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Introduction to Quadratic Equations and Complex Numbers

Mathematics Class XI Chapter 2: Complex Numbers and Quadratic Equations Introduction to Quadratic Equations and Complex Numbers

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The equation $ax^2+bx+c=0$ is known as quadratic equation in one variable where $D=b^2-4ac \geq 0$ i.e., we get the roots in real numbers.

Let’s take the equation $x^2=-1$ , where $x=\sqrt{-1}$ , this is not a real number, i.e., for a quadratic equation $ax^2+bx+c=0$ where $D=b^2-4ac$ < $0$ , we can’t get a real roots for this. So we can take $i=\sqrt{-1}$ , is the imaginary number to find out the root of the equation $x^2=-1$ .

Hence $D=b^2-4ac < 0$ , in the quadratic equation $ax^2+bx+c=0$ we get two imaginary roots.

Complex Number:

The number which is of the form $a+ib$ , where $a$ and $b$ are real numbers is defined to be a complex number.

We also take $z=a+ib$ , where $a$ is the real part, denoted by $Rez$ and $b$ is the imaginary part denoted by $Imz$ .

The set consists of all the complex numbers is called a complex number set or set of complex numbers and denoted as $\mathbb{C}$ .

For example,

$z=5+i7$ ,

Here $Re~Z=5, ~Im~Z=7$

Two complex numbers $z_1=a+ib$ and $z_2=c+id$ are equal iff $a=c$ and $b=d$

Example:

If $4x+i (3x-y)=3+i (-6)$ , where $x$ and $y$ are real numbers, then find the value of $x$ and $y$ .

Solution:

$4x+i(3x-y)=3+i(-6)$ (Given)

$\implies 4x=3, ~ 3x-y=-6$

$\implies x=\dfrac{3}{4} \implies 3\times \dfrac{3}{4}-y=-6 \implies y=\dfrac{33}{4}$

$\therefore x=\dfrac{3}{4},~~ y=\dfrac{33}{4}$

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