Quadratic Inequality

πŸ†Practice quadratic inequalities

The quadratic inequality shows us in which interval the function is positive and in which it is negative - according to the inequality symbol. To solve quadratic inequalities correctly, it is convenient to remember two things:

  1. Set of positivity and negativity of the function:
    Set of positivity - represents the XXs in which the graph of the parabola is above the XX axis, with YY value positive.
    Set of negativity - represents the XXs in which the graph of the parabola is below the XX axis, with YY value negative.
  2. Dividing by a negative term - reverses the sign of the inequality.

Method to solve the quadratic inequality:

  1. We will carry out the transposition of members and isolate the quadratic equation until one side equals 0. Remember that when we divide by a negative term, the inequality is reversed.
  2. Let's draw a diagram of the parabola - placing points of intersection with the XX axis and identifying the maximum and minimum of the parabola.
  3. Let's calculate the corresponding interval according to the exercise and the diagram.
    Quadratic equation >0∢>0∢ Set of positivity
    Quadratic equation <0∢<0∢ Set of negativity
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Test yourself on quadratic inequalities!

einstein

Solve the following equation:

\( x^2+4>0 \)

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Let's look at an example of a quadratic inequality

3x(x+1)<2(x2+15)+2x3x(x+1)<2(x^2+15)+2x

Solution:
We will progress step by step:

1) Let's transpose terms and isolate the quadratic equation until one side equals 00. Remember that when we divide by a negative term, the inequality is reversed.
In the first step, we will leave 00 on one side of the equation.
Note that, in this exercise, we must first solve what appears between parentheses.
We will open the parentheses and obtain:
3x2+3x>2x2+30+2x3x^2+3x>2x^2+30+2x

Now let's transpose terms and we will obtain:
X2+Xβˆ’30>0X^2+X-30>0

Magnificent. We have left 00 on one side. Let's continue to the second step.

2) Let's draw a diagram of the parabola - placing intersection points with the XX axis and identifying the maximum and minimum of the parabola.

Let's find the intersection points of the function with the XX axis:
According to the quadratic formula we will obtain:

​​​​​​​X=5,βˆ’6​​​​​​​X=5,-6

We will see that the extremity of the function is the minimum (smile) since the coefficient of X2X^2 is positive.
Let's draw a diagram:

A1-Desigualdad cuadrΓ‘tica

3) Let's calculate the corresponding interval according to the exercise and the diagram. Β 
In the exercise, we arrived at the following equation:
​​​​​​​X2+Xβˆ’30>0​​​​​​​X^2+X-30>0
That is, we are looking for the intervals in which the function is greater than 00. Its set of positivity.

We will ask ourselves: In which intervals is the function positive? At which XXs is the graph of the function above the XX axis?
The answer is when
X>5X>5
X<βˆ’6X<-6

And these are the solutions for the quadratic inequality.


Examples and exercises with solutions of quadratic inequality

Exercise #1

Solve the following equation:

x^2+4>0

Video Solution

Answer

All values of x x

Exercise #2

Solve the following equation:

x^2+9>0

Video Solution

Answer

All values of x x

Exercise #3

Solve the following equation:

-x^2-9>0

Video Solution

Answer

There is no solution.

Exercise #4

Solve the following equation:

-x^2+2x>0

Video Solution

Answer

0 < x < 2

Exercise #5

Solve the following equation:

x^2-3x+4<0

Video Solution

Answer

There is no solution.

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