# What is a quadratic inequality

## Quadratic inequalities

### What is a quadratic inequality?

### Introduction to inequality

You treat an inequality in much the same way as an equation. The equals sign is replaced by a **Inequality sign** (also **Comparison sign** called).

A **linear inequality** is a first-degree inequality and has the form $ mx + b \ le 0 $, for example. The **Comparison sign** can also be $ <$ or $ \ ge $ or $> $. You solve an inequality much like an equation. You apply equivalence transformations.

### What are equivalent transformations?

Under one **Equivalent conversion** do you understand transformations that are the truthfulness of a **equation** or **Inequality** Don `t change.

The word “equivalence transformation” comes from the Latin “aequus” for “equal” and “valere” for “to be worth”.

Equivalent transformations are the **Term Conversions**

- on one or both sides of an equation.
- as the
**addition**or**subtraction**of a term**both**Pages of the equation or inequality. - as the
**multiplication**or**division**with a term not equal to $ 0 $**both**Pages of the equation or inequality.

Note when multiplying and dividing by a number (or a term) in the case of inequalities:

- If the number (or term) is positive, the relation sign does not change.
- If the number (or the term) is negative, however, you have to reverse the relation sign.

You can see this in the following example of a linear inequality:

$ \ begin {array} {crclll} & 2x + 3 && (- 2) \ cdot 5 \ end {array} $

### Quadratic inequalities

A **quadratic inequality** has the general form $ ax ^ 2 + bx + c \ le 0 $. The **Comparison sign** can also be $ <$ or $ \ ge $ or $> $.

### Solve a general quadratic inequality.

How are you doing **Solving a square** in front? You perform the following four steps:

**Normal form**: Divide the entire inequality $ ax ^ 2 + bx + c \ le 0 $ by $ a $, the factor before the $ x ^ 2 $. If $ a $ is positive you have an inequality of the form $ x ^ 2 + \ frac {b} {a} x + \ frac {c} {a} \ le 0 $. To simplify the representation, the variables $ p $ and $ q $ were introduced, where $ p = \ frac {b} {a} $ and $ q = \ frac {c} {a} $. For the above inequality with positive $ a $, the inequality $ x ^ 2 + px + q \ le 0 $ results. If $ a $ is negative, the comparison sign must be reversed $ x ^ 2 + px + q \ ge 0 $. Such an inequality is called in**Normal form**.**Solve the equation in normal form**: Solve the equation $ x ^ 2 + px + q = 0 $ for example using the**p-q formula**. The solutions are $ x_1 $ and $ x_2 $.**Factorization**: Factor the left side of the inequality. So you get $ (x-x_1) \ cdot (x-x_2) \ le 0 $.**Factor determination**: Consider the two factors. The product of two factors is less than or equal to $ 0 $ if one of the two factors is greater than or equal to $ 0 $ and the other is less than or equal to $ 0 $.

### There are four cases of the inequality.

The fourth step leads, depending on the relation sign, to $ x ^ 2 + px + q \ le 0 $ or the inequality with $ <$ or $ \ ge $ or $> $ relation sign. Below you can see four different cases. The inequalities with the relation symbols $ <$ and $> $ are considered.

**Less than relation** $ (x-x_1) \ cdot (x-x_2) <0 $

- First case: either $ x-x_1 <0 $ and $ x-x_2> 0 $
- Second case: or $ x-x_1> 0 $ and $ x-x_2 <0 $
- This also applies to the
*Less than or equal to relation*.

**Greater than relation** $ (x-x_1) \ cdot (x-x_2)> 0 $

- Third case: either $ x-x_1> 0 $ and $ x-x_2> 0 $
- Fourth case: or $ x-x_1 <0 $ and $ x-x_2 <0 $
- Incidentally, this also applies to the
*Greater than or equal to relation*.

You can now practice solving a quadratic inequality using two examples.

### Application example for solving quadratic inequalities

### 1st example: Quadratic inequality in normal form

You should **quadratic inequality** Solve $ x ^ 2 + 4x-5 <0 $ in normal form.

Step 1: Bring the inequality into normal form. You don't need to take the first step in solving a quadratic inequality here. The quadratic inequality is already in **Normal form** in front.

Step 2: solve the equation. To solve the equation $ x ^ 2 + 4x-5 = 0 $ can **the p-q formula applied** become. Here $ p = 4 $ and $ q = -5 $. So you get

$ \ begin {array} {rcl} x_ {1,2} & = & - \ frac42 \ pm \ sqrt {\ left (\ frac42 \ right) ^ 2 - (- 5)} \ & = & - 2 \ pm \ sqrt {4 + 5} \ & = & - 2 \ pm \ sqrt {9} \ x_1 & = & - 2 + 3 = 1 \ x_2 & = & - 2-3 = -5 \ end {array} $

Step 3: Factor the left side of the inequality. You now write $ x ^ 2 + 4x-5 $ as the product $ (x-1) \ cdot (x + 5) $ and thus get the inequality $ (x-1) \ cdot (x + 5) <0 $.

Step 4: Solve the factorized inequality. Note that a product of two factors is negative if the two factors have different signs. This means that if one factor is negative, the other must be positive, and vice versa. So you get the following two cases:

- $ x-1> 0 $. Then $ x> 1 $ and $ x + 5 <0 $ or equivalent would be $ x <-5 $. There is no such thing as a $ x $ that can be greater than $ 1 $ and less than $ -5 $ at the same time.
- $ x-1 <0 $. Then $ x <1 $ and $ x + 5> 0 $ or equivalent would be $ x> -5 $. This means that $ -5

Now you can **Solution set** the quadratic inequality $ \ mathbb {L} = \ {x \ in \ mathbb {R} ~ | ~ -5

You can make this clear to yourself using a number line:

- Since the $ <$ or $> $ relation is being considered, $ -5 $ and $ 1 $ must be excluded. You can recognize this by the empty circle.
- Above the number line you can see the upper of the two cases. There are no solutions to this case.
- Below the number line you can see the lower case and (blue) the solution set.

### Example 2: Solving a general quadratic inequality.

Solve the quadratic inequality: $ - \ frac12x ^ 2 + 3x-4 <0 $

Multiply by $ -2 $. Note that you have to reverse the relation sign: $ x ^ 2-6x + 8> 0 $ This inequality is in normal form.

Find the solutions to the equation $ x ^ 2-6x + 8 = 0 $. You can get this with the help of the

**p-q formula**: $ x_1 = 2 $ and $ x_2 = 4 $.Factor the inequality $ (x-2) \ cdot (x-4)> 0 $.

- Solve the inequality: either $ x-2> 0 ~ \ Leftrightarrow ~ x> 2 $ and $ x-4> 0 ~ \ Leftrightarrow ~ x> 4 $, so $ x> 4 $. In the second case, $ x-2 <0 ~ \ Leftrightarrow ~ x <2 $ and $ x-4 <0 ~ \ Leftrightarrow ~ x <4 $, i.e. also $ x <2 $.

Here, too, you can make the solution set clear with the help of a number line. You can see the first of the two cases above and the second below the number line:

So it is $ \ mathbb {L} = \ {x \ in \ mathbb {R} ~ | ~ x <2 ~ \ text {or} ~ x> 4 \} $.

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