# What is a nonlinear equation

### What is a power equation?

An equation of the form $$ ax ^ 2 + b = c $$ is called a power equation.

Any natural number can be used where the exponent $$ 2 $$ is. You are only interested in the cases $$ x ^ 2 $$ and $$ x ^ 3 $$.

When solving these equations, proceed as always until you have reached $$ x ^ 2 = $$ ... or $$ x ^ 3 = $$ ... at the end of the transformation.

Taking a number with yourself is called **Potentiate**. $$ x * x = x ^ 2 $$ or $$ x * x * x = x ^ 3 $$

Hence the name of the equation.

The exponent is called **exponent**.

### Power equations with $$ x ^ 2 $$

**Simple example:**

$$ x ^ 2 = 9 $$

You think about which number multiplied by itself results in $$ 9 $$.

$$ x = 3 $$

However, the 3 is not the only solution, because $$ - 3 $$ is also possible.

$$ x = -3 $$

**Sample:**

$$ 3 * 3 = 9 $$ and $$ (- 3) * (- 3) = 9 $$.

The solution set contains two numbers.

$$ L = {- 3; 3} $$

In the course of your math career you will in this case **the second root** pull:

$$ x ^ 2 = 9 $$ $$ | sqrt ($$

### Power equations with $$ x ^ 3 $$

**Simple example:**

$$ x ^ 3 = 27 $$

You think about which number multiplied twice by itself results in $$ 27 $$.

$$3*3*3=27$$

$$ x = 3 $$

The solution set is $$ L = {3} $$.

Or is $$ - 3 $$ also possible? No, $$ 3 $$ is the only solution.

Because $$ (- 3) * (- 3) * (- 3) = - 27 $$.

You now know immediately what the solution for the equation $$ x ^ 3 = -27 $$ is.

The solution set is $$ L = {- 3} $$.

As your math career progresses, you will in this case **the third root** pull:

$$ x ^ 3 = 27 $$ $$ | root 3 () $$

You can recognize the 3rd root by the small three on the top left.

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### $$ x ^ 2 $$ and negative numbers

For equations of the form $$ x ^ 2 = -9 $$ you can write down that the equation **no solution** owns.

The solution set is empty: $$ L = {$$ $$} $$.

There is no single number that, when multiplied by itself, produces a negative number.

### Solve a complex power equation

$$ 3x * (x-1) + 5 + 3x = 17 $$ $$ | $$ Break the brackets

$$ 3x ^ 2-3x + 5 + 3x = 17 $$ $$ | $$ summarize

$$ 3x ^ 2 + 5 = 17 $$ $$ | -5 $$

$$ 3x ^ 2 = 12 $$ $$ |: 3 $$

$$ x ^ 2 = 4 $$ $$ | $$ Consider

Which number / s result when multiplied by themselves $$ 4 $$?

$$ x = 2 $$ and $$ x = -2 $$

$$ L = {- 2; 2} $$

### Equations with $$ x ^ 2 $$ and $$ x $$

$$ x ^ 2-2x = 3 $$

Divide the equation like this:

$$ x ^ 2-2x = 3 $$ $$ | + 2x $$

$$ x ^ 2 $$ stands alone on one side.

$$ x ^ 2 = 2x + 3 $$

### Solve by trying and creating a table:

Use numbers that make sense to you. $$ x ^ 2 $$ can only take the square numbers $$ {1, 4, 9, 16, 25, 36, 49, ...} $$.

Check number | $$ x ^ 2 $$ | $$ 2x + 3 $$ |
---|---|---|

$$-1$$ | $$1$$ | $$1$$ |

$$0$$ | $$0$$ | $$3$$ |

$$1$$ | $$1$$ | $$5$$ |

$$2$$ | $$4$$ | $$7$$ |

$$3$$ | $$9$$ | $$9$$ |

The numbers $$ - 1 $$ and $$ 3 $$ are the solutions.

$$ L = {- 1; 3} $$

You will solve this task later with a formula or graphically. Now such equations always have whole numbers as a solution.

The numbers $$ 1 $$ and $$ - 3 $$ are not solutions.

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### Equations with absolute bars

### Mathematical representation

$$ | x | = 2 $$

There are two cases that will be considered separately.

1st case: positive

$$ x = 2 $$

2nd case: negative

$$ x = -2 $$

The solution set is $$ L = {- 2; 2} $$.

### Cordial representation ☺

$$|$$♥$$|=2$$

There are two possibilities:

♥ loves me OR ♥ don't love me

(positive for me) (negative for me)

„♥ loves me ”gives the equation ♥ $$=2$$

„♥ don't love me ”gives the equation ♥ $$=-2$$

The solution set $$ L = {- 2; 2} $$ means:

- Life goes with you ♥ further.
- Life goes without ♥ further.

The **amount** is the **distance** a number for $$ 0 $$.

$$ | 2 | = 2 $$ and $$ | -2 | = 2 $$

So $$ | 2 | = | -2 | $$ also applies

The same goes for variables:

$$ | x | = x $$ and $$ | -x | = x $$

So $$ | x | = | -x | $$

Equations of the type $$ | x | = -2 $$ have no solution. $$ L = {$$ $$} $$

### There is not just a $$ x $$ in the amount?

Even then, there are two cases.

**Example:** $$ | x-4 | = 9 $$

1st case: $$ x-4 = 9 $$ $$ | + 4 $$

$$ x = 13 $$

2nd case: $$ x-4 = -9 $$ $$ | + 4 $$

$$ x = -5 $$

**1st sample:**

$$|13–4|=9$$

$$|9|=9$$

$$9=9$$

**2nd sample:**

$$|-5–4|=9$$

$$|-9|=9$$

$$9=9$$

The solution set is $$ L = {13; -5} $$.

It must **not always** give a negative and a positive solution.

The equation $$ | $$♥$$ - 5 | = 1 $$ has the solution set $$ L = {4; 6} $$.

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