# Quadratic Formula Review

## Overview

The quadratic formula is a useful method of factoring second order polynomials. It is commonly used when solving for quantities such as eigenvalues in ordinary differential equations. For reference the formula is shown below:

$$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$$

## Example

Find the roots of the polynomial \(12 x^{2}+10 x-8=4\)

This polynomial can be categorized as second order because the largest power of \(x\) is \(2\) (\(x^2\)). This indicates that there are two possible values of \(x\) that make the equation true.

First the equation must be rearranged to match the standard form:

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

This is accomplished by moving the 4 to the other side of the equation. Now we can equate the coefficients a, b, and c.

$$12 x^{2}+10 x-12=0$$

$$\begin{aligned} a &=12 \\ b &=10 \\ c &=-12 \end{aligned}$$

Now the coefficients are plugged into the above formula and the equation is simplified as follows

$$\begin{array}{c}{x=\frac{-10 \pm \sqrt{10^{2}-4(12)(-12)}}{2(12)}} \\ {x=\frac{-10 \pm \sqrt{100+576}}{24}} \\ {x=\frac{-10 \pm 26}{24}}\end{array}$$

The \(\pm\)operator here confirms that there are two solutions, one achieved by adding 26 and the other achieved by subtracting 26. These two answers are noted as \(x_1\) and \(x_2\)

$$\begin{array}{ll}{x_{1}=\frac{-10-26}{24}} & {x_{2}=\frac{-10+26}{24}} \\ {x_{1}=-\frac{3}{2}} & {x_{2}=\frac{2}{3}}\end{array}$$

The particular values of \(x\) that we have solved for represent zeros of the equation and can be used to find the factored form of the original polynomial.

$$\left(x+\frac{3}{2}\right)\left(x-\frac{2}{3}\right)=0$$

Here the calculated x values are set as zeros in factored form. The entire equation can be multiplied by 12 on both sides to get:

$$12\left(x+\frac{3}{2}\right)\left(x-\frac{2}{3}\right)=0$$

And if the 12 can be distributed inside the parenthesis as shown:

$$4 * 3\left(x+\frac{3}{2}\right) * 2\left(x-\frac{2}{3}\right)=0$$

$$4 *(3 x+2) *(2 x-3)=0$$

The resulting equation is the factored form of the original polynomial.

## Cases of the Quadratic Equation

It is worth noting that there are 3 distinct possibilities when calculating the result of the quadratic equation. The particular case that occurs in each problem depends on the quantity \(\left(b^{2}-4 a c\right)\) which lies inside the square root. This quantity is more commonly known as the discriminant.

### Case 1: \(\left(b^{2}-4 a c\right)>0\)

This case is similar to the example problem solved above. When the discriminant is positive, the equation results in two distinct real solutions because we can take the square root of a positive number.

### Case 2: \(\left(b^{2}-4 a c\right)=0\)

When the discriminant is equal to 0, the quadratic equation takes the form shown below:

$$x=\frac{-b \pm \sqrt{0}}{2 a}$$

Because the square root of 0 is also 0, it doesn't matter whether we add or subtract the discriminant - both give the same answer. In this case there are two real solutions to the equation, however both of them are the same. The factored polynomial will likely look something like:

$$(x+3)(x+3)=0$$

Where in this case x has two real values both of which are equal to -3.

### Case 3: \(\left(b^{2}-4 a c\right)<0\)

When the discriminant is negative, we have to take the square root of a negative number. As a result, the two solutions for x will be non-real numbers. Recall that the quantity *i* is defined as:

$$i=\sqrt{-1}$$

Hence the equation will have two imaginary solutions that are complex conjugates. Try to work along with the following example.

### Case 3 Example

$$5 x^{2}+8 x-5=0$$

$$\begin{array}{l}{a=5} \\ {b=8} \\ {c=5}\end{array}$$

$$\begin{array}{c}{x=\frac{-8 \pm \sqrt{8^{2}-4(5)(5)}}{2(5)}} \\ {x=\frac{-8 \pm \sqrt{-36}}{10}} \\ {x=-\frac{4}{5} \pm \frac{3 i}{5}}\end{array}$$

As expected, the negative discriminant results in two imaginary roots that are complex conjugates. For more information on working with these complex numbers, investigate imaginary number math.