Quadratic Equation - Definition, Formula and How to Solve

A quadratic equation is a second-order polynomial equation in a single variable. It is a second-degree algebraic expression and is of the form ax² + bx + c = 0. The term "quadratic" comes from the Latin word "quadratus" meaning square, which refers to the fact that the variable x is squared in the equation. In other words, a quadratic equation is an “equation of degree 2.” There are many scenarios where a quadratic equation is used. Further, a quadratic equation has numerous applications in physics, engineering, astronomy, etc.

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In this article, we will cover the concept of the quadratic equation. This concept falls under the broader category of complex numbers and quadratic equations, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Polynomial expression

An expression of the form $f(x)=a_0 x^n+a_1 x^{n-1}+a_2 x^{n-2}+\ldots+a_{n-1} x+a_n$, is called a polynomial expression.

Where $x$ is variable and $a_0, a_1, a_2, \ldots \ldots, a_n$ are constant, known as coefficients and $a_0 \neq 0, n$ is non-negative integer,

Degree
The highest power of the variable in the polynomial expression is called the degree of the polynomial. In $a_0 \cdot x^n+a_1 \cdot x^{n-1}+\ldots+a_n$, the highest power of x is n , so the degree of this polynomial is n . If coefficients are real numbers then it is called a real polynomial, and when they are complex numbers, then the polynomial is called a complex polynomial.

The root of polynomial:
If $f(x)$ is a polynomial, then $f(x)=0$ is called a polynomial equation.
The value of $x$ for which the polynomial equation, $f(x)=0$ is satisfied is called a root of the polynomial equation.
If $x=\alpha$ is a root of the equation $f(x)=0$, then $f(\alpha)=0$.
$\mathrm{Eg}, \mathrm{x}=2$ is a root of $\mathrm{x}^2-3 x+2=0$, as $x=2$ satisfies this equation.
A polynomial equation of degree $n$ has $n$ roots (real or imaginary).

Quadratic equation:
A polynomial equation in which the highest degree of a variable term is 2 is called a quadratic equation.
Standard form of quadratic equation is $a x^2+b x+c=0$
Where $\mathrm{a}, \mathrm{b}$, and c are constants (they may be real or imaginary) and called the coefficients of the equation and $a \neq 0$ (a is also called the leading coefficient).

$$
E g,-5 x^2-3 x+2=0, x^2=0,(1+i) x^2-3 x+2 i=0
$$

As the degree of the quadratic polynomial is 2 , so it always has 2 roots (number of real roots + number of imaginary roots $=2$ )
Roots of quadratic equation
The root of the quadratic equation is given by the formula:

$$
\mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{D}}}{2 \mathrm{a}}
$$

Where $D$ is called the discriminant of the quadratic equation, given by $D=b^2-4 a c$,

$$
\begin{aligned}
& \text { Proof: } \\
& a x^2+b x+c=0 \\
& \text { Take, 'a' common } \\
& a\left(x^2+\frac{b}{a} x+\frac{c}{a}\right)=0 \\
& a\left[\left(x+\frac{b}{2 a}\right)^2-\frac{b^2}{4 a^2}+\frac{c}{a}\right]=0 \\
& \left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2} \\
& \left(x+\frac{b}{2 a}\right)= \pm \frac{\sqrt{b^2-4 a c}}{2 a} \\
& x=-\frac{b}{2 a} \pm \frac{\sqrt{b^2-4 a c}}{2 a} \\
& x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}
\end{aligned}
$$

What is Discriminant?
The term $\left(b^2-4 a c\right)$ in the quadratic formula is known as the discriminant of a quadratic equation. The discriminant of a quadratic equation reveals the nature of roots.
Sum of roots:

$$
\alpha+\beta=\frac{-\mathrm{b}-\sqrt{\mathrm{D}}}{2 \mathrm{a}}+\frac{-\mathrm{b}+\sqrt{\mathrm{D}}}{2 \mathrm{a}}=\frac{-\mathrm{b}}{\mathrm{a}}
$$

Product of roots:

$$
\begin{aligned}
& \alpha \cdot \beta=\left(\frac{-b-\sqrt{\mathrm{D}}}{2 \mathrm{a}}\right) \cdot\left(\frac{-\mathrm{b}+\sqrt{\mathrm{D}}}{2 \mathrm{a}}\right) \\
& =\frac{\mathrm{b}^2-\mathrm{D}}{4 \mathrm{a}^2}=\frac{\mathrm{b}^2-\mathrm{b}^2+4 \mathrm{ac}}{4 \mathrm{a}^2}=\frac{4 \mathrm{ac}}{4 \mathrm{a}^2}=\frac{\mathrm{c}}{\mathrm{a}}
\end{aligned}
$$

The difference of root can also be found in the same way by manipulating the terms

$$
\alpha-\beta=\left|\frac{\sqrt{D}}{a}\right|
$$

Summary

Quadratic equations can be solved using various methods, each suitable for different forms of quadratic expressions. The quadratic formula is the most universally applicable method, providing a straightforward way to find the roots for any quadratic equation. This helps in analyzing two-degree polynomials graphically as well as algebraicly.

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Solved Examples Based On the Quadratic Equation:

Example 1: If $\alpha$ and $\beta$ are the roots of the equation $\frac{1}{i Z}-i Z=2(\sin \theta-i \cos \theta)$ where $0<\theta<\pi$ and $i=\sqrt{-1}$, and $\mathbf{z}$ is complex numbers, then the value of $|\alpha-i|+|\beta-i|$
Solution:

$$
\frac{1}{i Z}-i Z=2(\sin \theta-i \cos \theta)
$$

Multiply with "i" both side

$$
\begin{aligned}
& \frac{1}{Z}-i^2 Z=2\left(i \sin \theta-i^2 \cos \theta\right) \\
& \frac{1}{Z}+Z=2(\cos \theta+i \sin \theta) \\
& \frac{1}{Z}+Z=2 e^{i \theta}
\end{aligned}
$$

$$
\begin{aligned}
& Z^2-2 Z e^{i \theta}+1=0 \\
& Z=\frac{2 e^{i \theta} \pm \sqrt{4 e^{2 i \theta}-4}}{2} \\
& Z=e^{i \theta} \pm \sqrt{e^{2 i \theta}-1} \\
& Z=e^{i \theta} \pm \sqrt{e^{i \theta} \cdot 2 i \sin \theta} \\
& Z-i=e^{i \theta}-i \pm \sqrt{e^{i \theta} \cdot 2 i \sin \theta} \\
& Z-i=e^{i \theta}-e^{i \frac{\pi}{2}} \pm \sqrt{e^{i\left(\theta+\frac{\pi}{2}\right)} \cdot 2 \sin \theta}
\end{aligned}
$$

$$
\begin{aligned}
& Z-i=e^{i\left(\frac{\theta}{2}+\frac{\pi}{4}\right)} \cdot 2 i \sin \left(\frac{\theta}{2}-\frac{\pi}{4}\right) \pm e^{i\left(\frac{\theta}{2}+\frac{\pi}{4}\right)} \cdot \sqrt{2 \sin \theta} \\
& Z-i=e^{i\left(\frac{\theta}{2}+\frac{\pi}{4}\right)} \cdot\left[2 i \sin \left(\frac{\theta}{2}-\frac{\pi}{4}\right) \pm \sqrt{2 \sin \theta}\right] \\
& |Z-i|=1 \cdot \sqrt{4 \sin ^2\left(\frac{\theta}{2}-\frac{\pi}{4}\right)+2 \sin \theta} \\
& =\sqrt{2\left(1-\cos \left(\theta-\frac{\pi}{2}\right)\right)+2 \sin \theta} \\
& =\sqrt{2(1-\sin \theta)+2 \sin \theta} \\
& =\sqrt{2} \\
& |Z-i|=|\alpha-i|=|\beta-i| \\
& |\alpha-i|+|\beta-i|=2 \sqrt{2}
\end{aligned}
$$

Hence, the answer is $2 \sqrt{2}$.

Example 2: Value of ' $a^{\prime}$ for which $\left(a^2-1\right) x^2+(2 a+3) x+5=0$ represents a quadratic equation with real coefficients is
1) $\pm 1$
2) $R-\{+1,-1\}$
3) $R$
4) $R^{+}$

Solution:
As we have learned
Quadratic Equation with real Coefficients -
An equation of the form $a x^2+b x+c=0$
- whereinm
$a, b, c \in R$ and $a \neq 0$
for quadratic equation $a^2-1 \neq 0$
$a^2 \neq 1$
so $a \neq 1,-1$

Hence, the answer is the option 2.

Example 3: The roots of the equation $2 x^2-6 x+3=0$ are
Solution:

As we learned in

Roots of Quadratic Equation -
是

$$
\begin{aligned}
& \alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a} \\
& \beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}
\end{aligned}
$$

- wherein

$$
a x^2+b x+c=0
$$

is the equation

$$
\begin{aligned}
& a, b, c \in R, a \neq 0 \\
& \because \alpha, \beta=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& \because \alpha, \beta=\frac{6 \pm \sqrt{36-4(2)(3)}}{4} \\
& \because \alpha, \beta=\frac{6 \pm \sqrt{12}}{4}
\end{aligned}
$$

$$
=\because \alpha, \beta=\frac{3 \pm \sqrt{3}}{2}
$$

Hence, the answer is $\frac{3 \pm \sqrt{3}}{2}$.

Example 4: The sum of the roots of the equation, $x^2+|2 x-3|-4=0$, is
Solution:
As we have learned
Roots of Quadratic Equation -

$$
\begin{aligned}
& \alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a} \\
& \beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}
\end{aligned}
$$

wherein

$$
a x^2+b x+c=0
$$

is the equation

$$
\begin{aligned}
& a, b, c \in R, \quad a \neq 0 \\
& \text { Case (1) } x \geq 3 / 2 \\
& x^2+2 x-3-4=0 \\
& x^2+2 x-7=0
\end{aligned}
$$

$$
\Rightarrow x=\frac{-2 \pm \sqrt{4+28}}{2}=-1 \pm 2 \sqrt{2}
$$

Acceptable value $=2 \sqrt{2}-1$

$$
\begin{aligned}
& \text { case (2) } \\
& x<3 / 2 \\
& x^2-2 x+3-4=0 \\
& x^2-2 x-1=0 \\
& \Rightarrow x=\frac{2 \pm \sqrt{4+4}}{2}=1 \pm \sqrt{2}
\end{aligned}
$$

Acceptable value $1-\sqrt{2}$
Sum of roots $=1-\sqrt{2}+2 \sqrt{2}-1=\sqrt{2}$
Hence, the answer is $\sqrt{2}$.

Example 5: If $\alpha$ and $\beta$ are the roots of the equation $x^2-x+1=0$, then $\alpha^{2009}+\beta^{2009}=$

Solution: $\quad$

As we have learned
Roots of Quadratic Equation --

$$
\begin{aligned}
& \alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a} \\
& \beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}
\end{aligned}
$$

- wherein

$$
a x^2+b x+c=0
$$

is the equation

$$
a, b, c \in R, \quad a \neq 0
$$

Cube roots of unity -

$$
z=(1)^{\frac{1}{3}} \Rightarrow z=\cos \frac{2 k \pi}{3}+i \sin \frac{2 k \pi}{3}
$$

$\mathrm{k}=0,1,2$ so z gives three roots

$$
\Rightarrow 1, \frac{-1}{2}+i \frac{\sqrt{3}}{2}(\omega), \frac{-1}{2}-i \frac{\sqrt{3}}{2}\left(\omega^2\right)
$$

$$
\omega=\frac{-1}{2}+\frac{i \sqrt{3}}{2}, \omega^2=\frac{-1}{2}-\frac{i \sqrt{3}}{2}, \omega^3=1,1+\omega+\omega^2=0
$$

$1, \omega, \omega^2$ are cube roots of unity.

$$
\begin{aligned}
& \alpha, \beta=\frac{1 \pm \sqrt{-3}}{2}=\frac{1 \pm \sqrt{3 i}}{2}=-\left(\frac{-1 \pm \sqrt{3 i}}{2}\right)=-\omega,-\omega^2 \\
& \alpha^{2009}+\beta^{2009} \\
& -\omega^{2009}+\left(-\omega^2\right)^{2009} \\
& =-\omega^2-\omega=-\left(\omega^2+\omega\right)=-(-1)=1
\end{aligned}
$$

Hence, the answer is 1.