# Three forms of Quaadratic Equation

We can obtain three different representations of a quadratic function

1. Its standard representation: $f(x) = ax^2 + bx + c$.
The function f has two zeros, a double zero, or no zero according to whether its discriminant $b^2-4ac$ is positive, zero, or negative, respectively
2. Its representation in vertex form: $f (x) = a(x-p)^2 + q$. Here, the point $(p, q)$ is called the vertex of the graph.
The graph of f(x) has bilateral symmetry with respect to the vertical line defined by $x = p$.
If a > 0, f(x) is decreasing on (−∞, p] and therefore increasing on [p,∞).
If a < 0, then f(x) is increasing on (−∞, p] and therefore decreasing on [p,∞).
If a > 0, then f(x) achieves its minimum at p.
If a < 0, then f(x) achieves its maximum at p.
3. Its representation in factored form: $f (x) = a(x−\alpha)(x−\beta)$, where $\alpha, \beta$ are the zeros of f(x).Also $\alpha+ \beta=-\frac{b}{a}$ and $\alpha \beta=\frac{c}{a}$

Each of the three representations of a quadratic function reveals a different facet of the function. The quadratic formula is expressed in terms of (1), for standard notion, and (2) displays the line of symmetry of the graph of f(x) and also where it achieves its maximum or minimum. If the zeros of f(x) are our main interest, then (3) displays its zeros explicitly. Together, the three representations give a well-rounded picture of f(x); none gets it done alone.

#### Three forms of Quaadratic Equation

A solution of a quadratic equation is also called a root of the equation. The intuitive meaning of the root of a quadratic equation can be given pictorially, as follows.

The graph of y = 2x2 − x − 3 intersects the x-axis at (−1, 0) and (1.5, 0), and a simple computation confirms that −1 and 1.5 are roots of 2x2 − x − 3 = 0

When solving quadratic equations, we can use two methods:
1. Factoring
Not every quadratic equation can be solved by factoring. In this case, we need to use the quadratic formula.
$x= \frac{-b \pm \sqrt {b^2-4ac}}{2a}$

Prove that two roots of quadratic equations $ax^2 + bx + c = 0$ are $\frac{-b - \sqrt{b^2-4ac}}{2a}$ and $\frac{-b + \sqrt{b^2-4ac}}{2a}$
Proof
$ax^2 + bx + c = 0$
or $ax^2 + bx =-c$
Dividing both sides by a, we get
$x^2 + \frac{b}{a}x = - \frac{c}{a}$
Adding $\frac{b^2}{4a^2}$ on both sides we get
$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}= \frac{b^2}{4a^2}- \frac{c}{a}$
or $\left( x+\frac{b}{2a} \right )^2=\frac{b^2-4ac}{4a^2}$
Taking square roots we get
$x+\frac{b}{2a}=\frac{\pm \sqrt{b^2-4ac}}{2a}$
or $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
So, the roots of quadratic equations $ax^2 + bx + c = 0$ are $\frac{-b - \sqrt{b^2-4ac}}{2a}$ and $\frac{-b + \sqrt{b^2-4ac}}{2a}$

To solve a Quadratic equation using the formula, we use the following steps:
1. Put the quadratic equation into standard form : $ax^2 + bx + c = 0$
2. Write out the value for a, b, and c
3. Substitute value in the formula, solve and get the roots
4. Check each root in the original equation