# 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 $$\mathbb{R}$$.