D < 0 ? roots are imaginary.
(iv) If p + i q  is  one  root  of  a  quadratic equation, then the other root must be the conjugate
p – i q & vice versa.  (p, q ? R  &  i = 1 ? ) .
( b ) Consider the quadratic equation ax
2
+ bx + c = 0  where a, b, c ? Q & a ? 0 then ;
(i) If D is a perfect square, then roots are rational.
(ii) If ? = p + q is one root in this case, ( where p is rational & q is a surd) then other root will
be p – q .
Illustration 3 : If the coefficient of the quadratic equation are rational & the coefficient of x
2
is 1, then find the
equation one of whose roots is tan
8
?
.
Solution : We know that  tan 2 1
8
?
? ?
Irrational roots always occur in conjugational pairs.
Hence if one root is (–1 +
2
), the other root will be (–1 –
2
). Equation is
(x –(–1+
2
)) (x–(–1–
2
)) =0 ? x
2
+ 2x –1 = 0
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Illustration 4 : Find all the integral values of a for which the quadratic equation (x – a)(x – 10) + 1 = 0 has integral
roots.
Solution : Here the equation is x
2
– (a + 10)x + 10a + 1 = 0. Since integral roots will always be rational it
means D should be a perfect square.
From (i) D = a
2
– 20a + 96.
? D = (a – 10)
2
– 4 ? 4 = (a – 10)
2
– D
If D is a perfect square it means we want difference of two perfect square as 4 which is possible
only when (a –10)
2
= 4 and D = 0.
? (a – 10)

= ± 2 ? a = 12, 8 Ans.
Do yourself - 2 :
( i ) If 2 3 ? is a root of the equation x
2
+ bx + c = 0, where  b, c ? Q, find b, c.
(i i) For the following equations, find the nature of the roots (real & distinct, real & coincident or imaginary).
(a) x
2
– 6x + 10 = 0
(b)
2
x (7 3)x 6(1 3) 0 ? ? ? ? ?
(c) 4x
2
+ 28x + 49 = 0
(iii) If ?, m are real and ? ? m, then show that the roots of ( ? – m)x
2
– 5( ? + m)x – 2( ? – m) = 0 are real and
unequal.
4 . ROOTS UNDER PARTICULAR CASES :
2
+ bx + c = 0 has real roots and
( a ) If b = 0 ? roots are equal in magnitude but opposite in sign
( b ) If c = 0 ? one root is zero other is – b/a
( c ) If a = c ? roots are reciprocal to each other
( d ) If
a 0 c 0
a 0 c 0
? ? ?
?
? ?
?
? roots are of opposite signs
( e ) If
a 0, b 0,c 0
a 0, b 0,c 0
? ? ? ?
?
? ? ?
?
? both roots are negative.
( f ) If
a 0, b 0,c 0
a 0, b 0,c 0
? ? ? ?
?
? ? ?
?
? both roots are positive.
( g ) If sign of a = sign of b ? sign of c  ? Greater root in magnitude is negative.
( h ) If sign of b = sign of c ? sign of a  ? Greater root in magnitude is positive.
( i ) If a + b + c = 0 ? one root is 1 and second root is c/a or (–b–a)/a.
Illustration 5 : If equation
2
x bx k 1
ax c k 1
? ?
?
? ?
has roots equal in magnitude & opposite in sign, then the value of k
is -
(A)
a b
a b
?
?
(B)
a b
a b
?
?
(C)
a
1
b
? (D)
a
1
b
?
Solution : Let the roots are ? ? ?? ?? .
given equation is
(x
2
– bx)(k + 1) = (k – 1)(ax – c) {Considering, x ? c/a & k ? –1}
? x
2
(k + 1) – bx(k + 1) = ax (k – 1) – c(k – 1)
? x
2
(k + 1) – bx(k + 1) – ax (k – 1) + c(k – 1)  = 0
Now sum of roots = 0 ( ? ? ? ? ?? ? = 0)
? b(k + 1) + a(k – 1) = 0 ? k =
a b
a b
?
?
Ans. (B)
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*Illustration 6 : If roots of the equation (a – b)x
2
+ (c – a)x + (b – c) = 0 are equal, then a, b, c are in
(A) A.P. (B) H.P. (C) G.P. (D) none of these
Solution : (a – b)x
2
+ (c – a)x + (b – c) = 0
As roots are equal so
B
2
– 4AC = 0  ?? (c – a)
2
– 4(a – b)(b – c) = 0    ? (c – a)
2
– 4ab + 4b
2
+ 4ac – 4bc = 0
? (c – a)
2
+ 4ac – 4b(c + a) + 4b
2
= 0 ? (c + a)
2
– 2 . (2b)(c + a) + (2b)
2
= 0
? [c + a – 2b]
2
= 0 ? c + a – 2b = 0  ?? ? ?c + a = 2b
Hence  a, b, c are in A. P.
Alternative method :
? Sum of the coefficients = 0
Hence one root is 1 and other root is
b c
a b
?
?
.
Given that both roots are equal, so
1 =
b c
a b
?
?
?   a – b = b – c     ?? ? ? ?2b = a + c
Hence  a, b, c are in A.P. Ans. (A)
Do yourself - 3 :
( i ) Consider ƒ (x) = x
2
+ bx + c.
(a) Find c if x = 0 is a root of ƒ (x) = 0.
(b) Find c if
1
, ?
?
are roots of ƒ (x) = 0.
(c) Comment on sign of b & c, if 0 ? ? ? ? & | | | | ? ? ? , where ? ? ? ? are roots of ƒ (x) = 0.
5 . IDENTITY :
An equation which is true for every value of the variable within the domain is called an identity , for example :
5 (a – 3) =5a – 15, (a + b)
2
= a
2
+ b
2
+ 2ab for all a, b ? R.
Note : A quadratic equation cannot have three or more roots & if it has , it becomes an identity.
If ax
2
+ bx + c = 0 is an identity ? a = b = c = 0
Illustration 7 : If the equation ( ?
2
– 5 ? + 6)x
2
+ ( ?
2
– 3 ? + 2)x + ( ?
2
– 4) = 0 has more than two roots, then find
the value of ? ?
Solution : As the equation has more than two roots so it becomes an identity. Hence
?
2
– 5 ? + 6 = 0 ? ? = 2, 3
and ?
2
– 3 ? + 2 = 0     ? ? = 1, 2
and ?
2
– 4 = 0 ? ? = 2, –2
So ? = 2 Ans. ? = 2
6 . COMMON ROOTS OF TWO QUADR ATIC EQUATIONS :
( a ) Only one common root.
Let  ? be the common root of ax
2
+ bx + c = 0 & a'x
2
+ b'x + c' = 0  then
a ?
2
+ b ? + c = 0  &  a' ?
2
+ b' ? + c' = 0 . By Cramer’s Rule
2
1
bc ' b 'c a 'c ac ' ab ' a 'b
? ?
? ?
? ? ?
Therefore,
ca ' c 'a bc ' b 'c
ab ' a 'b a 'c ac '
? ?
? ? ?
? ?
So the condition for a common root is (ca' - c'a)
2
= (ab' - a'b) (bc'- b'c).
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30
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1 . INTRODUCTION :
The algebraic expression of the form ax
2
+ bx + c,
a 0 ?
is called a quadratic expression, because the highest t
order term in it is of second degree. Quadratic equation means, ax
2
+ bx + c = 0. In general whenever one says
zeroes of the expression ax
2
+bx + c, it implies roots of the equation ax
2
+ bx + c = 0, unless specified
otherwise.
A quadratic equation has exactly two roots which may be real (equal or unequal) or imaginary.
2 . SOLUTION OF QUADR ATIC EQUATION & RELATION BETWEEN ROOTS & CO-EFFICIENTS :
( a ) The general form of quadratic equation is ax
2
+ bx + c = 0, a ? 0.
The roots can be found in following manner :
2
b c
a x x
a a
? ?
? ?
? ?
? ?
= 0 ?
2
2
2
b c b
x 0
2a a 4a
? ?
? ? ? ?
? ?
? ?
2
2
2
b b c
x
2a a 4a
? ?
? ? ?
? ?
? ?
?
2
b b – 4ac
x
2a
? ?
?
This expression can be directly used to find the two roots of a quadratic equation.
( b ) The expression b
2
– 4 ac ? D is called the discriminant of the quadratic equation.
( c ) If & ? ? are the roots of the quadratic equation ax
2
+ bx + c = 0 , then :
(i) ? ? ? = – b/a (ii)  c / a ? ? ? (iii)  D /| a| ? ? ? ?
( d ) A  quadratic equation whose roots are & ? ? is ( x – ? ) ( x – ? ) = 0 i.e.
x
2
– ( ? ? ? ) x + ?? = 0 i.e. x
2
– (sum of roots) x + product of roots = 0.
Illustration 1 : If ? ? ? ? are the roots of a quadratic equation x
2
– 3x + 5 = 0, then the equation whose roots are
( ?
2
– 3 ? + 7) and ( ?
2
– 3 ? + 7) is -
(A) x
2
+ 4x + 1 = 0 (B) x
2
– 4x + 4 = 0 (C) x
2
– 4x – 1 = 0 (D) x
2
+ 2x + 3 = 0
Solution : Since ? ? ? ? are the roots of equation x
2
– 3x + 5 = 0
So ?
2
– 3 ? + 5 = 0
?
2
– 3 ? + 5 = 0
? ?
2
– 3 ?  = – 5
?
2
– 3 ?  = – 5
Putting in ( ?
2
– 3 ? + 7) & ( ?
2
– 3 ? + 7) .........(i)
– 5 + 7, – 5 + 7
? 2 and 2 are the roots.
? The required equation is
x
2
– 4x + 4 = 0. Ans. (B)
Illustration 2 : If ? and ? are the roots of ax
2
+ bx + c = 0, find the value of (a ? + b)
–2
+ (a ? + b)
–2
.
Solution : We know that ? + ? ?=
b
a
?   & ? ? =
c
a
(a ? + b)
–2
+ (a ? + b)
–2
= 2 2
1 1
(a b) (a b)
?
? ? ? ?
=
2 2 2 2 2 2
2 2 2
a b 2ab a b 2ab
(a ba ba b )
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
=
2 2 2 2
2 2 2
a ( ) 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ?
? ? ? ? ? ? ?
( ?
2
+ ?
2
can always be written as ( ? + ?)
2
– 2 ? ?)
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=
2 2 2
2 2 2
a ( ) 2 2ab( ) 2b
(a ab( ) b )
? ? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ?
=
2
2 2
2
2
2 2
b 2ac b
a 2ab 2b
a a
c b
a ab b
a a
? ? ? ? ?
? ? ?
? ? ? ?
? ?
? ?
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
=
2
2 2
b 2ac
a c
?
Alternatively :
Take b = – ( ? ?+ ?) a
(a ? + b)
–2
+ (a ? + b)
–2
=
2 2 2
1 1 1
a ( ) ( )
? ?
?
? ?
? ? ? ? ? ? ? ? ? ?
? ?
=
2 2
2 2 2
1
a
? ? ? ? ?
? ?
? ?
? ?
=
2 2
2 2 2 2
2
2
1 b 2ac b 2ac
a c a c
a .
a
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
Do yourself - 1 :
( i ) Find  the roots of following equations :
(a) x
2
+ 3x + 2 = 0 (b) x
2
– 8x + 16 = 0 (c) x
2
– 2x – 1 = 0
(i i) Find the roots of the equation a(x
2
+ 1) – (a
2
+ 1)x = 0, where a ? 0.
(iii) Solve :
2
6 x x
2
x 2
x 4
?
? ?
?
?
( i v) If the roots of 4x
2
+ 5k = (5k + 1)x differ by unity, then find the values of k.
3 . NATURE  OF  ROOTS :
( a ) Consider the quadratic equation ax
2
+ bx + c = 0 where a, b , c ? R & a ? 0 then ;
b D
x
2a
? ?
?
(i) D > 0 ? roots are real & distinct (unequal).
(ii) D = 0 ? roots are real & coincident (equal)
(iii) D < 0 ? roots are imaginary.
(iv) If p + i q  is  one  root  of  a  quadratic equation, then the other root must be the conjugate
p – i q & vice versa.  (p, q ? R  &  i = 1 ? ) .
( b ) Consider the quadratic equation ax
2
+ bx + c = 0  where a, b, c ? Q & a ? 0 then ;
(i) If D is a perfect square, then roots are rational.
(ii) If ? = p + q is one root in this case, ( where p is rational & q is a surd) then other root will
be p – q .
Illustration 3 : If the coefficient of the quadratic equation are rational & the coefficient of x
2
is 1, then find the
equation one of whose roots is tan
8
?
.
Solution : We know that  tan 2 1
8
?
? ?
Irrational roots always occur in conjugational pairs.
Hence if one root is (–1 +
2
), the other root will be (–1 –
2
). Equation is
(x –(–1+
2
)) (x–(–1–
2
)) =0 ? x
2
+ 2x –1 = 0
JEEMAIN.GURU
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Illustration 4 : Find all the integral values of a for which the quadratic equation (x – a)(x – 10) + 1 = 0 has integral
roots.
Solution : Here the equation is x
2
– (a + 10)x + 10a + 1 = 0. Since integral roots will always be rational it
means D should be a perfect square.
From (i) D = a
2
– 20a + 96.
? D = (a – 10)
2
– 4 ? 4 = (a – 10)
2
– D
If D is a perfect square it means we want difference of two perfect square as 4 which is possible
only when (a –10)
2
= 4 and D = 0.
? (a – 10)

= ± 2 ? a = 12, 8 Ans.
Do yourself - 2 :
( i ) If 2 3 ? is a root of the equation x
2
+ bx + c = 0, where  b, c ? Q, find b, c.
(i i) For the following equations, find the nature of the roots (real & distinct, real & coincident or imaginary).
(a) x
2
– 6x + 10 = 0
(b)
2
x (7 3)x 6(1 3) 0 ? ? ? ? ?
(c) 4x
2
+ 28x + 49 = 0
(iii) If ?, m are real and ? ? m, then show that the roots of ( ? – m)x
2
– 5( ? + m)x – 2( ? – m) = 0 are real and
unequal.
4 . ROOTS UNDER PARTICULAR CASES :
2
+ bx + c = 0 has real roots and
( a ) If b = 0 ? roots are equal in magnitude but opposite in sign
( b ) If c = 0 ? one root is zero other is – b/a
( c ) If a = c ? roots are reciprocal to each other
( d ) If
a 0 c 0
a 0 c 0
? ? ?
?
? ?
?
? roots are of opposite signs
( e ) If
a 0, b 0,c 0
a 0, b 0,c 0
? ? ? ?
?
? ? ?
?
? both roots are negative.
( f ) If
a 0, b 0,c 0
a 0, b 0,c 0
? ? ? ?
?
? ? ?
?
? both roots are positive.
( g ) If sign of a = sign of b ? sign of c  ? Greater root in magnitude is negative.
( h ) If sign of b = sign of c ? sign of a  ? Greater root in magnitude is positive.
( i ) If a + b + c = 0 ? one root is 1 and second root is c/a or (–b–a)/a.
Illustration 5 : If equation
2
x bx k 1
ax c k 1
? ?
?
? ?
has roots equal in magnitude & opposite in sign, then the value of k
is -
(A)
a b
a b
?
?
(B)
a b
a b
?
?
(C)
a
1
b
? (D)
a
1
b
?
Solution : Let the roots are ? ? ?? ?? .
given equation is
(x
2
– bx)(k + 1) = (k – 1)(ax – c) {Considering, x ? c/a & k ? –1}
? x
2
(k + 1) – bx(k + 1) = ax (k – 1) – c(k – 1)
? x
2
(k + 1) – bx(k + 1) – ax (k – 1) + c(k – 1)  = 0
Now sum of roots = 0 ( ? ? ? ? ?? ? = 0)
? b(k + 1) + a(k – 1) = 0 ? k =
a b
a b
?
?
Ans. (B)
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*Illustration 6 : If roots of the equation (a – b)x
2
+ (c – a)x + (b – c) = 0 are equal, then a, b, c are in
(A) A.P. (B) H.P. (C) G.P. (D) none of these
Solution : (a – b)x
2
+ (c – a)x + (b – c) = 0
As roots are equal so
B
2
– 4AC = 0  ?? (c – a)
2
– 4(a – b)(b – c) = 0    ? (c – a)
2
– 4ab + 4b
2
+ 4ac – 4bc = 0
? (c – a)
2
+ 4ac – 4b(c + a) + 4b
2
= 0 ? (c + a)
2
– 2 . (2b)(c + a) + (2b)
2
= 0
? [c + a – 2b]
2
= 0 ? c + a – 2b = 0  ?? ? ?c + a = 2b
Hence  a, b, c are in A. P.
Alternative method :
? Sum of the coefficients = 0
Hence one root is 1 and other root is
b c
a b
?
?
.
Given that both roots are equal, so
1 =
b c
a b
?
?
?   a – b = b – c     ?? ? ? ?2b = a + c
Hence  a, b, c are in A.P. Ans. (A)
Do yourself - 3 :
( i ) Consider ƒ (x) = x
2
+ bx + c.
(a) Find c if x = 0 is a root of ƒ (x) = 0.
(b) Find c if
1
, ?
?
are roots of ƒ (x) = 0.
(c) Comment on sign of b & c, if 0 ? ? ? ? & | | | | ? ? ? , where ? ? ? ? are roots of ƒ (x) = 0.
5 . IDENTITY :
An equation which is true for every value of the variable within the domain is called an identity , for example :
5 (a – 3) =5a – 15, (a + b)
2
= a
2
+ b
2
+ 2ab for all a, b ? R.
Note : A quadratic equation cannot have three or more roots & if it has , it becomes an identity.
If ax
2
+ bx + c = 0 is an identity ? a = b = c = 0
Illustration 7 : If the equation ( ?
2
– 5 ? + 6)x
2
+ ( ?
2
– 3 ? + 2)x + ( ?
2
– 4) = 0 has more than two roots, then find
the value of ? ?
Solution : As the equation has more than two roots so it becomes an identity. Hence
?
2
– 5 ? + 6 = 0 ? ? = 2, 3
and ?
2
– 3 ? + 2 = 0     ? ? = 1, 2
and ?
2
– 4 = 0 ? ? = 2, –2
So ? = 2 Ans. ? = 2
6 . COMMON ROOTS OF TWO QUADR ATIC EQUATIONS :
( a ) Only one common root.
Let  ? be the common root of ax
2
+ bx + c = 0 & a'x
2
+ b'x + c' = 0  then
a ?
2
+ b ? + c = 0  &  a' ?
2
+ b' ? + c' = 0 . By Cramer’s Rule
2
1
bc ' b 'c a 'c ac ' ab ' a 'b
? ?
? ?
? ? ?
Therefore,
ca ' c 'a bc ' b 'c
ab ' a 'b a 'c ac '
? ?
? ? ?
? ?
So the condition for a common root is (ca' - c'a)
2
= (ab' - a'b) (bc'- b'c).
JEEMAIN.GURU
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( b ) If both roots are same then
a b c
a ' b ' c '
? ? .
Illustration 8

: Find p and q such that px
2
+ 5x + 2 = 0 and 3x
2
+ 10 x +q = 0 have both roots in common.
Solution : a
1
= p, b
1
= 5, c
1
= 2
a
2
= 3, b
2
= 10, c
2
= q
We know that :
1 1 1
2 2 2
a b c
a b c
? ?
?
p 5 2
3 10 q
? ?
? p =
3
2
; q = 4
*Illustration 9 : The equations 5x
2
+ 12x + 13 = 0 and ax
2
+ bx + c = 0 (a,b,c ? R) have a common root, where
a, b, c are the sides of the ? ABC. Then find ?C .
(A) 45° (B) 60° (C) 90° (D) 30°
Solution : As we can see discriminant of the equation 5x
2
+ 12x + 13 = 0  is negative so roots of the
equation are imaginary. We know that imaginary roots always occurs in pair. So this equation can
not have single common roots with any other equation having real coefficients. So both roots are
common of the given equations.
Hence
a b c
(let)
5 12 13
? ? ? ?
then a = 5 ?, b = 12 ? , c = 13 ?
Now cosC =
2 2 2 2 2 2
a b c 25 144 169
0
2ab 2(5 )(12 )
? ? ? ? ? ? ?
? ?
? ?
? ?C = 90° Ans. (C)
Do yourself - 4 :
( i ) If x
2
+ bx + c = 0 & 2x
2
+ 9x + 10 = 0 have both roots, find b & c.
(i i) If x
2
– 7x + 10 = 0 & x
2
– 5x + c = 0 have a common root, find c.
(iii) Show that x
2
+ (a
2
– 2)x – 2a
2
= 0 and x
2
– 3x + 2 = 0 have exactly one common root for all a ? R.
7 . REMAINDER THEOREM :
If we divide a polynomial f(x) by (x – ?) the remainder obtained is f( ?). If f( ?) is 0 then (x – ?) is a factor of f(x).
Consider f(x) = x
3
– 9x
2
+ 23x – 15
f(1) = 0 ? (x – 1) is a factor of f(x).
f(x) = (x – 2)(x
2
– 7x + 9) + 3. Hence f(2) = 3 is remainder when f(x) is divided by (x – 2).
8 . SOLUTION OF R ATIONAL INEQUALITIES :
Let
f(x)
y
g(x)
?
be an expression in x where f(x) & g(x) are polynomials in x. Now, if it is given that
y > 0 (or < 0 or > 0 or < 0), this calls for all the values of x for which y satisfies the constraint. This solution set
can be found by following steps :
Step I : Factorize f(x) & g(x)  and generate the form :
1 2 k
p 1 2
n n n
1 2 k
m m m
1 2 p
(x a ) (x a ) ........(x a )
y
(x b ) (x b ) .......(x b )
? ? ?
?
? ? ?
where n
1
n
2
.........n
k
, m
1
,m
2
.......m
p
are natural numbers and a
1
,a
2
.......a
k
, b
1
,b
2
......b
p
are real
numbers. Clearly, here a
1
,a
2
......a
k
are roots of f(x)=0 & b
1
,b
2
,.....b
p
are roots of g(x) = 0.
JEEMAIN.GURU
```
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