Complex Number and Quadratic Equations - Solutions and Study Material
Complex Number : Let us denote by the symbol i. Then ,we have = -1. This means that I is a solution of the equation
A number of the form a + ib, where a and b are real numbers, is defined to be a complex .
Complex Number: 2 + 3i, , are complex numbers.
For the Complex Number z = a + ib, a is called the real part, denoted by Re z and b is called the imaginary part denoted by Im z of the complex number z. For example, if z = 2 + i5, then Re z = 2 and Im z = 5.
Two complex numbers = a + ib and = c + id are equal if a = c and b = d
Example: If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find the values of x and y.
Solution: We have , 4x + i (3x – y) = 3 + i (–6) ... (1)
Equating the real and the imaginary parts of (1), we get ,4x = 3, 3x – y = – 6,
which, on solving simultaneously, give x = , y =
Algebra of Complex Numbers: In this Section, we shall develop the algebra of complex numbers
Addition of two complex numbers Let = a + ib and = c + id be any two complex numbers .Then , the is define as follows :
= (a + c) + i(b + d) , which is again a complex number.
For Example (2 + 3i) +(-6 +5i) = (2-6) + I (3 + 5) = (-4 + 8 i)
The addition of complex number satisfy the following properties:
The closure law The sum of two complex numbers is a complex number, i.e., is a complex number for all complex numbers Let and Let
The commutative law For any two complex numbers and ,
The associative law For any three complex numbers ,
The existence of additive identity There exists the complex number 0 + i 0 (denoted as 0), called the additive identity or the zero complex number, such that, for every complex number z, z + 0 = z
The existence of additive inverse To every complex number z = a + ib, we have the complex number – a + i(– b) (denoted as – z), called the additive inverse or negative of z. We observe that z + (–z) = 0 (the additive identity)
The multiplication of complex numbers possesses the following properties, which we state without proofs.
The Closure Law: The product of two complex numbers is a complex number the product is a complex number for all complex numbers
The Commutative Law: For any two complex numbers
The Associative Law: For any three complex numbers
The Existence of Multiplicative Identity: There exists the complex number 1 + i 0 (denoted as 1), called the multiplicative identity such that z.1 = z, for every complex number z.
The Existence of Multiplicative Inverse: For every non-zero complex number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number
(denoted by ), called the multiplicative inverse of Z such that . = 1 (the multiplicative identity)
The Distributive Law: For any three complex number
Division of two complex numbers : Given any two complex numbers where , ≠ 0 , the quotient is defined by
For Example, let = 6 + 3 I and = 2 – I Then
=
Power of i: we know that
The Square Roots of a Negative Real Number: Note that
Therefore, the square roots of – 1 are i, – i. However, by the symbol we would
mean i only.
Now, we can see that i and –i both are the solutions of the equation
The Modulus and the Conjugate of a Complex Number : Let z = a + ib be a complex number. Then, the modulus of z, denoted by | z |, is defined to be the non-negative real number and the conjugate of z, denoted as z, is the complex number a – ib, i.e., z = a – ib
Example:
Observe that the multiplicative inverse of the non-zero complex number z is given by
Furthermore, the following results can easily be derived. For any two compex numbers z1 and z2 , we have,
provided 2z ≠ 0
provided ≠ 0.
Example:
Solution:
Quadratic Equations
We are already familiar with the quadratic equations and have solved them in the set of real numbers in the cases where discriminant is non-negative, i.e., ≥ 0,
Let us consider the following quadratic equation: with real coefficients a, b, c and a ≠ 0. Also, let us assume that the
Now, we know that we can find the square root of negative real numbers in the set of complex numbers. Therefore, the solutions to the above equation are available in the set of complex numbers which are given by
x =
Note: At this point of time, some would be interested to know as to how many roots does an equation have? In this regard, the following theorem known as the Fundamental theorem of Algebra is stated below (without proof).
“A polynomial equation has at least one root.”
As a consequence of this theorem, the following result, which is of immense importance, arrived at
“A polynomial equation of degree n has n roots.”
Example: solve
Solution:
Example: solve
Solution: = 1-4 = -3
Point to be Remember:
A number of the form a + ib, where a and b are real numbers, is called a complex number, a is called the real part and b is called the imaginary part of the complex number
= a + ib and = c + id . Then
(i) = (a + c) + i (b + d)
(ii) = (ac – bd) + i (ad + bc)
For any non-zero complex number z = a + ib (a ≠ 0, b ≠ 0), there exists the complex number
, denoted by called the multiplicative inverse of z such that (a + ib) = 1 + i0 =1
For any integer k,
The conjugate of the complex number z = a + ib, denoted by , is given by by = a + ib
The polar form of the complex number z = x + iy is r (cosθ + i sinθ), where ( the module of z) and (θ is known as the argument of z. The value of θ, such that – π < θ ≤ π, is called the principal argument of z
A polynomial equation of n degree has n roots.
The solutions of the quadratic equation where a, b, c ∈ R, a ≠ 0, b2 – 4ac < 0, are given by x =
Important Questions from Previous Year Exams
Time limit: 0
Quiz-summary
0 of 25 questions completed
Questions:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Information
Quadratic equation
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 25 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score
Your score
Categories
Not categorized0%
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Answered
Review
Question 1 of 25
1. Question
1 points
Evaluate the discriminant of
- 13x – 1 = 0
Correct
In any quadratic equation of the form
+ bx + c = 0
The discriminant D =
Discriminant D =
=
= 361
Incorrect
In any quadratic equation of the form
+ bx + c = 0
The discriminant D =
Discriminant D =
=
= 361
Question 2 of 25
2. Question
1 points
Evaluate the discriminant of
- 2x + c – Q + y = 0
Correct
In the given equation, b = -2, a = p, and c =c –q + y
Hence, discriminant D
=
=
= 4 – 4p (c – q + y)
= 4[1 – c + q – y]
Incorrect
In the given equation, b = -2, a = p, and c =c –q + y
Hence, discriminant D
=
=
= 4 – 4p (c – q + y)
= 4[1 – c + q – y]
Question 3 of 25
3. Question
1 points
Evaluate the discriminant of
= 0
Correct
In the given equation
= 0,
Hence, discriminant D
=
=
= 48 – 12d
Incorrect
In the given equation
= 0,
Hence, discriminant D
=
=
= 48 – 12d
Question 4 of 25
4. Question
1 points
Evaluate the discriminant of
= 0
Correct
In the given equation = 0
A = -5,
Hence discriminant D
=
=
=
Incorrect
In the given equation = 0
A = -5,
Hence discriminant D
=
=
=
Question 5 of 25
5. Question
1 points
Evaluate the discriminant of
7x – 25 + = 0 x ≠ 0
Correct
The given equation is
= 0
= 0
Now, a =7, b = -25, c = 10
Hence, discriminant D
=
=
= 625 – 280
= 345
Incorrect
The given equation is
= 0
= 0
Now, a =7, b = -25, c = 10
Hence, discriminant D
=
=
= 625 – 280
= 345
Question 6 of 25
6. Question
1 points
Evaluate the discriminant of
x + = 2.5
Correct
The given equation is
Now, a = 2, b = -5, c = 2.
Hence, discriminant D
=
=
= 9
Incorrect
The given equation is
Now, a = 2, b = -5, c = 2.
Hence, discriminant D
=
=
= 9
Question 7 of 25
7. Question
1 points
State the nature of the roots of the following equations without evaluating the same
k ≠ 0
Correct
Nature of the discriminant (D = ) determines the nature of the roots of a quadratic equation. Hence, in each of the given equations, it is to be determined that
D > 0 or = 0 or < 0.
In the given equation
Hence, determinant D
=
=
=
Since D > 0, so the roots are real and distinct (unequal).
Incorrect
Nature of the discriminant (D = ) determines the nature of the roots of a quadratic equation. Hence, in each of the given equations, it is to be determined that
D > 0 or = 0 or < 0.
In the given equation
Hence, determinant D
=
=
=
Since D > 0, so the roots are real and distinct (unequal).
Question 8 of 25
8. Question
1 points
State the nature of the roots of the following equations without evaluating the same
Correct
In the given equation
a = 9, b = 6, c = 1
Determinant D =
=
= 0
Since D = 0 so, the roots and real and equal
Incorrect
In the given equation
a = 9, b = 6, c = 1
Determinant D =
=
= 0
Since D = 0 so, the roots and real and equal
Question 9 of 25
9. Question
1 points
State the nature of the roots of the following equations without evaluating the same
Correct
In the given equation
, b = a, c= 1
Hence determinant D
=
=
=
Now, D > 0, the roots are not real.
Incorrect
In the given equation
, b = a, c= 1
Hence determinant D
=
=
=
Now, D > 0, the roots are not real.
Question 10 of 25
10. Question
1 points
State the nature of the roots of the following equations without evaluating the same
Correct
In the given equation
Hence, determinant D
=
=
= 400 > 0
Since D > 0, so the roots are real and distinct.
Incorrect
In the given equation
Hence, determinant D
=
=
= 400 > 0
Since D > 0, so the roots are real and distinct.
Question 11 of 25
11. Question
1 points
For what values of ‘m’ will the equation
has two real and equal roots
Correct
The given equation can be re-written as
(m + 4) + (m + 1) x + 1 = 0
Hence, discriminant D
=
=
=
=
= (m-5) (m+3)
For real and equal roots, D = 0
(m-5) (m+3) = 0
m = +5 or m = -3
Incorrect
The given equation can be re-written as
(m + 4) + (m + 1) x + 1 = 0
Hence, discriminant D
=
=
=
=
= (m-5) (m+3)
For real and equal roots, D = 0
(m-5) (m+3) = 0
m = +5 or m = -3
Question 12 of 25
12. Question
1 points
If in the equation , one root exceeds the other by 1, then establish a relation p and q.
Correct
Using the relation
Where, a = 1, b = -p, c = q, k = 1
Hence,
Hence, p and q are related by the equation
Incorrect
Using the relation
Where, a = 1, b = -p, c = q, k = 1
Hence,
Hence, p and q are related by the equation
Question 13 of 25
13. Question
1 points
If the roots of the equation are in the ratio of 3:4, then find a relation among a, b and c
Correct
Using the equation
where, Hence, if m = 3, n = 4
Putting the values, of m and n
Hence, a, b, and c are related by the equation
Incorrect
Using the equation
where, Hence, if m = 3, n = 4
Putting the values, of m and n
Hence, a, b, and c are related by the equation
Question 14 of 25
14. Question
1 points
For what values of ‘m’ will the roots of the equation be in the ratio of 2:3
Correct
Using the equation
For m = 2, n = 3, b = m – 2, a = 1, c = m + 2
Putting these values in the above equation, the required conditions is,
(2m + 1) (3m - 26) = 0
Hence, required values of m are
Incorrect
Using the equation
For m = 2, n = 3, b = m – 2, a = 1, c = m + 2
Putting these values in the above equation, the required conditions is,
(2m + 1) (3m - 26) = 0
Hence, required values of m are
Question 15 of 25
15. Question
1 points
If the roots of the equation are equal and opposite in sign, then find out the product of the roots.
Correct
The given equation is
= 0
Since the roots are equal and opposite in sign,
so coefficient of x = 0
p + q - 2r = 0
p + q = 2r.
No product of the roots
=
= pq - r(p + q)
=
=
Hence, product of the roots is
Incorrect
The given equation is
= 0
Since the roots are equal and opposite in sign,
so coefficient of x = 0
p + q - 2r = 0
p + q = 2r.
No product of the roots
=
= pq - r(p + q)
=
=
Hence, product of the roots is
Question 16 of 25
16. Question
1 points
If the roots of the equation are equal and real, then find the value of m.
Correct
The given equation is
Here,
Discriminant D =
=
=
Since the roots are real and equal, so D = 0
3 = 3m + 1
Hence, the value of m is
Incorrect
The given equation is
Here,
Discriminant D =
=
=
Since the roots are real and equal, so D = 0
3 = 3m + 1
Hence, the value of m is
Question 17 of 25
17. Question
1 points
Find the sum and the product of the roots of the equation (1 + 6m)
Correct
The given equation is
Now,
Sum of roots =
[See memory dig.]
=
=
Products of the roots =
[See memory dig.]
=
= m
Incorrect
The given equation is
Now,
Sum of roots =
[See memory dig.]
=
=
Products of the roots =
[See memory dig.]
=
= m
Question 18 of 25
18. Question
1 points
Find the value of k so that the sum of the roots of the equation is equal to the product of the roots
Correct
The given equation is
Sum of roots =
Product of roots =
From the condition given , sum of roots = Products of roots
k = 4
Incorrect
The given equation is
Sum of roots =
Product of roots =
From the condition given , sum of roots = Products of roots
k = 4
Question 19 of 25
19. Question
1 points
If are the roots of the equation , find the values of
Correct
Since are the roots of the equation
=
=
Incorrect
Since are the roots of the equation
=
=
Question 20 of 25
20. Question
1 points
If are the roots of the equation , find the values of
Correct
Since are the roots of the equation
=
=
Incorrect
Since are the roots of the equation
=
=
Question 21 of 25
21. Question
1 points
If are the roots of the equation , find the values of
Correct
Since are the roots of the equation
=
=
=
Incorrect
Since are the roots of the equation
=
=
=
Question 22 of 25
22. Question
1 points
If one root of the quadratic equation is 3, then find the other root.
Correct
The given equation is
Now, sum of roots =
=
Let, the two roots be
Hence, the other root is
Incorrect
The given equation is
Now, sum of roots =
=
Let, the two roots be
Hence, the other root is
Question 23 of 25
23. Question
1 points
Find the value of ‘m’ for which the roots
Correct
The given equation is
Now, sum of roots = ....(i)
Products of roots = .....(ii)
The given relation is
Putting the value of in equation ..(i)
Putting the value of in equation ...(ii)
m =
= - 42
Hence, the value of m is -42
Incorrect
The given equation is
Now, sum of roots = ....(i)
Products of roots = .....(ii)
The given relation is
Putting the value of in equation ..(i)
Putting the value of in equation ...(ii)
m =
= - 42
Hence, the value of m is -42
Question 24 of 25
24. Question
1 points
If
Correct
Since is a root of , so it must satisfy the equation
Hence,
Now. the equation is has equal roots so,
Decriminant D =
Hence, the value of q is
Incorrect
Since is a root of , so it must satisfy the equation
Hence,
Now. the equation is has equal roots so,
Decriminant D =
Hence, the value of q is
Question 25 of 25
25. Question
1 points
If are the roots of
in terms of k.
Correct
The given equation is
Now,
=
=
=
Incorrect
The given equation is
Now,
=
=
=
Important Videos on Complex Number and Quadratic Equations - Must Watch
Number Sets
Number Sets 1
Number Sets 2
Number Sets 3
Introduction to i and Imaginary Numbers
Calculating i Raised to Arbitrary Exponents
Complex Numbers
Complex Numbers Part 1
Complex Numbers Part 2
i as the Principal Root of -1 (a little technical)
IIT JEE Complex Numbers (part 1)
IIT JEE Complex Numbers (part 2)
IIT JEE Complex Numbers (part 3)
Algebra II: Imaginary and Complex Numbers
Imaginary Roots of Negative Numbers
Complex Conjugates Example
Adding Complex Numbers
Subtracting Complex Numbers
Multiplying Complex Numbers
Dividing Complex Numbers
Complex Roots from the Quadratic Formula
Please comment on Complex Number and Quadratic Equations - Solutions and Study Material
0 Comments