Quadratic Inequality

🏆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!

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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+X30>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=5,6​​​​​​​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 cuadrática

3) Let's calculate the corresponding interval according to the exercise and the diagram.  
In the exercise, we arrived at the following equation:
​​​​​​​X2+X30>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<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

Step-by-Step Solution

To solve this problem, let's examine the inequality x2+4>0 x^2 + 4 > 0 .

The expression x2+4 x^2 + 4 consists of two terms: x2 x^2 and 4 4 . Notice that:

  • The term x2 x^2 is always non-negative, which means x20 x^2 \geq 0 for any real number x x .
  • The constant term 4 4 is positive.

Combining these observations, we see that:

  • Since x2 x^2 is non-negative, x2+44 x^2 + 4 \geq 4 .
  • Therefore, x2+4 x^2 + 4 is always greater than zero, as adding 4 to a non-negative number will always yield a positive result.

Thus, there are no values of x x for which the expression x2+4 x^2 + 4 is zero or negative. Instead, the expression is always positive for all real numbers x x .

Therefore, the solution to the inequality x2+4>0 x^2 + 4 > 0 is all values of x x .

Answer

All values of x x

Exercise #2

Solve the following equation:

x^2+9>0

Video Solution

Step-by-Step Solution

Let's explore this problem step-by-step:

The inequality given is x2+9>0 x^2 + 9 > 0 .

1. To understand this inequality, we start by considering the expression x2 x^2 . We know that for any real number x x , x20 x^2 \geq 0 . This means x2 x^2 is always non-negative.

2. Since x20 x^2 \geq 0 for every real number, adding 9 to x2 x^2 will necessarily make the expression greater than zero, because a non-negative number plus a positive number gives a positive result: x2+99>0 x^2 + 9 \geq 9 > 0 .

3. Therefore, the inequality x2+9>0 x^2 + 9 > 0 holds true for all real numbers x x . There is no value of x x that makes the left side equal to or less than zero.

4. Thus, the solution to the inequality is that it holds for all values of x x .

Consequently, the correct choice from the options provided is:

  • All values of x x

Therefore, the solution is that the inequality x2+9>0 x^2 + 9 > 0 is true for all values of x x .

Answer

All values of x x

Exercise #3

Solve the following equation:

-x^2+2x>0

Video Solution

Step-by-Step Solution

To solve the inequality x2+2x>0-x^2 + 2x > 0, we begin by considering the corresponding equation x2+2x=0-x^2 + 2x = 0.

First, factor the quadratic equation:

  • Rearrange the terms: x2+2x=0-x^2 + 2x = 0 becomes x(2x)=0x(2 - x) = 0.
  • This gives us the roots x=0x = 0 and x=2x = 2.

These roots divide the number line into three intervals: x<0x < 0, 0<x<20 < x < 2, and x>2x > 2.

We need to test these intervals to determine where the inequality holds:

  • For x<0x < 0, choose a test point like x=1x = -1: the expression (1)2+2(1)=12=3-(-1)^2 + 2(-1) = -1 - 2 = -3, which is not greater than zero.
  • For 0<x<20 < x < 2, choose a test point like x=1x = 1: the expression (1)2+2(1)=1+2=1-(1)^2 + 2(1) = -1 + 2 = 1, which is greater than zero.
  • For x>2x > 2, choose a test point like x=3x = 3: the expression (3)2+2(3)=9+6=3-(3)^2 + 2(3) = -9 + 6 = -3, which is not greater than zero.

Thus, the inequality x2+2x>0-x^2 + 2x > 0 is satisfied for the interval 0<x<20 < x < 2.

Therefore, the solution to the inequality is 0<x<2\mathbf{0 < x < 2}, which corresponds to choice 2 in the given options.

Answer

0 < x < 2

Exercise #4

Solve the following equation:

-x^2-9>0

Video Solution

Step-by-Step Solution

To solve this quadratic inequality, x29>0 -x^2 - 9 > 0 , we will follow these steps:

  • Step 1: Identify the quadratic expression x29 -x^2 - 9 .
  • Step 2: Attempt transformation and determine when the expression x29 -x^2 - 9 , can be greater than zero.

Let's analyze the equation:

Rewrite the inequality:
x29>0-x^2 - 9 > 0

Add 9 to both sides:
x2>9-x^2 > 9

Multiply the entire inequality by 1-1 and remember to reverse the inequality sign:
x2<9x^2 < -9

Observe the inequality x2<9x^2 < -9:
Note that x2x^2, being a square of any real number, is always greater than or equal to zero.

As x2x^2 cannot be less than negative nine for any real number xx, the inequality has no solution in the realm of real numbers.

Therefore, the correct answer is:

There is no solution.

Answer

There is no solution.

Exercise #5

Solve the following equation:

x^2-8x+12>0

Video Solution

Step-by-Step Solution

Let's proceed to solve the inequality x28x+12>0 x^2 - 8x + 12 > 0 .

  • Start by factoring the quadratic: x28x+12 x^2 - 8x + 12 .
  • Identify the factors of 12 that sum to 8: 6 6 and 2 2 . This results in: (x6)(x2)=0 (x - 6)(x - 2) = 0 .

The factorization gives us the critical points x=6 x = 6 and x=2 x = 2 . These points divide the number line into three intervals: x<2 x < 2 , 2<x<6 2 < x < 6 , and x>6 x > 6 .

Now, we evaluate the sign of the product (x6)(x2) (x - 6)(x - 2) in each interval:

  • For x<2 x < 2 : Both (x6) (x - 6) and (x2) (x - 2) are negative, so their product is positive.
  • For 2<x<6 2 < x < 6 : (x2) (x - 2) is positive, (x6) (x - 6) is negative, so their product is negative.
  • For x>6 x > 6 : Both (x6) (x - 6) and (x2) (x - 2) are positive, so their product is positive.

The inequality (x6)(x2)>0 (x - 6)(x - 2) > 0 holds for x<2 x < 2 and x>6 x > 6 .

Thus, the solution to the inequality x28x+12>0 x^2 - 8x + 12 > 0 is x<2 x < 2 or x>6 x > 6 .

Therefore, the correct answer is x<2,6<x \boxed{x < 2, 6 < x} .

Answer

x < 2,6 < x

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