Learning Objectives

By the end of this section, you will be able to:

Before you get started, take this readiness quiz.

1. Solve: (2x−3=0).
If you missed this problem, review Example 2.2.
2. Solve: (2y^{2}+y=15).
If you missed this problem, review Example 6.45.
3. Solve (frac{1}{x^{2}+2 x-8}>0)
If you missed this problem, review Example 7.56.

We have learned how to solve linear inequalities and rational inequalities previously. Some of the techniques we used to solve them were the same and some were different.

We will now learn to solve inequalities that have a quadratic expression. We will use some of the techniques from solving linear and rational inequalities as well as quadratic equations.

We will solve quadratic inequalities two ways—both graphically and algebraically.

A quadratic equation is in standard form when written as (ax^{2}+bx+c=0). If we replace the equal sign with an inequality sign, we have a quadratic inequality in standard form.

Definition (PageIndex{1})

The standard form of a quadratic inequality is written:

(egin{array}{ll}{a x^{2}+b x+c<0} & {a x^{2}+b x+c leq 0} {a x^{2}+b x+c>0} & {a x^{2}+b x+c geq 0}end{array})

The graph of a quadratic function (f(x)=a x^{2}+b x+c=0) is a parabola. When we ask when is (a x^{2}+b x+c<0), we are asking when is (f(x)<0). We want to know when the parabola is below the (x)-axis.

When we ask when is (a x^{2}+b x+c>0), we are asking when is (f(x)>0). We want to know when the parabola is above the (y)-axis.

Example (PageIndex{1}) How to Solve a Quadratic Inequality Graphically

Solve (x^{2}−6x+8<0) graphically. Write the solution in interval notation.

Solution:

Step 1: Write the quadratic inequality in standard form.

The inequality is in standard form.

Step 2: Graph the function (f(x)=a x^{2}+b x+c) using properties or transformations.

We will graph using the properties.

Look at (a) in the equation.

(color{red}{a=1, b=-6, c=8})

(f(x)=x^{2}-6 x+8)

Since (a) is positive, the parabola opens upward.

The parabola opens upward.

(f(x)=x^{2}-6 x+8)

The axis of symmetry is the line (x=-frac{b}{2 a}).

Axis of Symmetry

(x=-frac{b}{2 a})

(egin{array}{l}{x=-frac{(-6)}{2 cdot 1}} {x=3}end{array})

The axis of symmetry is the line (x=3).

The vertex is on the axis of symmetry. Substitute (x=3) into the function.

Vertex

The vertex is ((3,-1)).

We find (f(0))

(y)-intercept

The (y)-intercept is ((0.8)).

We use the axis of symmetry to find a point symmetric to the (y)-intercept. The (y)-intercept is (3) units left of the axis of symmetry, (x=3). A point (3) units to the right of the axis of symmetry has (x=6).

Point symmetric to (y)-intercept

The point is ((6,8)).

We solve (f(x)=0).

(x)-intercepts

We can solve this quadratic equation by factoring.

The (x)-intercepts are ((2,0)) and ((4,0)).

We graph the vertex, intercepts, and the point symmetric to the (y)-intercept. We connect these (5) points to sketch the parabola.

Step 3: Determine the solution from the graph.

(x^{2}-6 x+8<0)

The inequality asks for the values of (x) which make the function less than (0). Which values of (x) make the parabola below the (x)-axis.

We do not include the values (2), (4) as the inequality is less than only.

The solution, in interval notation, is ((2,4)).

Exercise (PageIndex{1})

1. Solve (x^{2}+2 x-8<0) graphically
2. Write the solution in interval notation

1. Figure 9.8.4
2. ((-4,-2))

Exercise (PageIndex{2})

1. Solve (x^{2}-8 x+12 geq 0) graphically
2. Write the solution in interval notation

1. Figure 9.8.5
2. ((-infty, 2] cup[6, infty))

We list the steps to take to solve a quadratic inequality graphically.

### Solve a Quadratic Inequality Graphically

1. Write the quadratic inequality in standard form.
2. Graph the function (f(x)=ax^{2}+bx+c).
3. Determine the solution from the graph.

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the (x)-axis but note how the position of the parabola affects the solution.

Example (PageIndex{2})

Solve (-x^{2}-8 x-12 leq 0) graphically. Write the solution in interval notation.

Solution:

 The quadratic inequality in standard form. Graph the function The parabola opens downward. Find the line of symmetry. (egin{array}{l}{x=-frac{b}{2 a}} {x=-frac{-8}{2(-1)}} {x=-4}end{array}) Find the vertex. Vertex ((-4,4)) Find the (x)-intercepts. Let (f(x)=0). Factor: Use the Zero Product Property. Graph the parabola. (x)-intercepts ((-6,0), (-2.0)) Determine the solution from the graph. We include the (x)-intercepts as the inequality is "less than or equal to." ((-infty,-6] cup[-2, infty))

Exercise (PageIndex{3})

1. Solve (-x^{2}-6 x-5>0) graphically
2. Write the solution in interval notation

1. Figure 9.8.8
2. ((-1,5))

Exercise (PageIndex{4})

1. Solve (−x^{2}+10x−16≤0) graphically
2. Write the solution in interval notation

1. Figure 9.8.9
2. ((-infty, 2] cup[8, infty))

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the critical points for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the critical points to divide the number line into intervals and then determine whether the quadratic expression will be positive or negative in the interval. We then determine the solution for the inequality.

Example (PageIndex{3}) How to Solve Quadratic Inequalities Algebraically

Solve (x^{2}-x-12 geq 0) algebraically. Write the solution in interval notation.

Solution:

 Step 1: Write the quadratic inequality in standard form. The inequality is in standard form. Step 2: Determine the critical points--the solutions to the related quadratic equation. Change the inequality sign to an equal sign and then solve the equation. Step 3: Use the critical points to divide the number line into intervals. Use (-3) and (4) to divide the number line into intervals. Step 4: Above the number line show the sign of each quadratic expression using test points from each interval substituted from the original inequality. Test:(x=-5)(x=0)(x=5) (egin{array}{ccc}{x^{2}-x-12} & {x^{2}-x-12} & {x^{2}-x-12} {(-5)^{2}-(-5)-12} & {0^{2}-0-12} & {5^{2}-5-12} {18} & {-12} & {8}end{array}) Step 5: Determine the intervals where the inequality is correct. Write the solution in interval notation. The inequality is positive in the first and last intervals and equals (0) at the points (-4,3). The solution, in interval notation, is ((-infty,-3] cup[4, infty)).

Exercise (PageIndex{5})

Solve (x^{2}+2x−8≥0) algebraically. Write the solution in interval notation.

((-infty,-4] cup[2, infty))

Exercise (PageIndex{6})

Solve (x^{2}−2x−15≤0) algebraically. Write the solution in interval notation.

([-3,5])

In this example, since the expression (x^{2}−x−12) factors nicely, we can also find the sign in each interval much like we did when we solved rational inequalities. We find the sign of each of the factors, and then the sign of the product. Our number line would like this:

The result is the same as we found using the other method.

We summarize the steps here.

### Solve a Quadratic Inequality Algebraically

1. Write the quadratic inequality in standard form.
2. Determine the critical points—the solutions to the related quadratic equation.
3. Use the critical points to divide the number line into intervals.
4. Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Example (PageIndex{4})

Solve (x^{2}+6x−7≥0) algebraically. Write the solution in interval notation.

Solution:

 Write the quadratic inequality in standard form. Multiply both sides of the inequality by (-1). Remember to reverse the inequality sign. Determine the critical points by solving the related quadratic equation. Write the Quadratic Formula. Then substitute in the values of (a, b, c). Simplify. (x=frac{6 pm sqrt{8}}{2}) Simplify the radical. (x=frac{6 pm 2 sqrt{2}}{2}) Remove the common factor, (2). (egin{array}{l}{x=frac{2(3 pm sqrt{2})}{2}} {x=3 pm sqrt{2}} {x=3+sqrt{2}} quad x=3-sqrt{2} {x approx 1.6}quadquad::: xapprox 4.4end{array}) Use the critical points to divide the number line into intervals. Test numbers from each interval in the original inequality. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Exercise (PageIndex{7})

Solve (−x^{2}+2x+1≥0) algebraically. Write the solution in interval notation.

([-1-sqrt{2},-1+sqrt{2}])

Exercise (PageIndex{8})

Solve (−x^{2}+8x−14<0) algebraically. Write the solution in interval notation.

((-infty, 4-sqrt{2}) cup(4+sqrt{2}, infty))

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation (ax^{2}+bx+c=0). These two solutions then gave us either the two (x)-intercepts for the graph or the two critical points to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

For a quadratic equation of the form (ax^{2}+bc+c=0, a≠0).

The last row of the table shows us when the parabolas never intersect the (x)-axis. Using the Quadratic Formula to solve the quadratic equation, the radicand is a negative. We get two complex solutions.

In the next example, the quadratic inequality solutions will result from the solution of the quadratic equation being complex.

Example (PageIndex{5})

Solve, writing any solution in interval notation:

1. (x^{2}-3 x+4>0)
2. (x^{2}-3 x+4 leq 0)

Solution:

a.

 Write the quadratic inequality in standard form. 0) Determine the critical points by solving the related quadratic equation. Write the Quadratic Formula. Then substitute in the values of (a, b, c). Simplify. (x=frac{3 pm sqrt{-7}}{2}) Simplify the radicand. (x=frac{3 pm sqrt{7 i}}{2}) The complex solutions tell us theparabola does not intercept the (x)-axis.Also, the parabola opens upward. Thistells us that the parabola is completely above the (x)-axis. Complex solutions

We are to find the solution to (x^{2}−3x+4>0). Since for all values of (x) the graph is above the (x)-axis, all values of (x) make the inequality true. In interval notation we write ((−∞,∞)).

b. Write the quadratic inequality in standard form.

Determine the critical points by solving the related quadratic equation.

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the (x)-axis—no part of it is below the (x)-axis.

We are to find the solution to (x^{2}−3x+4≤0). Since for all values of (x) the graph is never below the (x)-axis, no values of (x) make the inequality true. There is no solution to the inequality.

Exercise (PageIndex{9})

Solve and write any solution in interval notation:

1. (-x^{2}+2 x-4 leq 0)
2. (-x^{2}+2 x-4 geq 0)
1. ((-infty, infty))
2. no solution

Exercise (PageIndex{10})

Solve and write any solution in interval notation:

1. (x^{2}+3 x+3<0)
2. (x^{2}+3 x+3>0)
1. no solution
2. ((-infty, infty))

## Key Concepts

• Solve a Quadratic Inequality Graphically
1. Write the quadratic inequality in standard form.
2. Graph the function (f(x)=ax^{2}+bx+c) using properties or transformations.
3. Determine the solution from the graph.
• How to Solve a Quadratic Inequality Algebraically
1. Write the quadratic inequality in standard form.
2. Determine the critical points -- the solutions to the related quadratic equation.
3. Use the critical points to divide the number line into intervals.
4. Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

## Solving Quadratic Inequalities – Explanation & Examples

Like equations have different forms, inequalities also exist in different forms, and quadratic inequality is one of them.

The solutions to quadratic inequality always give the two roots. The nature of the roots may differ and can be determined by discriminant (b 2 – 4ac).

The general forms of the quadratic inequalities are:

x 2 – 6x – 16 ≤ 0, 2x 2 – 11x + 12 > 0, x 2 + 4 > 0, x 2 – 3x + 2 ≤ 0 etc.

Next we outline a technique used to solve quadratic inequalities without graphing the parabola. To do this we make use of a sign chart A model of a function using a number line and signs (+ or −) to indicate regions in the domain where the function is positive or negative. which models a function using a number line that represents the x-axis and signs (+ or −) to indicate where the function is positive or negative. For example,

The plus signs indicate that the function is positive on the region. The negative signs indicate that the function is negative on the region. The boundaries are the critical numbers, −2 and 3 in this case. Sign charts are useful when a detailed picture of the graph is not needed and are used extensively in higher level mathematics. The steps for solving a quadratic inequality with one variable are outlined in the following example.

### Example 3

It is important to note that this quadratic inequality is in standard form, with zero on one side of the inequality.

Step 1: Determine the critical numbers. For a quadratic inequality in standard form, the critical numbers are the roots. Therefore, set the function equal to zero and solve.

− x 2 + 6 x + 7 = 0 − ( x 2 − 6 x − 7 ) = 0 − ( x + 1 ) ( x − 7 ) = 0 x + 1 = 0 or x − 7 = 0 x = − 1 x = 7

The critical numbers are −1 and 7.

Step 2: Create a sign chart. Since the critical numbers bound the regions where the function is positive or negative, we need only test a single value in each region. In this case the critical numbers partition the number line into three regions and we choose test values x = − 3 , x = 0 , and x = 10 .

Test values may vary. In fact, we need only determine the sign (+ or −) of the result when evaluating f ( x ) = − x 2 + 6 x + 7 = − ( x + 1 ) ( x − 7 ) . Here we evaluate using the factored form.

f ( − 3 ) = − ( − 3 + 1 ) ( − 3 − 7 ) = − ( − 2 ) ( − 10 ) = − N e g a t i v e f ( 0 ) = − ( 0 + 1 ) ( 0 − 7 ) = − ( 1 ) ( − 7 ) = + P o s i t i v e f ( 10 ) = − ( 10 + 1 ) ( 10 − 7 ) = − ( 11 ) ( 3 ) = − N e g a t i v e

Since the result of evaluating for −3 was negative, we place negative signs above the first region. The result of evaluating for 0 was positive, so we place positive signs above the middle region. Finally, the result of evaluating for 10 was negative, so we place negative signs above the last region, and the sign chart is complete.

Step 3: Use the sign chart to answer the question. In this case, we are asked to determine where f ( x ) ≥ 0 , or where the function is positive or zero. From the sign chart we see this occurs when x-values are inclusively between −1 and 7.

Using interval notation, the shaded region is expressed as [ − 1 , 7 ] . The graph is not required however, for the sake of completeness it is provided below.

Indeed the function is greater than or equal to zero, above or on the x-axis, for x-values in the specified interval.

### Example 4

Begin by finding the critical numbers, in this case, the roots of f ( x ) = 2 x 2 − 7 x + 3 .

2 x 2 − 7 x + 3 = 0 ( 2 x − 1 ) ( x − 3 ) = 0 2 x − 1 = 0 or x − 3 = 0 2 x = 1 x = 3 x = 1 2

The critical numbers are 1 2 and 3. Because of the strict inequality > we will use open dots.

Next choose a test value in each region and determine the sign after evaluating f ( x ) = 2 x 2 − 7 x + 3 = ( 2 x − 1 ) ( x − 3 ) . Here we choose test values −1, 2, and 5.

f ( − 1 ) = [ 2 ( − 1 ) − 1 ] ( − 1 − 3 ) = ( − ) ( − ) = + f ( 2 ) = [ 2 ( 2 ) − 1 ] ( 2 − 3 ) = ( + ) ( − ) = − f ( 5 ) = [ 2 ( 5 ) − 1 ] ( 5 − 3 ) = ( + ) ( + ) = +

And we can complete the sign chart.

The question asks us to find the x-values that produce positive results (greater than zero). Therefore, shade in the regions with a + over them. This is the solution set.

Sometimes the quadratic function does not factor. In this case we can make use of the quadratic formula.

### Example 5

Find the critical numbers.

Identify a, b, and c for use in the quadratic formula. Here a = 1 , b = − 2 , and c = − 11 . Substitute the appropriate values into the quadratic formula and then simplify.

x = − b ± b 2 − 4 a c 2 a = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( − 11 ) 2 ( 1 ) = 2 ± 48 2 = 2 ± 4 3 2 = 1 ± 2 3

Therefore the critical numbers are 1 − 2 3 ≈ − 2.5 and 1 + 2 3 ≈ 4.5 . Use a closed dot on the number to indicate that these values will be included in the solution set.

Here we will use test values −5, 0, and 7.

f ( − 5 ) = ( − 5 ) 2 − 2 ( − 5 ) − 11 = 25 + 10 − 11 = + f ( 0 ) = ( 0 ) 2 − 2 ( 0 ) − 11 = 0 + 0 − 11 = − f ( 7 ) = ( 7 ) 2 − 2 ( 7 ) − 11 = 49 − 14 − 11 = +

After completing the sign chart shade in the values where the function is negative as indicated by the question ( f ( x ) ≤ 0 ).

Try this! Solve: 9 − x 2 > 0 .

It may be the case that there are no critical numbers.

### Example 6

To find the critical numbers solve,

Substitute a = 1 , b = − 2 , and c = 3 into the quadratic formula and then simplify.

x = − b ± b 2 − 4 a c 2 a = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( 3 ) 2 ( 1 ) = 2 ± − 8 2 = 2 ± 2 i 2 2 = 1 + i 2

Because the solutions are not real, we conclude there are no real roots hence there are no critical numbers. When this is the case, the graph has no x-intercepts and is completely above or below the x-axis. We can test any value to create a sign chart. Here we choose x = 0 .

Because the test value produced a positive result the sign chart looks as follows:

We are looking for the values where f ( x ) > 0 the sign chart implies that any real number for x will satisfy this condition.

The function in the previous example is graphed below.

We can see that it has no x-intercepts and is always above the x-axis (positive). If the question was to solve x 2 − 2 x + 3 < 0 , then the answer would have been no solution. The function is never negative.

Try this! Solve: 9 x 2 − 12 x + 4 ≤ 0 .

### Example 7

Find the domain: f ( x ) = x 2 − 4 .

Recall that the argument of a square root function must be nonnegative. Therefore, the domain consists of all real numbers for x such that x 2 − 4 is greater than or equal to zero.

It should be clear that x 2 − 4 = 0 has two solutions x = ± 2 these are the critical values. Choose test values in each interval and evaluate f ( x ) = x 2 − 4 .

f ( − 3 ) = ( − 3 ) 2 − 4 = 9 − 4 = + f ( 0 ) = ( 0 ) 2 − 4 = 0 − 4 = − f ( 3 ) = ( 3 ) 2 − 4 = 9 − 4 = +

Shade in the x-values that produce positive results.

Answer: Domain: ( − ∞ , − 2 ] ∪ [ 2 , ∞ )

### Key Takeaways

• Quadratic inequalities can have infinitely many solutions, one solution or no solution.
• We can solve quadratic inequalities graphically by first rewriting the inequality in standard form, with zero on one side. Graph the quadratic function and determine where it is above or below the x-axis. If the inequality involves “less than,” then determine the x-values where the function is below the x-axis. If the inequality involves “greater than,” then determine the x-values where the function is above the x-axis.
• We can streamline the process of solving quadratic inequalities by making use of a sign chart. A sign chart gives us a visual reference that indicates where the function is above the x-axis using positive signs or below the x-axis using negative signs. Shade in the appropriate x-values depending on the original inequality.
• To make a sign chart, use the function and test values in each region bounded by the roots. We are only concerned if the function is positive or negative and thus a complete calculation is not necessary.

### Part A: Solutions to Quadratic Inequalities

Determine whether or not the given value is a solution.

4 x 2 − 12 x + 9 ≤ 0 x = 3 2

5 x 2 − 8 x − 4 < 0 x = − 2 5

Given the graph of f determine the solution set.

### A stuntman will jump off a 20 m building.

A high-speed camera is ready to film him between 15 m and 10 m above the ground.

When should the camera film him?

We can use this formula for distance and time:

(Note: if you are curious about the formula, it is simplified from d = d0 + v0t + ½a0t 2 , where d0=20, v0=0, and a0=&minus9.81, the acceleration due to gravity.)

### First, let us sketch the question:

The distance we want is from 10 m to 15 m:

And we know the formula for d :

### Now let's solve it!

First, let's subtract 20 from both sides:

Now multiply both sides by &minus(1/5). But because we are multiplying by a negative number, the inequalities will change direction . read Solving Inequalities to see why.

To be neat, the smaller number should be on the left, and the larger on the right. So let's swap them over (and make sure the inequalities still point correctly):

Lastly, we can safely take square roots, since all values are greater then zero:

We can tell the film crew:

"Film from 1.0 to 1.4 seconds after jumping"

## Worked example 19: Solving quadratic inequalities with fractions

### Solving the equation

To solve this equation we multiply both sides of the equation by ((x+3)(x-3)) and simplfy: egin frac<2> imes (x+3)(x-3) &= frac<1> imes (x+3)(x-3) 2(x-3) &= x+3 2x-6 &= x+3 x &= 9 end

### Solving the inequality

It is very important to recognise that we cannot use the same method as above to solve the inequality. If we multiply or divide an inequality by a negative number, then the inequality sign changes direction. We must rather simplify the inequality to have a lowest common denominator and use a table of signs to determine the values that satisfy the inequality.

### Determine the lowest common denominator and simplify the fraction

Keep the denominator because it affects the final answer.

### Determine the critical values of (x)

From the factorised inequality we see that the critical values are (x=-3), (x=3) and (x=9).

### Complete a table of signs

From the table we see that the function is less than or equal to zero for (x < -3) or (3 < x leq 9). We do not include (x = -3) or (x = 3) in the solution because of the restrictions on the denominator.

### Write the final answer and represent on a number line

Supercharge your high school students' solving skills with our printable quadratic inequalities worksheets. Represent the inequality as an equation, moving the terms to one side and equating it to zero, factor the equation and find the zeros to obtain break points or critical points, graph them on a number line, and determine the interval. Solving quadratic inequalities algebraically, graphically, and completing the table of signs, graphing the parabola and shading the solution region based on the inequality are the exercises presented in these pdfs. You now have the chance to test your skills with our free quadratic inequalities worksheets.

With either -1 or 1 as leading coefficients and integer break points, these printable quadratic inequalities worksheets get high school students transforming quadratic inequalities into quadratic equations and factoring them to solve the inequality.

Triumph in solving quadratic inequalities whose leading coefficients are integers by finding the factors, equating them to zero, determining their roots, and finding the solution range in this compilation of quadratic inequalities worksheet pdfs.

Instruct high school students to express the quadratic inequalities in standard form by moving the expressions on one side and equating to 0, factor them, determine their roots, to find the critical point and figure out the solution interval.

Factor the quadratic inequalities and list out the intervals in the table of signs. Take a test value from each interval and apply on both the factors. Complete the table with the resultant plus or minus sign. Select the solution range that satisfies the inequality.

Check if the quadratic inequality is inclusive or strict. Graph the parabola y = f(x) for the quadratic inequality f(x) &le 0 or f(x) &ge 0. Find the vertex and identify the values of x for which the part of the parabola will either be negative or positive depending on the inequalities.

High school students plot x-intercepts, figure out the axis of symmetry and the vertex of the parabola, determine the direction, and illustrate the inequality using dotted or solid lines. Shade the parabola below or above the x-axis, inside or outside the parabola based on the solution.

STEPS IN SOLVING QUADRATIC INEQUALITIES. There are 4 steps.

STEP 1. Transform the given inequality into standard form: f(x) = ax^2 + bx + c 0).
Example: Solve (2x - 3)(x + 4) > -5.
In Step 1, transform it into standard form: f(x) = 2x^2 + 5x - 7 5.
In step 1, transform it into standard form f(x) = 3x^2 - 7x - 5 > 0.

STEP 2. Solve the quadratic equation f(x) = 0 to get the 2 real roots x1 and x2. You can use any of
the 4 methods (factoring, completing the square, quadratic formula, and graphing) or the new Diagonal Sum Method (Amazon e-book 2010).
Before proceeding solving, make sure that the quadratic equation has 2 real roots. How? Find out if the discriminant D = b^2 - 4ac is positive (> 0). There is no need to calculate the exact value of D. Just use mental math to make sure that D > 0. If D 0), based on the 2 real roots obtained from Step 2. You can use one of these 3 methods to solve quadratic inequalities:
1. Using the number line and test points method
2. Using the algebraic method.
3. Using the graphing method.

STEP 4. Express the answers, or solution set, in terms of intervals. You must master on how to write interval symbols. For examples:
(a, b): open interval between a and b the 2 end points are not included in the solution set.
[a, b]: closed interval the end points a and b are included in the solution set.
(-infinity, b]: half closed interval: the end point b is included in the solution set.

1. THE NUMBER LINE AND TEST POINT METHOD.
The 2 real roots x1 and x2, obtained from Step 2, are plotted on the number line. They divide the number line into one segment and 2 rays. Always use the origin O as test point. Substitute x = 0 into the quadratic inequality in standard form. If it is true, then the origin is located on the true segment (or the true ray). If one ray is a part of the solutions set, then the other ray is also a part of the solution set, due to the symmetrical property of the parabola graph.
Example 1. Solve: 5x^2 - 34x 0 if a is negative (-) and f(x) 0), opposite to the sign of a = -6, between the 2 real roots 1/2 and 1.
Example 4: The trinomial f(x) = 3x^2 - 4x - 7 is negative ( 0) in this interval, opposite in sign to the constant a.

3. THE GRAPHING METHOD.
You can solve a quadratic inequality f(x) 0) by graphing the quadratic function f(x). When the parabola graph of f(x) is above the x-axis, f(x) is positive. When the parabola is above the x-axis, f(x) is negative.
You don't have to accurately graph the parabola. Bases on the 2 real roots obtained from solving the equation f(x) = 0, you may just roughly sketch the parabola. Pay attention to if the parabola is upward (a positive) or downward (a negative).
By this method, you may solve a system of 2 (or 3) quadratic inequalities by graphing the two (or 3) parabolas on the same coordinate grid, if possible.

REMARK 1. When the inequality has an additional equal sign (greater, lesser, or equal to), the end points are automatically included into the solution set.

REMARK 2. If the Discriminant D = b^2 - 4ac is negative (D 0.
Solution. The Discriminant D = 64 - 460 0), regardless of the values of x. The inequality is always true.
Example 6. Solve: f(x) = -4x^2 + 9x - 7 > 0.
Solution. D = 81 - 112 0.
Solution. Use the Diagonal Sum Method to solve f(x) = 0. Roots have opposite signs. There are 2 probable root-pairs: (-1/5, 9/1),(-3/1, 3/5). The diagonal sum of the second pair is: -15 + 3 = -12 = -b. The 2 real roots are -3 and 3/5. Use the origin O as test point to solve f(x) > 0. Substitute x = o into the inequality. It shows: -9 > 0. It is not true, then the origin O is not located on the solution set. The solution set should be the 2 rays as expressed by the 2 open intervals (-infinity, -3) and (3/5, +infinity).
Example 8. Solve: f(x) = 8x^2 - 42x - 11

## Objectives

After studying this lesson, you will be able to:

• Solve a quadratic equation with real coefficients by factorization and by using quadratic formula
• Find relationship between roots and coefficients
• Form a quadratic equation when roots are given
• Differentiate between a linear equation and a linear inequality
• State that a planl region represents the solution of a linear inequality
• Represent graphically a linear inequality in two variables
• Show the solution of an inequality by shading the appropriate region
• Solve graphically a system of two or three linear inequalities in two variables

## 4. Plot the parabola corresponding to the quadratic function.

4. Plot the parabola corresponding to the quadratic function.

You do not have to make an exact plot as I did here. A sketch will be sufficient to determine the solution. What is important is that you can easily determine for which values of x the graph is below zero, and for which it is above. Since this is an upward opening parabola we know that the graph is below zero in between the two roots we just found and it is above zero when x is smaller than the smallest root we found, or when x is larger than the largest root we found.

When you have done this a couple of times you will see that you do not need this sketch anymore. However, it is a good way to get a clear view on what you are doing and therefore it is recommended to make this sketch.

5. Determine the solution of the inequality.

Now we can determine the solution by looking at the graph we just plotted. Our inequality was x^2 +4x -5 > 0.

We know that in x = -5 and x = 1 the expression is equal to zero. We must have that the expression is larger than zero and therefore we need the regions left from the smallest root and right of the largest root. Our solution will then be:

Make sure to write "or" and not "and" because then you would suggest that the solution would have to be an x that is both smaller than -5 and larger than 1 at the same time, which is of course impossible.

If instead we would have to solve x^2 +4x -5 < 0 we would have done the exact same until this step. Then our conclusion would be that x has to be in the region between the roots. This means:

Here we have only one statement because we only have one region of the plot we want to describe.

Remember that a quadratic function does not always have two roots. It might happen that it has only one, or even zero roots. In that case we are still able to solve the inequality.

### What If the Parabola Has No Roots?

In the case that the parabola does not have any roots there are two possibilities. Either it is an upwards opening parabola that lies entirely above the x-axis. Or it is a downwards opening parabola that lies entirely under the x-axis. Therefore the answer to the inequality will either be that it is satisfied for all possible x, or that there is no x such that the inequality is satisfied. In the first case every x is a solution, and in the second case there is no solution.

If the parabola has only one root we are basically in the same situation with the exception that there is exactly one x for which equality holds. So if we have an upwards opening parabola and it has to be larger than zero still every x is a solution except for the root, since there we have equality. This means that if we have a strict inequality the solution is all x, except for the root. If we do not have a strict inequality the solution is all x.

If the parabola has to be smaller than zero and we have strict inequality there is no solution, but if the inequality is not strict there is exactly one solution, which is the root itself. This is because there is equality in this point, and everywhere else the constraint is violated.

Analogously, for a downward opening parabola we have that still all x are a solution for a non-strict inequality, and all x except for the root when the inequality is strict. Now when we have a larger than constraint, there is still no solution, but when we have a larger than or equal to statement, the root is the only valid solution.

These situations might seem difficult, but this is where plotting the parabola can really help you to understand what to do.

In the picture, you see an example of an upward opening parabola that has one root in x=0. If we call the function f(x), we can have four inequalities:

Inequality 1 does not have a solution, since in the plot you see that everywhere the function is at least zero.

Inequality 2, however, has as solution x=0, since there the function is equal to zero, and inequality 2 is a non-strict inequality that allows equality.