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Quadratic equations are a fundamental part of elementary algebra. The general form of a qaudratic equation is $ax^2+bx+c=0$, where $x$ is the unkown variable and $a$, $b$ and $c$ are known numbers. These numbers are called the coefficients of the equation.

“It’s called a quadratic, which means squared, equation because the unknown variable $x$ is squared. It is also sometimes called a second degree polynomial equation, representing the highest power in the equation.”

Graphically, quadratic equations are parabolics and the solutions or ‘roots’ to a quadratic equation are the intersections of the parabolic with the x-axis.
There are several ways to solve a quadratic equation such as

1. Factoring
2. Completing the square

quadratic equations are not always presented in the standard form of $ax^2+bx+c=0$. In real life situations this is rarely the case. Putting it in the standard form greatly simplifies the problem at hand. It is therefore important to develop the skills to recognize a quadratic equation in disguise and solve it. Here are some examples of quadratic equations in disguise:

In disguise In standard form
$x^{2/3}+2x^{1/3}+3=0$ Substitute $x^{1/3}$ with $u$:
$u^2+2u+3=0$
$\frac{1}{2x+3} +x-1=0$ Multiply both sides of the equation with $\frac{1}{2x+3}$:
$1+(x-1)(2x+3)=0$
$2x^2+x-2=0$
$\frac{1}{log(x)} +log(x)-2=0$ Substitute $log(x)$ by $u$ and multiply both sides by $u$:
$1+(u-2)u=0$
$u^2 -2u+1=0$
$x^4+8x^2+16=0$ Substitute $x^2$ by $u$:
$u^2+8u+16=0$
$\frac{x+2}{x^2-x}=4$ $x+2=4x^2-4x$
$4x^2-5x-2=0$

## Solving quadratics by taking square roots

To start developing the skills to solve quadratic equations in disguise like these, it is important to learn how (and why) to solve simple quadratic equations by taking square roots. These basic skills are needed in solving most quadratic equations.

Hidden or disguised quadratic equations can often easily be solved by substitution (see the example of the table above) to rewrite the equation to the standard form. One way to do this is factoring.

## Factoring

Factoring is the opposite of expanding, in which you remove all the brackets in an equation. Here are some examples of factoring the quadratic equation.

In standard form In factors
$x^2+x-12=0$ $(x+4)(x-3)=0$
$3x^2+9x-12=0$ $3(x+4)(x-1)=0$
$3x^2+5x-2=0$ $(x+2)(3x-1)=0$

It is easy to see that once you’ve put it in factor form, you can easily find the roots (the solutions) of the quadratic equation. Either one of the factors should be zero to obtain zero. This is what makes this method so good, you put in a little effort to factor the quadratic equation but once you’ve found the factors, you basically found the solutions.

While expanding is fairly easy, factoring can be quite hard and requires a lot of practice. Take a look at the table above. The first example $x^2+x-12=0$ is called a monic quadratic equation. This is a fancy way of saying $a=1$. These are the easiest forms to factor. The second example $3x^2+9x-12=0$ is a monic quadratic in disguise, since all factors can be divided by a factor 3, which we simply put in front of both factors. The last example is called a non-monic quadratic, since $a \neq 1$.

Here we will show you several different forms and methods of factoring.

## Completing the square

The factoring method generally only works for nice integer solutions. Once the solutions become a little more diffucult like ratio’s for example it can take you a very long time to find the solutions. For these cases we can use the method of competing the square. Historically, this is the most commonly used method for solving quadratic equations. It is a frequantly used thechnique in math to rewrite terms in for example sums and integrals and put it in standard forms which can be solved. Consider, for example the following integral:

$$\int{\frac{dx}{x^2-6x+13}}$$

A commonly used method for solving integrals of this form is to used partial fractions. However, sometimes the denominator does not factor nicely, as in this case. When there is also no obvious substitution we can complete the square as follows

$$x^2-6x+13=(x^2-6x+9)+4=(x-3)^2+4$$

So let’s solve the integral by rewriting the denominator.

$$\int{\frac{dx}{(x-3)^2+4}}$$

By using the identities for trigoniometric integrals we can easily solve this integral

$$\frac{1}{2}\arctan\left(\frac{x-3}{2}\right)+C$$

So you see, this is a nice skill to have and sometimes the only way to work your way out of a mathematic problem. Also, this method can be applied to every quadratic equation, even for imaginairy cases.