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.

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