A polynomial equation of second degree is called quadratic equation. In other words, an equation of the form $ax^2+bx+c=0\;\;\;(a≠0)$ is called a quadratic equation.

Theorem: A quadratic equation has two and only two roots.

Let $ax^2+bx+c=0\;\;\;(a≠0)$ be a quadratic equation. Then, $\begin{array}{l} & x^2+\frac{b}{a}x+\frac{b}{a} &= 0 \\ \text{or,} & \left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2 + \frac{c}{a} &= 0 \\ \text{or,} & \left(x+\frac{b}{2a}\right)^2 &= \left(\frac{b}{2a}\right)^2 – \frac{c}{a} \\ \text{or,} & \left(x+\frac{b}{2a}\right)^2 &=\frac{b^2+4ac}{4a^2} \\ \text{or,} & \left(x+\frac{b}{2a}\right) &= \pm\frac{\sqrt{b^2-4ac}}{2a} \\ \text{or,} & x &= -\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} \\ \therefore & x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{array}$

$\therefore \alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}$ $\text{and,}\;\;\;\beta=\frac{-b-\sqrt{b^2-4ac}}{2a}$ are the two roots of the quadratic equation.

If possible, suppose that $\alpha$, $\beta$ and $\delta$ are three distinct roots of the quadratic equation $ax^2+bx+c=0\;\;\;(a≠0)$ Then, $a\alpha^2+b\alpha+c=0\;\;\text{__(1)}$ $a\beta^2+b\beta+c=0\;\;\text{__(2)}$ $a\delta^2+b\delta+c=0\;\;\text{__(3)}$

Solving $\text{(1)}$ and $\text{(2)}$, $\begin{array}{l} a\alpha^2 &+b\alpha &+c &= 0 \\ a\beta^2 &+b\beta &+c &= 0 \\ – & – & – \\ \hline \end{array}$ $\begin{array}{l} & a(\alpha^2-\beta^2)+b(\alpha-\beta) &= 0 \\ \text{or,} & (\alpha-\beta)(a\alpha+a\beta+b) &= 0 \\ \therefore & a\alpha+a\beta+b &=0\;\; \text{__(4)}\end{array}$ $(\because \alpha≠\beta)$

Solving $\text{(2)}$ and $\text{(3)}$, $\begin{array}{l} a\beta^2 &+b\beta &+c &= 0 \\ a\delta^2 &+b\delta &+c &= 0 \\ – & – & – \\ \hline \end{array}$ $\begin{array}{l} & a(\beta^2-\delta^2)+b(\beta-\delta) &= 0 \\ \text{or,} & (\beta-\alpha) (a\beta+a\delta+b) &= 0 \\ \therefore & a\beta+a\delta+b &= 0 \;\; \text{__(5)} \end{array}$ $(\because \beta≠\delta)$

Solving $\text{(4)}$ and $\text{(5)}$, $\begin{array}{l} a\alpha &+ a\beta &+ b &= 0 \\ a\beta &+ a\delta &+ b &= 0 \\ – & – & – \\ \hline \end{array}$ $\begin{array}{l} & a(\alpha-\delta) &= 0 \\ \therefore & \alpha-\delta &= 0 \;\;\; (\because a≠0) \end{array}$

which is a contradiction to our assumption that $\alpha$, $\beta$ and $\delta$ are distinct. Hence, quadratic equation cannot have more than two roots.

Nature of the Roots of a Quadratic Equation

Let $\alpha$ and $\beta$ be the two roots of a quadratic equation $ax^2+bx+c=0\;\;\;(a≠0)$ $\text{Then,}\;\;\;\alpha=\frac{-b+\sqrt{b^2-4ac}}{2a}$ $\text{and,}\;\;\;\beta=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Here, $a$, $b$ and $c$ are rational numbers. The value of the expression under the radical sign i.e. $b^2-4ac$ depends on the coefficients $a$, $b$ and $c$. Hence, the following three cases arise:

Case I:

If $b^2-4ac>0$ (the quantity under the radical sign is positive) then the roots are real and unequal.

If $b^2-4ac$ is positive as well as a perfect square then the roots are rational and unequal, provided that $a$, $b$ and $c$ are rational.

Case II:

If $b^2-4ac=0$, then the roots are real and equal, each being equal to $-\frac{b}{2a}$.

Case III:

If $b^2-4ac<0$, then the roots are imaginary and unequal.

Hence, by computing the value of $b^2-4ac$, we can determine the nature of the quadratic equation. This quantity is therefore known as the discriminant of the quadratic equation.

Now, the following two conclusions can be derived from the above discussions.

In a quadratic equation with rational coefficients, irrational roots always occur in conjugate pair.

If $b^2-4ac>0$ and not a perfect square, then $\sqrt{b^2-4ac}$ will be irrational. Since $\sqrt{b^2-4ac}$ occurs in both roots $\alpha$ and $\beta$, let $-\frac{b}{2a}=p\;\;\;\text{and}\;\;\;\frac{\sqrt{b^2-4ac}}{2a}=\sqrt{q}$ Then, we get, $\alpha=p+\sqrt{q}\;\;\;\text{and}\;\;\;\beta=p-\sqrt{q}$

Thus, both of the roots are irrational, each being the conjugate of the other.

Therefore, in a quadratic equation with rational coefficients, the irrational roots occur in conjugate pair.

In a quadratic equation with real coefficients, imaginary (complex) roots always occur in conjugate pair.

If $b^2-4ac<0$, then $\sqrt{b^2-4ac}$ will be imaginary. Since $\sqrt{b^2-4ac}$ occurs in both roots $\alpha$ and $\beta$, let $-\frac{b}{2a}=p\;\;\;\text{and}\;\;\;\frac{\sqrt{b^2-4ac}}{2a}=iq$ Then, we get, $\alpha=p+iq\;\;\;\text{and}\;\;\;\beta=p-iq$

Thus, both of the roots are imaginary (complex), each being the conjugate of the other.

Therefore, in a quadratic equation with real coefficients, the imaginary (complex) roots occur in conjugate pair.