Definition and general solving of quadratic equations

An equation as $x^2+3x-10=0$ is said to be a quadratic equation or a second degree equation because the exponent of $x$ (which is the unknown) is $2$ (an equation such as, for example, $4x^3+2x+10=0$ would not be of the second degree, but of the third).

The general form of an equation of this type is:

$$ax^2+bx+c=0$$

Where $x$ is the unknown and $a$, $b$, $c$ are any numbers.

The formula that allows us to solve this type of equations is the following one:

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

In this operation a sign $\pm$ appears, and the fact is that, in principle, a quadratic equation can have two different solutions, one of them is obtained when we use $+$ and other one when we use $ -$.

We are going to apply this formula to the equation $x^2+3x-10=0$.

We write the values of $a$, $b$ and $c$:

$$a= 1, b= 3 \mbox{ and } c=-10$$

and we replace them in the formula:

$$\displaystyle x=\frac{-3 \pm \sqrt{3^2-4 \cdot 1 \cdot (-10)}}{2 \cdot 1}=\frac{-3 \pm \sqrt{9+40}}{2}=\frac{-3\pm \sqrt{49}}{2}=$$

$$=\frac{-3 \pm 7}{2}$$

And we get two different solutions:

$$\displaystyle \frac{-3+7}{2}=\frac{4}{2}=2 \\ \frac{-3-7}{2}=\frac{-10}{2}=-5$$

Therefore, the proposed equation has the solutions $2$ and $-5$.

In most of the textbooks the solutions are indicated by writing a subscript in the letter $x$, so that in our case we would have:$$x_1=2 \\ x_2=-5$$

The solutions of the equation are called roots. It is the same to say that $2$ and $-5$ are the solutions, than it is to say that the roots of the equation $x^2+3x-10=0$ are $2$ and $-5$.

Let's see other examples:

Solve the equation $6x^2-5x-4=0$. $a=6$, $b=-5$ and $c=-4$

$$\displaystyle x=\frac{-(-5) \pm \sqrt{(-5)^2-4 \cdot 6 \cdot (-4)}}{2 \cdot 6}= \frac{5 \pm \sqrt{25+96}}{12}=\frac{5 \pm 11}{12}=$$

$$\displaystyle=\left\{\begin{matrix} x_1=\frac{4}{3} \\ x_2=-\frac{1}{2}\end{matrix}\right.$$

Find the solutions of the equation $x^2+x-2=0$. $a=1$, $b=1$ and $c=-2$

$$\displaystyle x=\frac{-1 \pm \sqrt{1^2-4 \cdot 1 \cdot (-2)}}{2 \cdot 1}= \frac{-1 \pm \sqrt{9}}{2}=\frac{-1 \pm 3}{2}=\left\{\begin{matrix} x_1=1 \\ x_2=-2\end{matrix}\right.$$

Which are the roots of $2x^2-5x-1=0$? $a=2$, $b=-5$ and $c=-1$

$$\displaystyle x=\frac{5 \pm \sqrt{5^2-4 \cdot 2 \cdot (-1)}}{2 \cdot 2}= \frac{5 \pm \sqrt{25+8}}{4}=\frac{5 \pm \sqrt{33}}{4}= \\ =\left\{\begin{matrix} x_1=2.69 \\ x_2=-0.19\end{matrix}\right.$$

Solve $x^2-16=0$. $a=1$, $b=0$ and $c=-16$

$$\displaystyle x=\frac{0 \pm \sqrt{0-4 \cdot (-16) }}{2}= \frac{\pm 8}{2}=\left\{\begin{matrix} x_1=4 \\ x_2=-4\end{matrix}\right.$$

Find the roots of $2x^2-4x=0$. $a=2$, $b=-4$ and $c=0$.

$$\displaystyle x=\frac{4 \pm \sqrt{16-4\cdot 2 \cdot 0 }}{2 \cdot 2}=\left \{ \begin{matrix} x_1=2 \\x_2=0 \end{matrix}\right. $$

Sometimes the terms of the equation are grouped in a different way, as in $5-x=3x^2$ in which case we only need to move everything to the first member $-3x^2-x+5=0$

In other cases it is possible that the unknown is not represented using the letter $x$, as in $3k^2-8k+5=0$, but this does not change things.

The solutions for this equation are:

$$\begin{matrix} k_1=1 \\ k_2= \displaystyle \frac{5}{3}\end{matrix}$$

It is important to remember that the square root of a negative number does not exist within the set of the real numbers. If we find a case like this we will say that the equation has no solutions in $\mathbb{R}$.

$x^2+2x+5=0$. $a=1$, $b=2$ and $c=5$.

$$\displaystyle x=\frac{-2 \pm \sqrt{4-20}}{2 \cdot 2}= \frac{-2 \pm \sqrt{-16}}{4}$$

This equation has no solutions in $\mathbb{R}$.