# Definition of Equation. Roots of the Equation.

The equation is two expressions separated by an equal sign (=). We will mainly deal with equations that contain one or more variables.

Examples of equations:

• ${2}+{3}={5}$ (no variables)
• ${4}{x}+{3}={1}$ (variable in left expression)
• ${15}={{x}}^{{2}}+{2}{x}$ (variable in right expression)
• ${{x}}^{{3}}+{4}-{2}{{x}}^{{2}}=\frac{{1}}{{x}}+{{2}}^{{x}}$ (variable in both expressions)

Roots of the equation are such values of the variable, that turn equation into correct equality.

Example 1. Determine, whether 2 and 3 are roots of the equation ${15}={{x}}^{{2}}+{2}{x}$.

To check it, we just plug values instead of the variable.

Check for 2:

${15}={{2}}^{{2}}+{2}\cdot{2}$

${15}\ne{8}$

We haven't obtained correct equality, so 2 is not root of the equation.

Check for 3:

${15}={{3}}^{{2}}+{2}\cdot{3}$

${15}={15}$

Since equality is correct, then 3 is root of the equation.

Example 2. Determine, whether $-{1}$ and ${2}$ are roots of the equation ${{y}}^{{3}}-{3}={3}{y}-{{y}}^{{2}}$.

To check it, we just plug values instead of the variable.

Check for $-{1}$:

${{\left(-{1}\right)}}^{{3}}-{3}={3}\cdot{\left(-{1}\right)}-{{\left(-{1}\right)}}^{{2}}$

$-{4}=-{4}$

Since equality is correct, then $-{1}$ is root of the equation.

Check for ${2}$:

${{2}}^{{3}}-{3}={3}\cdot{2}-{{2}}^{{2}}$

${5}\ne{2}$

Since equality is incorrect, then ${2}$ is not root of the equation.

Note, that equation can have no roots.

To solve the equation means to find all its roots.

In general, it is quite hard to solve equation. Often, roots can be found only approximately (with the help of computer). But we will consider types of equations, that can be easily solved.

Exercise 1. Determine, whether ${2}$ and $-{2}$ are roots of the equation ${2}{x}+{4}={0}$.

Answer: $-{2}$ is root, ${2}$ is not.

Exercise 2. Determine, whether $\frac{{1}}{{2}}$ and $-\sqrt{{{5}}}$ are roots of the equation ${10}{a}-{5}={2}{{a}}^{{3}}-{{a}}^{{2}}$.

Exercise 3. Determine, whether ${1}$ and $\sqrt{{{3}}}$ are roots of the equation ${{y}}^{{4}}-{2}{{y}}^{{2}}+{4}={0}$.