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

Question:1 Solve each of the following equations: $x^2+3=0$

Answer:

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 $\pm\sqrt3i$

Question:2 Solve each of the following equations: $2x^2+x+1=0$

Answer:

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

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

Question:3 Solve each of the following equations: $x^2+3x+9=0$

Answer:

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

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

Question:4 Solve each of the following equations: $-x^2+x-2=0$

Answer:

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

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

Question:5 Solve each of the following equations: $x^2+3x+5=0$

Answer:

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 $\frac{-3\pm\sqrt{11}i}{2}$

Question:6 Solve each of the following equations: $x^2-x+2=0$

Answer:

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

Question:7 Solve each of the following equations: $\sqrt{2}x^2+x+\sqrt{2}=0$

Answer:

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

Question:8 Solve each of the following equations: $\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}=0$

Answer:

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 $\frac{\sqrt2\pm\sqrt{34}i}{2\sqrt3}$

Question:9 Solve each of the following equations: $x^2+x+\frac{1}{\sqrt{2}}=0$

Answer:

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:

$x^2+\frac{x}{\sqrt{2}}+1=0$

Answer:

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

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