NCERT Solutions for Exercise 5.3 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

NCERT Solutions for Exercise 5.3 Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

Edited By Safeer PP | Updated on Jul 06, 2022 06:33 PM IST

NCERT solutions for exercise 5.3 Class 11 Maths chapter 5 solves quadratic equations with discriminant less than zero. Exercise 5.3 Class 11 Maths gives ten quadratic equations to be solved. Try solving the Class 11 Maths chapter 5 exercise 5.3 without looking at the answers. If any doubt arises while solving Class 11th Maths chapter 5 exercise 5.3, revise the concepts and give it a try and finally can verify the steps and answers using NCERT solutions for Class 11 Maths chapter 5 exercise 5.3. With Class 11 Maths ch 5 ex 5.3, there are 3 more exercises in the chapter complex numbers and quadratic equations, the same are listed below.

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Given equation is
$x^2+3=0$
Now, we know that the roots of the quadratic equation is given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case value of a = 1 , b = 0 and c = 3
Therefore,
$\frac{-0\pm \sqrt{0^2-4.1.(3)}}{2.1}= \frac{\pm\sqrt{-12}}{2} = \frac{\pm2\sqrt3i}{2}=\pm\sqrt3i$
Therefore, the solutions of requires equation are $\pm\sqrt3i$

Given equation is
$2x^2+x+1=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case value of a = 2 , b = 1 and c = 1
Therefore,
$\frac{-1\pm \sqrt{1^2-4.2.1}}{2.2}= \frac{-1\pm\sqrt{1-8}}{4} = \frac{-1\pm\sqrt{-7}}{4}=\frac{-1\pm\sqrt7i}{4}$
Therefore, the solutions of requires equation are

$\frac{-1\pm\sqrt7i}{4}$

Given equation is
$x^2+3x+9=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case value of a = 1 , b = 3 and c = 9
Therefore,
$\frac{-3\pm \sqrt{3^2-4.1.9}}{2.1}= \frac{-3\pm\sqrt{9-36}}{2} = \frac{-3\pm\sqrt{-27}}{2}=\frac{-3\pm3\sqrt3i}{2}$
Therefore, the solutions of requires equation are

$\frac{-3\pm3\sqrt3i}{2}$

Given equation is
$-x^2+x-2=0$
Now, we know that the roots of the quadratic equation is given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case value of a = -1 , b = 1 and c = -2
Therefore,
$\frac{-1\pm \sqrt{1^2-4.(-1).(-2)}}{2.(-1)}= \frac{-1\pm\sqrt{1-8}}{-2} = \frac{-1\pm\sqrt{-7}}{-2}=\frac{-1\pm\sqrt7i}{-2}$
Therefore, the solutions of equation are

$\frac{-1\pm\sqrt7i}{-2}$

Given equation is
$x^2+3x+5=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case value of a = 1 , b = 3 and c = 5
Therefore,
$\frac{-3\pm \sqrt{3^2-4.1.5}}{2.1}= \frac{-3\pm\sqrt{9-20}}{2} = \frac{-3\pm\sqrt{-11}}{2}=\frac{-3\pm\sqrt{11}i}{2}$
Therefore, the solutions of the equation are $\frac{-3\pm\sqrt{11}i}{2}$

Given equation is
$x^2-x+2=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case value of a = 1 , b = -1 and c = 2
Therefore,
$\frac{-(-1)\pm \sqrt{(-1)^2-4.1.2}}{2.1}= \frac{1\pm\sqrt{1-8}}{2} = \frac{1\pm\sqrt{-7}}{2}=\frac{1\pm\sqrt{7}i}{2}$
Therefore, the solutions of equation are $\frac{1\pm\sqrt{7}i}{2}$

Given equation is
$\sqrt{2}x^2+x+\sqrt{2}=0$
Now, we know that the roots of the quadratic equation is given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a = \sqrt 2 , b =1 \ and \ c = \sqrt2$
Therefore,
$\frac{-1\pm \sqrt{1^2-4.\sqrt2.\sqrt2}}{2.\sqrt2}= \frac{-1\pm\sqrt{1-8}}{2\sqrt2} = \frac{-1\pm\sqrt{-7}}{2\sqrt2}=\frac{-1\pm\sqrt{7}i}{2\sqrt2}$
Therefore, the solutions of the equation are $\frac{-1\pm\sqrt{7}i}{2\sqrt2}$

Given equation is
$\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a = \sqrt 3 , b =-\sqrt2 \ and \ c = 3\sqrt3$
Therefore,
$\frac{-(-\sqrt2)\pm \sqrt{(-\sqrt2)^2-4.\sqrt3.3\sqrt3}}{2.\sqrt3}= \frac{\sqrt2\pm\sqrt{2-36}}{2\sqrt3} = \frac{\sqrt2\pm\sqrt{-34}}{2\sqrt3}$ $=\frac{\sqrt2\pm\sqrt{34}i}{2\sqrt3}$
Therefore, the solutions of the equation are $\frac{\sqrt2\pm\sqrt{34}i}{2\sqrt3}$

Given equation is
$x^2+x+\frac{1}{\sqrt{2}}=0$
Now, we know that the roots of the quadratic equation is given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a = 1 , b =1 \ and \ c= \frac{1}{\sqrt2}$
Therefore,
$\frac{-1\pm \sqrt{1^2-4.1.\frac{1}{\sqrt2}}}{2.1}= \frac{-1\pm\sqrt{1-2\sqrt2}}{2} = \frac{-1\pm\sqrt{-(2\sqrt2-1)}}{2}$ $=\frac{-1\pm\sqrt{(2\sqrt2-1)}i}{2}$
Therefore, the solutions of the equation are

$\frac{-1\pm\sqrt{(2\sqrt2-1)}i}{2}$

Question:10 Solve each of the following equations:

Given equation is
$x^2+\frac{x}{\sqrt{2}}+1=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a = 1 , b =\frac{1}{\sqrt2} \ and \ c= 1$
Therefore,
$\frac{-\frac{1}{\sqrt2}\pm \sqrt{(\frac{1}{\sqrt2})^2-4.1.1}}{2.1}= \frac{-\frac{1}{\sqrt2}\pm\sqrt{\frac{1}{2}-4}}{2} = \frac{-\frac{1}{\sqrt2}\pm\sqrt{-\frac{7}{2}}}{2}$ $=\frac{-1\pm\sqrt{7}i}{2\sqrt2}$
Therefore, the solutions of the equation are

$\frac{-1\pm\sqrt{7}i}{2\sqrt2}$

More About NCERT Solutions for Exercise 5.3 Class 11 Maths Chapter 5

All the solutions of exercise 5.3 Class 11 Maths are complex numbers with a non zero imaginary part in them. The steps involved in solving quadratic equations are familiarised in Class 10 NCERT Maths book. The same steps are followed for solving the questions in Class 11 Maths chapter 5 exercise 5.3. The only difference is that the discriminant is less than zero, so there will be an imaginary part in the solution. One example is given just before Class 11 Maths chapter 5 exercise 5.3.

Benefits of NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.3

• The questions discussed in the NCERT exercises are important for grasping the concepts well.

• There are 10 questions for practice in the Class 11 Maths chapter 5 exercise 5.3. All the questions in NCERT syllabus Class 11 Maths chapter 5 exercise 5.3 are important from the exam point of view.

Subject Wise NCERT Exampler Solutions

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1. Which Indian mathematician is known for the work ‘Ganitasara Sangraha’?

Ganitasara Sangraha is the work of Indian mathematician Mahavira.

2. Bijaganita written in 1150 is the work of which Indian mathematician?

3. A polynomial equation is given. The degree of the equation is n. What is the number of roots?

The given polynomial equation with degree n has n number of roots. For example, a cubic polynomial equation will have 3 roots.

4. What conclusion can be made if the discriminant of a quadratic equation is zero?

If the discriminant of the given quadratic equation=0 implies that it has only one solution or have two real and equal solutions.

5. What can be inferred if the discriminant of a quadratic equation is positive

The conclusion is that the given quadratic equation has two distinct solutions that are real.

If the discriminant is <0 the roots of the equation has an imaginary part in them. Such types of questions are given in the NCERT solutions for exercise 5.3 Class 11 Maths chapter 5.

7. Homany examples are given for the Class 11 NCERT mathematics topic 5.6?

One example is given. Students are already familiar with solutions of quadratic equations in the lower classes

8. How many problems are solved in NCERT solutions for exercise 5.3 Class 11 Maths chapter 5?

10 questions of Class 11 Maths chapter 5 exercise 5.3 are solved

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