# Roots of a Quadratic Equation

A value that makes the quadratic expression zero is called the root of a quadratic equation.

A quadratic equation is a second degree polynomial. Hence, it has two roots for which the quadratic expression is equal to zero. For example, $ax^2+bx+c = 0$ is a quadratic equation in standard algebraic form. The following two are the roots which make the quadratic expression zero.

## Formula

$x \,=\, \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$

### Proof

The two roots of the general form quadratic equation can be derived in mathematics by solving the quadratic equation $ax^2+bx+c = 0$.

01

#### Basic simplification for the Perfect square conversion

Move the constant term $c$ to right hand side of the equation.

$\implies ax^2 + bx = -c$

Divide both sides by the literal coefficient of the $x^2$.

$\implies$ $\dfrac{ax^2 + bx}{a} = -\dfrac{c}{a}$

$\implies$ $\dfrac{ax^2}{a} + \dfrac{bx}{a} = -\dfrac{c}{a}$

$\implies$ $\Bigg(\dfrac{a}{a}\Bigg)x^2 + \Bigg(\dfrac{b}{a}\Bigg)x = -\dfrac{c}{a}$

$\implies$ $\require{cancel} \Bigg(\dfrac{\cancel{a}}{\cancel{a}}\Bigg)x^2 + \Bigg(\dfrac{b}{a}\Bigg)x = -\dfrac{c}{a}$

$\implies$ $x^2 + \Bigg(\dfrac{b}{a}\Bigg)x = -\dfrac{c}{a}$

02

#### Transforming the Equation as a square of a binomial

The left hand side expression can be transformed as a square of a binomial by some adjustments.

$\implies$ $x^2 + 1 \times \Bigg(\dfrac{b}{a}\Bigg)x$ $\,=\,$ $-\dfrac{c}{a}$

The multiplying factor $1$ can be written as the quotient of $2$ by $2$ because the division of $2$ by $2$ is $1$.

$\implies$ $x^2 + \Bigg(\dfrac{2}{2}\Bigg) \times \Bigg(\dfrac{b}{a}\Bigg)x$ $\,=\,$ $-\dfrac{c}{a}$

$\implies$ $x^2 + 2\Bigg(\dfrac{b}{2a}\Bigg)x$ $\,=\,$ $-\dfrac{c}{a}$

There is a chance to transform the left hand side expression as a square of an addition based binomial. So, add $\Bigg(\dfrac{b}{2a}\Bigg)^2$ and subtract it by the same term.

$\implies$ $x^2 + 2\Bigg(\dfrac{b}{2a}\Bigg)x + \Bigg(\dfrac{b}{2a}\Bigg)^2 -\Bigg(\dfrac{b}{2a}\Bigg)^2$ $\,=\,$ $-\dfrac{c}{a}$

Move the negative term to right hand side.

$\implies$ $x^2 + 2\Bigg(\dfrac{b}{2a}\Bigg)x + \Bigg(\dfrac{b}{2a}\Bigg)^2$ $\,=\,$ $\Bigg(\dfrac{b}{2a}\Bigg)^2 -\dfrac{c}{a}$

$\implies$ $x^2 + \Bigg(\dfrac{b}{2a}\Bigg)^2 + 2x\Bigg(\dfrac{b}{2a}\Bigg)$ $\,=\,$ $\Bigg(\dfrac{b}{2a}\Bigg)^2 -\dfrac{c}{a}$

The left hand side expression represents the expansion of the square of the sum of the two terms. It can be simplified by the a+b whole square formula.

$\implies {\Bigg(x + \dfrac{b}{2a}\Bigg)}^2$ $\,=\,$ $\dfrac{b^2}{4a^2} -\dfrac{c}{a}$

$\implies {\Bigg(x + \dfrac{b}{2a}\Bigg)}^2$ $\,=\,$ $\dfrac{b^2 -4ac}{4a^2}$

03

#### Solving the value of x

The square of the left hand side expression becomes square root to right hand side expression.

$\implies$ $x + \dfrac{b}{2a}$ $\,=\,$ $\pm \sqrt{\dfrac{b^2 -4ac}{4a^2}}$

$\implies$ $x + \dfrac{b}{2a}$ $\,=\,$ $\pm \sqrt{\dfrac{b^2 -4ac}{{(2a)}^2}}$

$\implies$ $x + \dfrac{b}{2a}$ $\,=\,$ $\pm \dfrac{\sqrt{b^2 -4ac}}{2a}$

$\implies$ $x = -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 -4ac}}{2a}$

$\,\,\, \therefore \,\,\,\,\,\, x = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$

04

#### Roots of the Equation

Therefore, $x$ is equal to $\dfrac{-b + \sqrt{b^2 -4ac}}{2a}$ and $x$ is also equal to $\dfrac{-b -\sqrt{b^2 -4ac}}{2a}$. For these two values, the quadratic expression is equal to zero.

##### Representation

If the two roots are represented by $\alpha$ and $\beta$, the two roots of the quadratic equation are written as follows.

$\alpha = \dfrac{-b + \sqrt{b^2 -4ac}}{2a}$

$\beta = \dfrac{-b -\sqrt{b^2 -4ac}}{2a}$

But remember, $\alpha$ and $\beta$ are the values of variable $x$ for the quadratic equation $ax^2+bx+c = 0$.