Complex Numbers Quadratic Equations

COMPLEX NUMBERS & QUADRATIC EQUATIONS

1. INTEGRAL POWERS OF IOTA (i)
Definition
A number of the form where and is called a complex number and ‘i’ is called iota.
A complex number is usually denoted by z and the set of complex number is denoted by C.
i.e.,
For example, etc. are complex numbers.
(i) Euler was the first mathematician to introduce the symbol i (iota) for the square root of – 1 with property He also called this symbol as the imaginary unit.
(ii) For any positive real number a, we have

(iii) The property is valid only if at least one of a and b is non-negative. If a and b are both negative then .
Integral powers of iota (i)
Since hence we have , and . To find the value of first divide n by 4. Let q be the quotient and r be the remainder.
i.e., where

In general we have the following results , , where n is any integer.

2. REAL AND IMAGINARY PARTS OF A COMPLEX NUMBER
If x and y are two real numbers, then a number of the form is called a complex number. Here ‘x’ is called the real part of z and ‘y’ is known as the imaginary part of z. The real part of z is denoted by Re(z) and the imaginary part by Im(z).
If z = 3 – 4i, then Re(z) = 3 and Im(z) = – 4.

A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and purely imaginary if its real part is zero i.e., Re(z) = 0.

Illustration -1
Which of the following numbers is not a complex number?
and 1 – 5i
Solution
can be written as , which is of the form a + ib. Thus, is a complex number.
is not of the form a + ib. Thus, is not a complex number.
1 − 5i is of the form a + ib. Thus, 1 − 5i is a complex number.

 

 

Illustration -2
Express the following in the form of a + bi:
(i) (ii)
Solution
(i)
(ii)

Illustration -3
What are the real and imaginary parts of the complex number ?
Solution
The complex number can be written as , which is of the form a + ib.
Re z = a = and Im z = b =

3. ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS
Let two complex numbers bc and
Addition ( ) :
Subtraction :
Multiplication :
Division :
(where at least one of c and d is non-zero)
(Rationalization)

Properties of algebraic operations on complex numbers
Let and are any three complex numbers then their algebraic operations satisfy following properties :
(i) Addition of complex numbers satisfies the commutative and associative properties
i.e., and
(ii) Multiplication of complex numbers satisfies the commutative and associative properties.
i.e., and
(iii) Multiplication of complex numbers is distributive over addition
i.e., and

 

 

 

Illustration -4
If Z1 = 3 − i and Z2 = 1 + 2i, then write the complex number (Z1 + 2Z2 − 4) in the form a + ib and determine the values of a and b.
Solution
We have Z1 = 3 − i and Z2 = 1 + 2i
Z1 + 2Z2 − 4 = (3 − i) + 2 (1 + 2i) − 4
= 3 − i + 2 + 4i − 4
= 1 + 3i
Which is of the form a + ib
 a = 1 and b = 3

Illustration -5
What is the additive inverse of ?
Solution
Let ; Additive inverse of

Illustration -6
Simplify the following:

Solution

= (– 1)17 (i)17 + (i–1)8
= – (i)16+1 + (i8)–1
= –i + [i–4×2]–1 [i–4k+1 =i]
= –i + (1)–1 [i–4k=1]
= –i + 1
= (1 – i)

Illustration -7
If x + iy = (2 + 5i) (7 + i), then what are the values of x and y?
Solution
x + iy = (2 + 5i)(7 + i)
x + iy = (2 × 7 – 5 × 1) + i(2 × 1 + 5 × 7)
x+iy = (14 – 5) + i(2 + 35)
x + iy = 9 + i(37)
On equating the real and imaginary parts, we obtain
x = 9 and y = 37

 

 

Illustration -8
What is the value of ?
Solution
We know that

[i2 = – 1]

Illustration -9
What is the multiplicative inverse of 5 − 9i?
Solution
Let z = a + ib = 5 − 9i
Accordingly, a = 5 and b = −9
We know that

Thus, is the multiplicative inverse of 5 − 9i.

Illustration -10
Write the complex number in the form of a + ib.
Solution

[i2 = –1]
[i2 = –1]

Illustration -11
Find the real values of  such that is
(i) purely real and (ii) purely imaginary.
Solution
G.E.

(i) G.E. is purely real  Imaginary part of G.E. = 0

General solution is .
(ii) G.E. is purely imaginary  Real part of G.E. is zero.
G.S. is

Illustration -12
Find the value of x3 + 7×2 – x + 16, when x = 1+2i.
Solution
We have, x = 1 + 2i

Now, 

Hence, the value of the given polynomial when x = 1 + 2i is –17 + 24i

4. EQUALITY OF TWO COMPLEX NUMBERS
Two complex numbers and are said to be equal if and only if their real and imaginary parts are separately equal.
i.e., and
Complex numbers do not possess the property of order i.e., is not defined. For example, the statement makes no sense.

Illustration -13
For what values of x and y, z1 = (x + 1) − 10i and z2 = 19 + i(y − x) represent equal complex numbers?
Solution
Two complex numbers are equal if their corresponding real and imaginary parts are equal.
For the given complex numbers,
x + 1 = 19 and y − x = −10
⇒ x = 18 and y − 18 = −10
⇒ x = 18 and y = 8
Thus, the values of x and y are 18 and 8 respectively.

5. CONJUGATE COMPLEX NUMBER
If there exists a complex number  R, then its conjugate is defined as .

Hence, we have and .
Geometrically, the conjugate of z is the reflection or point image of z in the real axis.

 

6. PROPERTIES OF CONJUGATE
If and are existing complex numbers, then we have the following results:
(i)
(ii)
(iii)
(iv) In general
(v)
(vi)
(vii) purely real
(viii) purely imaginary
(ix) purely real
(x)

7. RECIPROCAL OF A COMPLEX NUMBER
For an existing non-zero complex number , the reciprocal is given by .
Illustration -14
Determine the conjugate and multiplicative inverse of .
Solution
Let
Accordingly, conjugate, and
Now, the multiplicative inverse is given by

Illustration -15
What is the conjugate of ?
Solution
Let
In order to find the conjugate of z, we first write it in the form of a + ib.
(By the multiplication of complex numbers)
On multiplying the numerator and the denominator with (7 + i), we obtain

Now,
Thus, the conjugate of the given complex number is .

Illustration -16
Find the conjugate of the following complex numbers
i) 3 + i ii) iii) iv) v)
Solution
We know that the complex conjugate is obtained by reversing the sign of imaginary part of the complex number. Hence,
i) Let z = 3 + i,
ii) Let , then
iii) Let , then
iv) Let then
v) Let .

 

Illustration -17
Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 –24i.
Solution
Here, we have (x – iy) (3 + 5i) is the conjugate of – 6 – 24i,

(conjugate of –6 –24i = –6 + 24i)
……….[1]
Equating real and imaginary parts of (1), we get 3x + 5y = – 6 ……….[2]
and 5x – 3y = 24 ……….[3]
on solving (2) and (3), we get x = 3, y = – 3

 

 

8. MODULUS OF A COMPLEX NUMBER
Modulus of a complex number is defined by a positive real number given by where a, b real numbers. Geometrically |z| represents the distance of point P from the origin, i.e. |z| = OP.

If the corresponding complex number is known as unimodular complex number. Clearly z lies on a circle of unit radius having centre (0, 0).

9. PROPERTIES OF MODULUS
(i) and |z|  .
(ii) and
(iii)
(iv)
(v) .
In general
(vi)
(vii)
(viii)
or
(ix) is purely imaginary or

(x)
(Law of parallelogram)

Illustration -18
What is the modulus of z = (1 + i)10?
Solution
Modulus,
It can be written as

Continuing in this manner, we can write
(10 times)

Now,

Illustration -19
The modulus of the complex number
Solution
Here, we have (x – iy) (3 + 5i) is the conjugate of – 6 – 24i,

(conjugate of –6 –24i = –6 + 24i)
……….[1]
Equating real and imaginary parts of (1), we get 3x + 5y = – 6 ……….[2]
and 5x – 3y = 24 ……….[3]
on solving (2) and (3), we get x = 3, y = – 3

10. ARGUMENT OF COMPLEX NUMBER
Let be any complex number. If this complex number is represented geometrically by a point P, then the angle made by the line OP with real axis is known as argument or amplitude of z and is expressed as

arg Also, argument of a complex number is not unique, since if be a value of the argument, so also is where .

11. PRINCIPAL VALUE OF ARG (Z)
The value of the argument, which satisfies the inequality is called the principal value of argument, where (acute angle)

and principal values of argument z will be and as the point z lies in the 1st ,
2nd , 3rd and 4th quadrants respectively.

12. PROPERTIES OF ARGUMENTS
(i) or 1 or – 1)
In general , (ii)
(iii) or 1 or – 1)
(iv) or 1 or – 1)
(v) or 1 or – 1)
(vi) If then , where
(vii)
(viii)
(ix)
(x) (If z is purely imaginary)
(xi)
where and
(xii) The general value of is .

13. VALUE OF ARGUMENTS OF COMPLEX NUMBERS
Complex number Value of arguments
+ve Re (z) 0
–ve Re (z)
+ve Im (z)
–ve Im (z)
– (z)
respectively
(iz)
–(iz)

n. arg (z)

arg (z1) + arg (z2)

arg (z1) – arg (z2)

 

 

Illustration -20
If z and w are two non zero complex numbers such that | z | = | w | and amp z + amp
w = , then prove that z = –
Solution

If amp z =  ….(i)
amp = – 
amp =  –  ….(ii)
Hence amp z + amp (– ) = 
but w = –  z = –

Illustration -21
If z = + then amp. z equals
Solution
z1 = + 2i ; z2 = 1 + i;
Hence z =
amp z = amp R – [8 amp + 6 amp ] = 0 – = + = 2+
 amp z =

Illustration -22
If z = then find the principal value of arg z = ?
Solution
amp z = (tan–11 + tan–12 + tan–13) – [ tan–1(–1) + tan–1(–1/2) + tan–1(–1/3)] + 2k
=
Hence amp Z = –

Illustration -23
If z = 1+i 3, then find the principal value of arg(z) is
Solution
Here x = 1, y = 3 both are positive ;
Therefore tan  =  tan  =

Illustration -24
For any two complex numbers z1 and z2 we have |z1|2 + |z2|2 = |z1 + z2|2, then is
(A) 2 (B) 3 (C) 1 (D) 0
Solution

if and only if real part of z1¬ = 0.

14. SQUARE ROOT OF A COMPLEX NUMBER
Let be a complex number,
Then , for b> 0
, for b < 0.
To find the square root of replace i by – i in the above results.

15. LOGARITHM OF A COMPLEX NUMBER

Illustration -25
Find the square root of i.
Solution
Let
…..(2)
Now

Solving (1) and (3), we get ,
From (2) : 2xy is positive  x and y are of same sign.

Hence
Note. Similarly

 

Illustration -26
Find the square root of (–11 – 60i)
Solution
Let
…..(2)
Now,
Solving (1) and (3), we get x2 = 25 and y2 = 36  x =  5 and y =  6
From (2) : 2 xy is negative  x and y are opposite signs.
Hence

Illustration -27
Find the square roots of
Solution
Let Then,

and
On solving we get and and
Hence,

 

 

Illustration -28
Find the square roots of 8  15i is
Solution
Here y = 15 < 0
 = 
=  (5  3i).

16. VARIOUS REPRESENTATIONS OF A COMPLEX NUMBER

1. GEOMETRICAL REPRESENTATION (CARTESIAN REPRESENTATION)
The complex number is represented by a point P whose coordinates are referred to rectangular axes and which are called real and imaginary axis respectively. This plane is called argand plane or argand diagram or complex plane or Gaussian plane.

Distance of any complex number from the origin is called the modulus of complex number and is denoted by |z|, i.e., .
Angle of any complex number with positive direction of x–axis is called amplitude or argument of z.
i.e.,
2. Trigonometrical (Polar) representation
In  OPM, let , then and Hence z can be expressed as
where r = |z| and  = principal value of argument of z.
For general values of the argument

Sometimes is written in short as .

Illustration -29
Represent the complex number in polar form.
Solution
Let z = r (cosθ + i sinθ) be the polar form of the complex number .
and r sinθ = −1
On squaring and adding, we obtain
r2(cos2 + sin2) =
r2= 3 + 1 = 4
r=  2
r = 2 (r cannot be negative)
Now,
and
Here, cosθ is positive and sinθ is negative. Hence, θ lies in quadrant IV.

Thus, the required polar form of the given complex number is

Illustration -30
What are the modulus and the argument of the complex number ?
Solution
Let r(cos + i sin) =
Which gives,
and
On squaring and adding, we obtain
r2(cos2 + sin2) =

r = 1
r = 1( r > 0)
Now, and
Here, both cosθ and sinθ are negative.
Hence, θ lies in quadrant III.

Thus, the modulus and argument of the given complex number are 1 and respectively

17. VECTOR REPRESENTATION
If P is the point (a, b) on the argand plane corresponding to the complex number .
Then ,
 and
arg (z) = direction of the vector

18. EULERIAN REPRESENTATION (EXPONENTIAL FORM)
Since we have = and thus z can be expressed as , where and
arg (z).

 

Illustration -31
If ei = cos  + i sin  then for the  ABC, eiA . eiB . eiC is
(A) –i (B) 1
(C) –1 (D) none of these
Solution
eiA.eiB.eiC = ei(A + B + C) = cos (A + B + C) + i sin (A + B + C) = –1

Illustration -32
Real part of is
(A) (B) (C) (D)
Solution
(A)

 Real part of is

Illustration -33
The amplitude of is equal to
(A) (B) (C) (D)

Solution
(B)
Let

.

Illustration -34
If then lies in
(A) I quadrant (B) II quadrant (C) III quadrant (D) IV quadrant
Solution
(C)

Now


lies in III quadrant.

 

Illustration -35
If , then in value of is
(A) –i (B) 1 (C) –1 (D) None of these

Solution
(C)
[ ]
= .

 

19. APPLICATION OF COMPLEX NUMBERS IN GEOMETRY

1. DISTANCE FORMULA
The distance between two points and is given by = |affix of Q – affix of P|

2. SECTION FORMULA
If R(z) divides the line segment joining and in the ratio then
(i) For internal division
(ii) For external division

3. EQUATION OF THE PERPENDICULAR BISECTOR
If and are two fixed points and is moving point such that it is always at equal distance from and then locus of is perpendicular bisector of PQ

i.e., PR = QR or

After solving,

4. EQUATION OF A STRAIGHT LINE
(i) Parametric form
Equation of a straight line joining the point having affixes and is
(ii) Non-parametric form
Equation of a straight line joining the points having affixes and is
 .
(iii) (a) Three points and are collinear if,

(b) If three points are collinear then slope of AB = slope of BC = slope of AC .
Concyclic points
Four points
are concyclic if and only if is purely real.
General equation of a straight line
The general equation of a straight line is of the form , where a is complex number and b is real number.

Slope of a line
The complex slope of the line is and real slope of the line is .

Length of perpendicular
The length of perpendicular from a point to the line is given by or
5. TRIANGLE
Let ABC be a triangle with vertices and then
i) Centroid of the is

ii) Incentre of the is
where,
iii) Circumcentre of the is

also
iv) Orthocentre of the is
also

or z =
If is an equilateral then
i)
ii)
iii)
iv) If is circumcentre of then

Note: i) If then the origin and form an equilateral triangle.
ii) If then the origin and form an isosceles triangle.
iii) If are the vertices of an isosceles right angle triangle and right angled at then

 

Area of a Triangle
i) The area of triangle whose vertices are is .
ii) The area of triangle whose vertices are is .
iii) The area of triangle whose vertices are is , where is complex cube roots of unity

6. CIRCLE
Equation of a circle

(i) The equation of a circle whose centre is at point having affix and radius r is
(ii) If the centre of the circle is at origin and radius r, then its equation is .
(iii) represents interior of a circle and represents exterior of the circle .
General equation of a circle
The general equation of the circle is where a is complex number and .
 Centre and radius are – a and respectively.
Equation of circle in diametric form
If end points of diameter represented by and and be any point on the circle then, .
which is required equation of circle in diametric form.
Illustration -36
Let = c, b  0 be a line in the complex plane, where is the complex conjugate of b. If z1 is the reflection of a point z2 through the line, then is equal to
(A) (B) b (C) 0 (D) c
Solution
The given line = c, …(1)
b  0 is the perpendicular bisector of the line joining the points z1 and z2.
Then |z – z1|2 = |z – z2|2
(z–z1) =(z–z2)
z + (z2 – z1)+ |z1|2 – |z2|2 =0 …(2)
(1) and (2) are the same. Therefore
= k(say)
Now +  c = k(z2 – z1) + kz2 k(|z2|2  |z1|2)
= k = 0  .

Illustration -37
The centre and radius of the circle is
(A)  1  i , 3 (B) 1 + i , 3 (C) 2  i , 4 (D) 2 + i , 4

Solution
Given equation can be rewritten as
So, it represents a circle with centre at –1 –i and
radius = .

Illustration -38
If z1, z2 and z3 ( in anticlockwise sense) represents the vertices of a triangle, find the centroid, incentre, circumcentre and the orthocentre of the triangle.

Solution
Let G be the centriod and let the line joining A and G meets the line BC at the point D. We have
BD = DC
D 
G divides AD internally in ratio 2 : 1
 G  

Let I be the incentre and let the line connecting A and I meet the line BC at D1. We have

and =

=

Let ‘O’ be the circum-centre and let the line connecting A and O meet the
line BC at D2.
We have
and
 D2 = and
O =
=

Let ‘P’ be the orthocentre and let the line connecting the points A and P meet the line BC and D3.
We have,
and
 D3 =
and P =
=

 

Illustration -39
Let  and  be two fixed non-zero complex numbers and z a variable complex number. If the two straight lines and are mutually perpendicular, then prove that = 0

Solution
Slope of the first line = – ; Slope of the second line = –
Lines will be perpendicular if – + = 0
 = 0.

Illustration -40
Find the centre and radius of the circle .

Solution
Given equation can be rewritten as
So, it represents a circle with centre at –1 –i and
radius = .

Illustration -41
If z1, z2, z3 are complex numbers such that , show that the points represented by z1, z2, z3 lie on a circle passing through the origin.

 

Solution
Since P(z1), Q(z2), R(z3) and S(z4) are concyclic points,
PSQ = PRQ
 arg. = arg
 arg = 0
 = real

If z4 = 0 + i0, then = real …(1)
We have from which z3 = …(2)
From (1) and (2), = real  = real
 = real, which is true.
Therefore z1, z2, z3 and the origin are concyclic.

Illustration -42
The complex numbers z = x + iy which satisfy the equation lie on
(A) x–axis (B) the straight line y = 5
(C) a circle passing through the origin (D) None of these.

Solution

 z would lie on the right bisector of the line segment connecting the points 5i and – 5i .
Thus z would lie on the x–axis.
Hence (A) is the correct answer.

Illustration -43
The points z1, z2, z3, z4 in the complex plane are the vertices of a parallelogram taken in order if and only if
(A) z1+z4=z2+z3 (B) z1+z3=z2+z4
(C) z1+z2=z3+z4 (D) None of these.

 

Solution
In parallelogram, diagonals bisects each other, thus mid–point of AC and BD should be same .
  z1+ z3 = z2 +z4
Hence (B) is the correct answer.

Illustration -44
If z = x + iy, and then w = 1 implies that in the complex plane
(A) z lies on the imaginary axis (B) z lies on the real axis
(C) z lies on the unit circle (D) None of these.

Solution

As |w| = 1  | z – i| = |1 – iz| = |z + i| ( as 1/i = –i)
 z lies on the right bisector of the line segment connecting the points i and –i.
Thus ‘z’ lies on the real axis.
Hence (B) is correct answer.

Illustration -45
If the area of the triangle on the complex plane formed by the points z, iz and z + iz is 50 square units, then |z| is
(A) 5 (B) 10
(C) 15 (D) none of these
Solution
|z|2 = 50  |z|2 = 100  |z| = 10
Hence (B) is the correct answer.

19. ROTATION THEOREM
Rotational theorem i.e., angle between two intersecting lines. This is also known as coni method.

Let and be the affixes of three points A, B and C respectively taken on argand plane.
Then we have
and =
and let arg arg
and arg
Let ,
= arg arg
= arg arg = arg
or angle between AC and AB = arg

Complex number as a rotating arrow in the argand plane

Let ..…(i)
be a complex number representing a point P in the argand plane.
Then and
Now consider complex number

or {From (i)}
Clearly the complex number represents a point Q in the argand plane, when and .
Clearly multiplication of z with rotates the vector through angle in anticlockwise sense. Similarly multiplication of z with will rotate the vector in clockwise sense.
(i) If and are the affixes of the points A,B and C such that and .
Therefore, .

Then will be obtained by rotating through an angle  in anticlockwise sense, and therefore,
or or
(ii) If A, B and C are three points in argand plane such that and then use the rotation about A to find , but if use coni method.
(2) If four points are con-cyclic then
= real
or

 

 

Illustration -46
The complex number z =1+i is rotated through an angle 3/2 in anticlockwise direction about the origin and stretched by additional unit, then the new complex number is
(A) – – i (B) – i
(C) 2– i (D) none of these

Solution
[D]
If z1 be the new complex number then |z1| = |z| + = 2
Also  z1 = z. 2
= 2(1 + i) ( 0 – i) = – 2i +2 = 2( 1– i).

Illustration -47
Rotate A (7,6) about B(4,2) by 90º anti clockwise. Find its new position A2

Solution
We need to find the position vector of A2 or where O is the origin.
Point B will act as tail and point A will act as head. Note that = 7 + 6i while = 4+2i
Hence Vector to be rotated is = – = (7+6i) – (4+2i) = 3+4i
The new vector = cis 90º = (3+4i) (0 + i) = -4 +3i The position vector of A2 or
= + = (4+2i) + (-4+3i) = 0 + 5i hence the new position A2 is (0, 5)

Illustration -48
Consider a square ABCD such that z1, z2, z3 and z4 represent its vertices A, B, C and D respectively. Express ‘z3’ and ‘z4’ in terms of z1 and z2 .

Solution

 z3 = z1 +( z2 – z1)( 1+i)
Similarly ,
 z4 = z1 +i( z2 – z1)

20. TRIANGLE INEQUALITIES
In any triangle, sum of any two sides is greater than the third side and difference of any two side is less than the third side. By applying this basic concept to the set of complex numbers we are having the following results.
(1)
(2)
(3)
(4)
Note:
i) If then
ii) If then greatest value of is
iii) The least value of is
Illustration -49
If |z – 3i| = 1 then show that 2 |z| 4.
Solution
Using triangle inequality we can say that
| | z1 | – | z2 | | | z1 + z2 | | z1 | + | z2 | …..(1)
Let z1 =z while z2 =-3i .Also let |z| =r. Note that |-3i| = 3.
From (1) we can conclude that | r – 3| 1 r + 3
Solving both sides of above inequality we get 2 r 4.
Try to interpret this result geometrically. Hint: |z – 3i| = 1 represents a circle)

Illustration -50
For any complex number z, find the minimum value of .
Solution
We have, for ,
Thus, minimum value of is 2.
Try to interpret this result geometrically.
(Hint: represents sum of distances of a variable point z from origin and from (0,2))

Illustration -51
If then find the greatest value of .
Solution

Hence, the greatest value of is 6
Try to interpret this result geometrically.
(Hint: represents a circular disc)

 

21. DE’ MOIVRE’S THEOREM
(1) If is any rational number, then .
(2) If
……
then
where .
(3) If and n is a positive integer, then ,
where .

Deductions
If then
(i)
(ii)
(iii)
(iv)
This theorem is not valid when n is not a rational number or the complex number is not in the form of

Illustration -52

(A) (B) (C) (D) None of these
Solution
(c)

(where )
(Using De Moivre’s theorem)
=
Putting , 1
we get
and
Therefore

 

 

Illustration -53
If , then can take the value
(A) (B) (C) (D) i
Solution

(using De Moivre’s theorem)

Illustration -54
The value of is
(A) (B)
(C) (D)
Solution
(B)
[By de-movire’s theorem].

22. STANDARD LOCI IN THE ARGAND PLANE
If z is a variable point and are two fixed points in the argand plane, then
(i)  Locus of z is the perpendicular bisector of the line segment joining and .
(ii) = constant
 Locus of z is an ellipse
(iii)
 Locus of z is the line segment joining and
(iv)
 Locus of z is a straight line joining and but z does not lie between and .
(v)
 Locus of z is a hyperbola.
(vi)  Locus of z is a circle with and as the extremities of diameter.
(vii)  Locus of z is a circle.
(viii) arg  Locus of z is a segment of circle.
(ix) arg =  Locus of z is a circle with and as the vertices of diameter.
(x) arg = 0 or  Locus of z is a straight line passing through and .

Illustration -55
When is purely imaginary, the locus described by the point in the Argand diagram is a
(A) Circle of radius (B) Circle of radius
(C) Straight line (D) Parabola
Solution
(A)
Given that Im
Let  =

If it is purely imaginary then real part must be equal to zero.
 
Which is a circle and its radius is given by

Therefore Argand diagram is circle of radius .

Illustration -56
If then the locus described by the point in the Argand diagram is a
(A) Straight line (B) Circle
(C) Parabola (D) None of these
Solution
(B)

Putting 


 .
Which is the equation of a circle.

Illustration -57
The locus of the points z which satisfy the condition arg = is
(A) A straight line (B) A circle
(C) A parabola (D) None of these
Solution
(B)
We have
Therefore
Hence

 
Which is obviously a circle.

Illustration -58
If the imaginary part of is –2, then the locus of the point representing in the complex plane is
(A) A circle (B) A straight line (C) A parabola (D) None of these
Solution
(B)
We have

But it is given that imaginary part of is – 2
 . Which is a straight line.

23. ROOTS OF A COMPLEX NUMBER
1. nth ROOTS OF COMPLEX NUMBER z1/n
Let be a complex number. By using De’moivre’s theorem nth roots having n distinct values of such a complex number are given by

where

2. PROPERTIES OF THE ROOTS OF z1/n
(i) All roots of z1/n are in geometrical progression with common ratio
(ii) Sum of all roots of z1/n is always equal to zero.
(iii) Product of all roots of
(iv) Modulus of all roots of z1/n are equal and each equal to or
(v) Amplitude of all the roots of z1/n are in A.P. with common difference
(vi) All roots of z1/n lies on the circumference of a circle whose centre is origin and radius equal to Also these roots divides the circle into n equal parts and forms a polygon of n sides.

 

3. THE nth ROOTS OF UNITY
The nth roots of unity are given by the solution set of the equation

where .

4. PROPERTIES OF nth ROOTS OF UNITY
(i) Let the nth roots of unity can be expressed in the form of a series i.e., Clearly the series is G.P. with common ratio i.e.,
(ii) The sum of all n roots of unity is zero i.e.,
(iii) Product of all n roots of unity is
(iv) Sum of pth power of n roots of unity

(v) The n, nth roots of unity if represented on a complex plane locate their positions at the vertices of a regular polygon of n sides inscribed in a unit circle having centre at origin, one vertex on positive real axis.

5. CUBE ROOTS OF UNITY
Cube roots of unity are the solution set of the equation 

 , where
Therefore roots are or .

Alternative



If one of the complex root is then other root will be or vice-versa.

Properties of cube roots of unity
(i)
(ii)
(iii)
(iv) The cube roots of unity, when represented on complex plane, lie on vertices of an equilateral triangle inscribed in a unit circle having centre at origin, one vertex being on positive real axis.
(v) A complex number for which or can always be expressed in terms of
(vi) Cube root of – 1 are .

6. FOURTH ROOTS OF UNITY
The four, fourth roots of unity are given by the solution set of the equation
 
Fourth roots of unity are vertices of a square which lies on coordinate axes.

Illustration -59
The two numbers such that each one is square of the other, are
(A) (B) (C) (D)
Solution
(D)
Since and

Illustration -60
If is a cube root of unity, then =
(A) 1 (B) 0 (C) 2 (D) 4

Solution
(D)
If is a complex cube root of unity then and , therefore

24. Polynomial
Algebraic expression containing many terms of the form , n being a non-negative integer is called a polynomial. i.e., , where x is a variable, are constants and
Example : , .
(1) Real polynomial: Let be real numbers and x is a real variable.
Then is called real polynomial of real variable x with real coefficients.
Example : etc. are real polynomials.
(2) Complex polynomial: If be complex numbers and x is a varying complex number.
Then is called complex polynomial of complex variable x with complex coefficients.
Example : etc. are complex polynomials.
(3) Degree of polynomial: Highest power of variable x in a polynomial is called degree of polynomial.
Example : is a n degree polynomial.
is a 3 degree polynomial.
is single degree polynomial or linear polynomial.
is an odd linear polynomial.
A polynomial of second degree is generally called a quadratic polynomial. Polynomials of degree 3 and 4 are known as cubic and biquadratic polynomials respectively.
(4) Polynomial equation : If f(x) is a polynomial, real or complex, then f(x) = 0 is called a polynomial equation.

25. QUADRATIC EQUATION
An equation of the form ax2 + bx + c = 0, where a  0 and a, b, c are real numbers, is called a quadratic equation. The numbers a, b, c are called the coefficients of the quadratic equation.
Example :
Quadratic equations are of two types
(1) Purely quadratic equation : A quadratic equation in which the term containing the first degree of the unknown quantity is absent is called a purely quadratic equation.
i.e. where a, c  C and a  0
(2) Adfected quadratic equation : A quadratic equation which contains terms of first as well as second degrees of the unknown quantity is called an adfected quadratic equation.
i.e. where a, b, c  C and a  0, b  0.
Note:
An equation of degree n has n roots, real or imaginary.
Surd and imaginary roots always occur in pairs in a polynomial equation with real coefficients i.e. if 2 – 3i is a root of an equation, then 2 + 3i is also its root. Similarly if is a root of given equation, then is also its root.
An odd degree equation has at least one real root whose sign is opposite to that of its last term (constant term), provided that the coefficient of highest degree term is positive.
Every equation of an even degree whose constant term is negative and the coefficient of highest degree term is positive has at least two real roots, one positive and one negative.

26. ROOT OF A QUADRATIC EQUATION
The values of variable x which satisfy the quadratic equation is called roots of quadratic equation.
Solution of quadratic equation
(1) Factorization method
Let .
Then and will satisfy the given equation.
Hence, factorize the equation and equating each factor to zero gives roots of the equation.
Example :
 ;

(2) Sri Dharacharya method : By completing the perfect square as

Adding and subtracting , which gives, .
Hence the quadratic equation (a0) has two roots, given by ,
Every quadratic equation has two and only two roots.
A root of the quadratic equation is a number  (real or complex) such that a2 + b + c = 0. The roots of the quadratic equation are given by x = .

24. DISCRIMINANT OF A QUADRATIC EQUATION
The quantity D (D= b2  4ac) is known as the discriminant of a quadratic equation.
• The quadratic equation has real and equal roots if and only if D = 0
i.e. b2  4ac = 0.
• The quadratic equation has real and distinct roots if and only if D > 0
i.e. b2  4ac > 0
The quadratic equation has complex roots with non-zero imaginary parts if and only if D < 0
i.e. b2  4ac < 0. If p + iq (p and q being real) is a root of the quadratic equation where i = , then p – iq is also a root of the quadratic equation.

28. IDENTITY
If the quadratic equation is satisfied by more than two distinct numbers (real or complex), then it becomes an identity i.e. a = b = c = 0. For example
is satisfied by three value of x which are a, b and c. Hence this is an identity in x.
Quadratic Identity
will be an identity ( or can have more than two solutions) if a =0, b =0, c= 0,.

Illustration – 61
Find the roots of x2 – 32x – 900 = 0.
Solution
x2 – 50x + 18x – 900 = 0
x(x – 50) + 18(x – 50) = 0
(x – 50) (x + 18) = 0
x = 50, –18

 

Illustration – 62
Solve the quadratic equation .
Solution
The given quadratic equation is .
The discriminant of this equation is
b2 − 4ac =
Thus, the solution of the given equation is

Illustration – 63
Find the roots of

Solution

i.e.
or (using factorization method)

;

29. NATURE OF ROOTS OF QUADRATIC EQUATIONS
In quadratic equation , the term is called discriminant of the equation, which plays an important role in finding the nature of the roots. It is denoted by  or D.
(1) If a, b, c  R and a  0, then : (i) If D < 0, then equation has non-real complex roots.
(ii) If D > 0, then equation has real and distinct roots, namely , and then …..(i)
(iii) If D = 0, then equation has real and equal roots
and then …..(ii)
To represent the quadratic expression in form (i) and (ii), transform it into linear factors.
(iv) If D  0, then equation has real roots.
(2) If a, b, c  Q, a  0, then :
(i) If D > 0 and D is a perfect square  roots are unequal and rational.
(ii) If D > 0 and D is not a perfect square  roots are irrational and unequal.
(3) Conjugate roots : The irrational and complex roots of a quadratic equation always occur in pairs. Therefore
(i) If one root be then other root will be .
(ii) If one root be then other root will be .
(4) If D1 and D2 be the discriminants of two quadratic equations, then
(i) If , then
(a) At least one of and . (b) If then
(ii) If , then
(a) At least one of and . (b) If then .
Nature of the roots of two Quadratic Equations
If 1 and 2 are the discriminants of two quadratic equations and , such that
 then there will be at least two real roots for the equation or .
 If , then there will be atleast two imaginary roots for the equation or .
 If , then the equation will have two real roots and two imaginary roots.

 

 If , then the equation has either four real roots or no real roots.
 If such that and or and then the equation will have two equal roots and two distinct roots. .
 If where and or and then the equation will have two equal real roots.
 If such that and then the equation will have two pairs of equal roots .

30. ROOTS UNDER PARTICULAR CONDITIONS
For the quadratic equation .
(1) If  roots are of equal magnitude but of opposite sign.
(2) If  one root is zero, other is .
(3) If  both roots are zero.
(4) If  roots are reciprocal to each other.
(5) If  roots are of opposite signs.
(6) If  both roots are negative, provided .
(7) If  both roots are positive, provided .
(8) If sign of a = sign of b  sign of c  greater root in magnitude, is negative.
(9) If sign of b = sign of c  sign of a  greater root in magnitude, is positive.
(10) If  one root is 1 and second root is c/a.
(11) If , then equation will become an identity and will be satisfied by every value of x.
(12) If and b, c  I and the root of equation are rational numbers, then these roots must be integers.
NOTE:
If an equation has only one change of sign, it has one +ve root and no more.
If all the terms of an equation are +ve and the equation involves no odd power of x, then all its roots are complex.

Illustration – 64
Find the nature of the root of the equation 2×2 – 6x + 7 = 0
Solution
D = (–6)2 – 4(2) (7)
= 36 – 56 = –20 < 0
given equation have no real roots
Illustration -65
Find the nature of the root of the equation (x – a)(x – b) = c2
Solution
Given roots of x2 – (a + b) x + ab – c2 = 0
Here D = (a + b)2 – 4(ab – c2)
= (a + b)2 – 4ab + c2
= (a – b)2 + c2 0
the roots of given equation are always real

Illustration -66
If the roots of the equation are equal then the value of m is
Solution



31. FORMATION OF THE QUADRATIC EQUATION WITH ROOTS ARE  AND 

A quadratic equation whose roots are  and  is given by
 i.e.

Illustration – 67
Let us find the quadratic equation whose roots are 3 and – 2.
Solution
The required quadratic equation is
= 0
x2 – x – 6 = 0.

Illustration – 68
Find the quadratic equation whose roots are and .
Solution
We have sum of roots = 4, and
Product of roots = 1;
The required quadratic equation is
x2 – 4x + 1 = 0 .

Illustration – 69
Frame equation whose roots are 2 i .
(A) x2– 4x + 7 = 0 (B) x2– 3x + 2 = 0 (C) 2×2 + 4x +3 = 0 (D) 3×2– 5x + 9 = 0

Solution
A
Sum of roots is 4.
Product of roots = =4+3= 7
equation : x2– 4x+7 = 0

32. RELATION BETWEEN THE ROOTS AND COEFFICIENTS

If  and  are the roots of quadratic equation , (a  0) then
Sum of roots
Product of roots
If roots of quadratic equation (a  0) are  and  then
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)

Illustration – 70
Find the sum of the following quadratic equation 5×2 + 6x + 7 = 0.
(A) (B) (C) (D)
Solution
C
Given quadratic equation is
Here, a =5, b = 6, c = 7
Sum of roots =

Illustration – 71
Find the equation whose roots are reciprocals of the roots of 5×2 + 6x + 7 = 0 .
Solution
i.e., 7×2 + 6x + 5 = 0

Illustration – 72
Find the equation whose roots are equal but opposite in sign to the roots of 2×2 + 3x + 4 = 0.
Solution
i.e., 2×2 – 3x + 4 = 0

Illustration – 73
Find the equation whose roots, are 3 times the roots of the equation x2 – 5x + 6 = 0.
Solution
i.e., x2 – 15x + 54 = 0.

Illustration – 74
If  and  are the roots of ax2 + bx + c = 0, then the values of and are
(A) (B) (C) (D) None of these
Solution
A
Since are the roots of ax2 + bx + c = 0
or .

Also .

33. SYMMETRIC FUNCTION OF THE ROOTS

A function of and is said to be a symmetric function, if it remains unchanged when and are interchanged.
For example, is a symmetric function of and whereas is not a symmetric function of and .
In order to find the value of a symmetric function of and , express the given function in terms of and . The following results may be useful.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

Illustration -75
If sum of the roots of the equation ax2 + bx + c =0 is equal to the sum of the squares of their reciprocals, show that bc2 , ca2 , ab2 are in AP

Solution
Let α and β be the roots of the equation ax2 + bx + c =0 .
Then, α + β= -b/a ………..(1)
α β= c/a ………….(2)
Now, α + β = 1/α 2 + 1/β2
or, α + β = (α2 +β2)/( α β)2
or, α + β =( (α +β)2 – 2αβ ) / ( α β)2 ……..(3)
Now put the values of (α+ β) and ( α β) from (1) and (2) in equation (3).
Thus, we get
-b/a = (b2 -2ac)/ c2
or, -bc2 = b2a -2ca2
or, bc2 + ab2 = 2ca2
Hence bc2 , ca2 , ab2 are in A.P.
Proved.

 

 

 

 

Illustration -76
If α and β are the roots of the equation x2 +px +q =0 and x2n + pn xn + qn =0, where n is an even integer, prove that α/β , β/α are the roots of the equation xn +1 + (x+1)n =0.
Solution
α and β are the roots of the equation x2 +px +q =0
So, α + β =-p………(1)
α β =q ……………..(2)
Since α and β are the roots of the equation x2n + pn xn + qn =0
α2n + pn αn + qn =0
or,( αn )2 + pn αn + qn =0……….(3)
and,
( βn )2 + pn βn + qn =0…………..(4)
From (3) and (4) we see that are the roots of y2 + pn y + qn =0
So, αn + βn =-pn………(5)
α nβn =qn ……………..(6)
From (1), we have
α + β =-p
or, (α + β)n =(-p)n = pn (n is even)
or, (α + β)n =-(-pn )= -(αn + βn ) (from 5)
or, αn + βn + (α + β)n = 0 ……..(7)
Dividing (7) by αn , we get
(α/ β)n + 1 + ((α/ β) +1 )n =0 ………(8)
Dividing (7) by βn , we get
( β/ α)n + 1 + (( β/α) +1 )n =0 ………(9)
From (8) and (9) we see that α/ β and β/ α are the roots of
xn +1 + (x+1)n =0.

34. COMMON ROOTS OF QUADRATIC EQUATIONS
1. ONLY ONE ROOT IS COMMON
Let  be the common root of quadratic equations and .
 ,
By Crammer’s rule :
or
 ,
 The condition for only one root common is
2. BOTH ROOTS ARE COMMON
Then required condition is .
Note: To find the common root of two equations, make the coefficient of second degree term in the two equations equal and subtract. The value of x obtained is the required common root. Two different quadratic equations with rational coefficient cannot have single common root which is complex or irrational as imaginary and surd roots always occur in pair.
3. PROPERTIES OF ROOTS OF THE EQUATION
• If a and c are of the same sign then is +ve and hence roots have same signs
• If a and c are of the opposite signs then is -ve and hence roots have opposite sign
• If then the roots are reciprocal to each other.
• If both the roots are -ve then a, b, c will have the same sign
• If both roots are +ve then a, c will have the same sign different from the sign of b
• If a + b +c = 0 then the roots are 1 and
• If a + c = b then the roots are -1 and
• If the roots are in the ratio m : n then (or)
• If one root is square of the other then
• If one root is equal to the nth power of the other root then .
• If roots differ by k then .

Illustration -77
If and have a common root then the value of a is

Solution
Condition for the common roots is

Illustration -78
If , have a common root then the condition is
Solution
Let be the common root
….(1)
….(2)
(1) – (2)
substitute in (1)
Note: If and have the same roots then
Illustration -79
If the equation and have both roots common, then the value of is
Solution

since the roots of (1) and (2) are equal,

Illustration -80
If x2 – 6x + 5 = 0 and x2 – 3kx + 35 = 0 have a common root find k.
Solution
Let be common root
by eliminate ,

900 = 45(– k + 2) (k – 14) 20 = –k2 + 16k – 28
k2 – 16k + 48 = 0 (k – 4) (k – 12) = 0
k = 4 or 12

Illustration -81
If x2 + bx + c = 0, x2 + cx + b = 0 (b c), have a common root then show that b + c + 1 = 0.
Solution
Let be the common root
… (1)
… (2)
(1) – (2)
substitute in (1)

Illustration -82
Find the value of p if the equations and have a common roots.
Solution
Let a be a common root, then
…(1)

and

Illustration -83
If one root of the equation is double the other, then the relation between a, b, c is
Solution
Condition is given

35. PROPERTIES OF QUADRATIC EQUATION
(1) If f(a) and f(b) are of opposite signs then at least one or in general odd number of roots of the equation lie between a and b.

 

 

(2) If then there exists a point c between a and b such that , .

 

 

As is clear from the figure, in either case there is a point P or Q at where tangent is parallel to x-axis i.e. at .
(3) If  is a root of the equation then the polynomial is exactly divisible by or is factor of .
(4) If the roots of the quadratic equations , are in the same ratio then .
(5) If one root is k times the other root of the quadratic equation then .

Illustration -84
Find the values of the parameter l for which the roots of are
(i) opposite in sign
(ii) equal in magnitude but opposite in sign.
(iii) positive
(iv) negative
(v) one root is greater than 3 and other is smaller than 3.

Solution

Also,
(i) This means that O lies between the roots of the equation
and i.e U and

.
(ii) This means that the sum of the roots is zero
and
i.e and l = 1.
Which does not belong to
is null se .
(iii) This means that both the roots are greater zero

U

(-5, -1]
(iv) This implies that both the roots are less than zero

U [4, )

(v) This means 3 lies between the roots
and
and
and

36. THE SIGN OF THE EXPRESSION

Let , be a quadratic expression. Since,
……(i)
The following is true from equation (i)
(i) for all values of if and only if and .
(ii) if and only if and .
In this case , if and only if
(iii) If and (<0), then

(iv) If , then has a minimum (maximum) value at and this value is given by
.

SIGN OF QUADRATIC EXPRESSION
Let or
Where a, b, c  R and a  0, for some values of x, f(x) may be positive, negative or zero. This gives the following cases :
(i) a > 0 and D < 0, so for all i.e., is positive for all real values of x.
(ii) a < 0 and D < 0, so for all x  R i.e., f(x) is negative for all real values of x.
(iii) a > 0 and D = 0, so for all x  R i.e., f(x) is positive for all real values of x except at vertex, where .
(iv) a < 0 and D = 0, so for all x  R i.e. f(x) is negative for all real values of x except at vertex, where .
(v) a > 0 and D > 0, let have two real roots  and  , then for all and for all .
(vi) and , let have two real roots  and  . Then for all and for all
Illustration -85
Discuss the sign of the expression x2 – 5x + 4 when x is real.
Solution
given expression x2 – 5x + 4
D = 25 – 24 = 1
x2 – 5x + 4 = (x – 1) (x – 4)
If 1 < x < 4, then x2 – 5x + 4 < 0
If x < 1 or x > 4 then x2 – 5x + 4 > 0
If x = 1 or x = 4 then x2 – 5x + 4 = 0

Illustration -86
If the equation 3×2 + 2(b2 + 1) x + (b2  3b + 2) = 0 possess roots of opposite sign, then b lies in
(A) (, 1) (B) (, 0) (C) (1, 2) (D)
Solution
(C)
3×2 + 2(b2 + 1) x + (b2  3b + 2) = 0 will have two roots of opposite sign if it has real roots and the product of the roots is negative.
i.e., 4(b2 + 1)2  12(b2  3b + 2)  0 and .
Both these conditions are fulfilled
if b2  3b + 2 < 0 i.e., if (b  1) (b  2) < 0  1 < b < 2.
Alternative
Here a > 0, equation possess roots of opposite sign if f(0) < 0  b2  3b + 2 < 0
 (b  1) (b  2) < 0 ;  b  (1, 2).

37. MAXIMUM OR MINIMUM VALUE OF QUADRATIC EXPRESSION

Maximum and minimum value of quadratic expression can be found out by two methods :
1. DISCRIMINANT METHOD
In a quadratic expression .
(a) If a > 0, quadratic expression has least value at . This least value is given by .
(b) If a < 0, quadratic expression has greatest value at . This greatest value is given by .
(c) If (or) , , then the minimum and maximum values of are given by
* If x is real then the maximum and minimum values of are and

2. GRAPHICAL METHOD
Vertex of the parabola is ,
i.e., ,  ,
Hence, vertex of is

(a) For a > 0, f(x) has least value at . This least value is given by .
(b) For a < 0, f(x) has greatest value at . This greatest value is given by .

Illustration -87
Find the maximum or minimum values of the following expressions on R : 2x + 5 – 3×2
Solution
Given expression 2x + 5 – 3×2
a = –3 < 0, b = 2, c = 5
given expression have maximum value.
maximum value =
=

Illustration -88
If x is real, then the maximum value of 5 + 4x  x2 is
(A) 5 (B) 6 (C) 9 (D) 1
Solution
(C)
Let y = 5 + 4x  x2 , then x2  4x + y  5 = 0
x is real
 16  4(y  5)  0 ;
 4y  36
 y  9
 maximum value of y = 9.

Illustration -89
Find the maximum and minimum values of for real values of x.
Solution
Let = y  y(x2 + x + 1) = x2 – x + 1
(y – 1)x2 + (y + 1)x + (y – 1) = 0. On solving for x, we get
x= =
Now for real values of x we will get real values of y, hence x to be real
.

38. LOCATION OF THE ROOTS

1. LOCATING THE ROOTS OF QUADRATIC EQUATION UNDER GIVEN CONDITIONS
• Location of real roots of the equation whose roots are with respect to one real number K.
• If both roots are greater than k then and
• If both roots are less than k, then and
• If k lies between the roots then with respect to two real numbers .
• If then

• If then

• If then
i) and ii)

2. POSITION OF ROOTS OF A QUADRATIC EQUATION.
Let , where a, b, c  R be a quadratic expression and be real numbers such that . Let ,  be the roots of the equation i.e. . Then , where D is the discriminant of the equation.
i. Condition for a number k (If both the roots of f(x) = 0 are less than k)

(i) (roots may be equal) (ii) (iii) , where
ii. Condition for a number k (If both the roots of f(x) = 0 are greater than k)

(i) (roots may be equal) (ii) (iii) , where
iii. Condition for a number k (If k lies between the roots of f(x) = 0)

i) (ii) , where

iv. Condition for numbers k1 and k2 (If exactly one root of f(x) = 0 lies in the interval (k1, k2))

(i) (ii) , where .

v. Condition for numbers k1 and k2 (If both roots of f(x) = 0 are confined between k1 and k2)

(i) (roots may be equal) (ii) (iii)
(iv) , where and

vi. Condition for numbers k1 and k2 (If k1 and k2 lie between the roots of f(x) = 0)

(i) (ii) (iii) , where

Illustration -90
If both roots of the quadratic equation are less than 5, then find the interval in which k lies

Solution

.
. or

from (1), (2) and (3)

Illustration -91
If (y2 – 5y + 3)(x2 +x +1) < 2x for all x  R, then the interval in which y lies is
(A) (B) ( -, -2] (C) (D) ( 1, 4)
Solution
(A)
( y2 – 5y +3) (x2 + x +1) < 2x  x  R
 y2 – 5y + 3 <
Let = p  px2 + ( p –2)x +p = 0
Since x is real, ( p –2)2 – 4p2  0
 -2  p 
Minimum value of is -2
So, y2 – 5y +3 < -2  y2 – 5y + 5 < 0
 y  .

39. GRAPH OF A QUADRATIC EXPRESSION
The expression ax2 + bx + c is said to be a real quadratic expression in x where a, b, c are real
and a  0.
(i) a > 0 and b2  4ac < 0
 f(x) > 0  x  R.

 

(ii) a > 0 and b2 – 4ac = 0
 f(x)  0  x  R.

(iii) a > 0 and b2  4ac > 0.
Let f(x) = 0 has two real roots  and  ( < ).
Then f(x) > 0  x  (, )  (, )
and f(x) < 0  x (, )

(iv) a < 0 and b2  4ac < 0
 f(x) < 0  x  R.

(v) a < 0 and b2 – 4ac = 0
 f(x)  0  x  R.

(vi) a < 0 and b2  4ac > 0
Let f(x) = 0 have two real roots  and  ( < ).
Then f(x) < 0  x  (, )(, )
And f(x) > 0  x  (, )

Notes:
If a > 0, then minima of f(x) occurs at x = -b/2a and if a < 0 , then maxima of f(x) occurs at x = b/2a
If f(x) = 0 has two distinct real roots, then a. f(d) < 0 if and only if d lies between the roots and
a. f(d) > 0 if and only if d lies outside the roots.

Working Rule to Find the Range of a Rational Expression in x, Where x is Real
• Put the given rational expression equal to y and form the quadratic equation in x.
• Find the discriminant D of the quadratic equation obtained in step 1.
• Since x is real, therefore, put D  0. We get an inequation in y.
• Solve the above inequation for y. The range of y so obtained determines the range attained by the given rational expression.
Illustration -91
Consider the graph of f(x) = ax2 +bx +c in the adjacent figure. We can conclude that

(A) c > 0 (B) a< 0 (C) b < 0 (D) a+ b+ c < 0

Solution
(A), (B), (C)
Clearly figure represents a downward parabola having its vertex in the second quadrant.
 a < 0 ,  b/2a < 0  b< 0
also , ax2 + bx + c = 0 has roots of opposite signs
 c/a < 0  c > 0.

Illustration -92
The interval in which ‘a’ lies when the graph of the function
is strictly above the x-axis is

Solution

Illustration -93
For the quadratic equation x2 – (m – 3)x + m = 0, find the value of m for which
(i) one root is smaller than 2 and the other is greater than 2
(ii) both roots are greater than 2
(iii) both roots lie in (1, 2)
(iv) exactly one root lie in (1, 2)

Solution

Let f(x) = x2 – (m – 3)x + m and D is = (m – 1)(m – 9)

(i) (a) D> 0 and (b) f (2) < 0
i.e., m < 1 or m > 9 and, m > 10
 m  (10, )

(ii) The required necessary and sufficient conditions are
D  0Þ m  1 or m  9………(1)
f(2) > 0 Þ m < 10………..(2)
> 2Þ m > 7……………(3)
From (1),(2) and (3) m  [9, 10)

(iii) D  0Þ m1 or m9 ……(1)
a. f(1) > 0Þ 4>0 mR ……(2)
a. f(2) > 0Þ m < 10 ……(3)
1 < < 2Þ m>5 and m<7 ……(4)

Taking intersection of these four conditions , we get m  .
(iv) D > 0 Þ m < 1 or m > 9 ……(1)
f(1) . f(2) < 0Þ m > 10 ……(2)
Taking intersection of these two conditions, we get m  (10, )

40. QUADRATIC INEQUALITIES AND SOLUTION OF INEQUATIONS

1. QUADRATIC INEQUATIONS
Let f(x) = ax2 + bx + c be a quadratic expression. Then inequations of the type f(x)  0 or f(x)  0 are known as quadratic inequations. The study of these can be easily done by taking the corresponding quadratic expression and by applying the basic results of quadratic expression.

Illustration -94
The value of x in the inequation
(A) x  (2, ) (B) x  (1, 2)
(C) x  (1, ) (D) none of these
Solution
(A)
We have < (x + 2). This inequation is equivalent to the system
  
Combining all three inequation of system, we get x > 2 i.e. x  (2, ).

2. WAVY CURVE METHOD
Let …..(i)
Where and are fixed natural numbers satisfying the condition
First we mark the numbers on the real axis and the plus sign in the interval of the right of the largest of these numbers, i.e. on the right of . If is even then we put plus sign on the left of and if is odd then we put minus sign on the left of . In the next interval we put a sign according to the following rule :
When passing through the point the polynomial f(x) changes sign if is an odd number and the polynomial f(x) has same sign if is an even number. Then, we consider the next interval and put a sign in it using the same rule. Thus, we consider all the intervals. The solution of is the union of all intervals in which we have put the plus sign and the solution of is the union of all intervals in which we have put the minus sign.
3. RATIONAL ALGEBRAIC INEQUATIONS
I. Values of rational expression P(x)/Q(x) for real values of x, where P(x) and Q(x) are quadratic expressions : To find the values attained by rational expression of the form for real values of x, the following algorithm will explain the procedure :
Algorithm
Step I: Equate the given rational expression to y.
Step II: Obtain a quadratic equation in x by simplifying the expression in step I.
Step III: Obtain the discriminant of the quadratic equation in Step II.
Step IV: Put Discriminant  0 and solve the inequation for y. The values of y so obtained determines the set of values attained by the given rational expression.
II. Solution of rational algebraic inequation: If P(x) and Q(x) are polynomial in x, then the inequation and are known as rational algebraic inequations.
To solve these inequations we use the sign method as explained in the following algorithm.
Algorithm
Step I: Obtain P(x) and Q(x).
Step II: Factorize P(x) and Q(x) into linear factors.
Step III: Make the coefficient of x positive in all factors.
Step IV: Obtain critical points by equating all factors to zero.
Step V: Plot the critical points on the number line. If there are n critical points, they divide the number line into (n + 1) regions.
Step VI: In the right most region the expression bears positive sign and in other regions the expression bears positive and negative signs depending on the exponents of the factors.
III. Lagrange’s theorem
Let f(x) be a function defined on [a b] such that
(i) f(x) is continuous on [a b] and
(ii) f(x) is differentiable on (a, b), then c  (a, b), such that
IV Lagrange’s identity
If then :

 

 

 

 

Illustration -95
Let .
Find intervals, where f(x) is positive or negative.
Solution

Illustration -96
Solve for x
Solution

> 0

The solution from the number line is

 

 

 

Illustration -97
Find the values of x for which
Solution
Let and
Now in g(x) the coefficient of is positive and is positive
Hence,