# Which is the best quadratic equation trick

### For reading

If a quadratic equation only has terms with the variable (mostly $$ x $$) and should be $$ = 0 $$, there is a simple trick for solving it.

#### Examples

- $$ x ^ 2 + x = 0 $$
- $$ 2x ^ 2 + 1.5x = 0 $$
- $$ 2x -x ^ 2 = 0 $$

#### Trick: Exclude x = factorize

Solve the equation $$ x ^ 2 + 2x = 0 $$.

Factor out x. So you get 2 factors. That's why the step is called **Factoring**.

Exclude $$ x ^ 2 + 2x = 0 $$ $$ | $$

$$ x (x + 2) = 0 $$

There is now a product on the left. A product becomes 0 when one or the other part is 0.

1st case: $$ x = 0 $$

So 1st solution: $$ x_1 = 0 $$

2nd case: $$ x + 2 = 0 $$

So 2nd solution: $$ x_2 = -2 $$

Solution set: $$ L = {0; -2} $$

You solve equations that only contain terms with variables by factoring them out. Then apply: A product is 0 if at least one of the factors is 0.

**Zero product rate**: The product $$ a · b $$ of two terms a and b is zero if and only if at least one of the factors a or b is zero.

Shorter: From $$ a b = 0 $$ follows $$ a = 0 $$ or $$ b = 0 $$ or both equal 0.

### First transform, then factor

You first transform equations that have a factor before the $$ x ^ 2 $$.

#### 1st example

$$ - 4 $$$$ x ^ 2-8x = 0 $$ $$ | $$ $$: $$$$ (- 4) $$

$$ x ^ 2 + 2x = 0 $$

$$ x (x + 2) = 0 $$

1st case: $$ x = 0 $$

So 1st solution: $$ x_1 = 0 $$

2nd case: $$ x + 2 = 0 $$

So 2nd solution: $$ x_2 = -2 $$

Solution set: $$ L = {0; -2} $$

#### 2nd example

$$ (1) / (3) x ^ 2 = 2x $$ $$ | · 3 $$

$$ x ^ 2 = 6x $$ | $$ - 6x $$

$$ x ^ 2-6x = 0 $$

$$ x (x-6) = 0 $$

1st case: $$ x = 0 $$

So 1st solution: $$ x_1 = 0 $$

2nd case: $$ x-6 = 0 $$

So 2nd solution: $$ x_2 = 6 $$

Solution set: $$ L = {0; 6} $$

If a quadratic equation only contains quadratic and linear terms, you can solve the equation by rearranging and factoring.

The equation is divided by $$ - 4 $$ so that the coefficient of the square term becomes $$ 1 $$.

A second solution is possible if you exclude $$ - 4x $$.

$$ - 4x ^ 2-8x = 0 $$

$$ - 4x (x + 2) = 0 $$

1st solution: $$ - 4x = 0 $$ with $$ x_1 = 0 $$

2nd solution: $$ x + 2 = 0 $$ with $$ x_2 = -2 $$

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### Equations of the form $$ x ^ 2 + px = 0 $$, $$ p in RR $$

This is the general solution:

### Solution steps

Equation: | $$ x ^ 2 + px = 0 $$ |

Factorize: | $$ x (x + p) = 0 $$ |

Zero product rate: | $$ x = 0 $$ or $$ x + p = 0 $$ |

Solutions: | $$ x_1 = 0 $$ and $$ x_2 = -p $$ |

Solution set: | $$ L = {0 ;-p} $$ |

#### example

Equation: | $$ x ^ 2-2.5 x = 0 $$ |

Factorize: | $$ x (x-2.5) = 0 $$ |

Zero product rate: | $$ x = 0 $$ or $$ x-2.5 = 0 $$ |

Solutions: | $$ x_1 = 0 $$ and $$ x_2 = 2.5 $$ $$ p = -2.5 $$, i.e. $$ - p = + 2.5 $$ |

Solution set: | $$ L = {0; 2.5} $$ |

**A** Solution of these quadratic equations is always 0!

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