Definition and general solving of quadratic equations

An equation as $x^{2} + 3 x - 10 = 0$ 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, $4 x^{3} + 2 x + 10 = 0$ would not be of the second degree, but of the third).

The general form of an equation of this type is:

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

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:

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

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 $-$ .

Example

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

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

$a = 1, b = 3 and c = - 10$

and we replace them in the formula:

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

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

And we get two different solutions:

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

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 $x^{2} + 3 x - 10 = 0$ are $2$ and $- 5$ .

Let's see other examples:

Example

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

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

$= {\begin{matrix} x_{1} = \frac{4}{3} \\ x_{2} = - \frac{1}{2} \end{matrix}$

Example

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

$x = \frac{- 1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot (- 2)}}{2 \cdot 1} = \frac{- 1 \pm \sqrt{9}}{2} = \frac{- 1 \pm 3}{2} = {\begin{matrix} x_{1} = 1 \\ x_{2} = - 2 \end{matrix}$

Example

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

$x = \frac{5 \pm \sqrt{5^{2} - 4 \cdot 2 \cdot (- 1)}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 + 8}}{4} = \frac{5 \pm \sqrt{33}}{4} = = {\begin{matrix} x_{1} = 2.69 \\ x_{2} = - 0.19 \end{matrix}$

Example

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

$x = \frac{0 \pm \sqrt{0 - 4 \cdot (- 16)}}{2} = \frac{\pm 8}{2} = {\begin{matrix} x_{1} = 4 \\ x_{2} = - 4 \end{matrix}$

Example

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 2 \cdot 0}}{2 \cdot 2} = {\begin{matrix} x_{1} = 2 \\ x_{2} = 0 \end{matrix}$

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

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

The solutions for this equation are:

$\begin{matrix} k_{1} = 1 \\ k_{2} = \frac{5}{3} \end{matrix}$

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 $R$ .

Example

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

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

This equation has no solutions in $R$ .