Math 2 MM2A4d
Solving quadratic inequalities algebraically
You have already learned how to solve quadratic equations in multiple ways. For example, solve the quadratic equation . You can do this with the quadratic formula, or you can use the zero product property. Either way, the two zeros for this equation are
7 and -2.
Sometimes you may have to solve an inequality. For example, rather than the example above, , you may have . This means that rather than being asked for the two values for x that will give you 0, you are asked for all of the values for x that will make the equation greater than 0. You can solve this algebraically or graphically. Let’s look at how to solve quadratic inequalities algebraically.
Example 1:
STEP 1: Solve the inequality as if it were equal to 0.
Factor: (x – 1)(x + 4)
Set both factors equal to 0: x – 1 = 0 x + 4 = 0
+ 1 + 1 – 4 – 4
x = 1 x = – 4
STEP 2: Put both zeros on a number line.
Once both numbers are on a number line, three sections are created. In an inequality you will either shade the section in between the two zeros, or you will shade in the other two sections. You will NEVER have all three sections colored in.
STEP 3: Test a number in each section to determine which sections will be colored in.
* Pick a number in the section most to the left and plug it into the inequality:
f(– 6) =
36 – 18 – 4 > 0
14 > 0
Because 14 is greater than 0, this section will be colored in.
* Pick a number in the section between the two zeros and plug it into the inequality:
f(0) =
0 + 0 – 4 > 0
– 4 > 0
Because – 4 is NOT greater than 0, this section is not colored in.
* Pick a number in the section to the right and plug it into the inequality:
f( 2) =
4 + 6 – 4 > 0
6 > 0
Because 6 is greater than 0, this section should be colored in.
STEP 4: Bring it all together.
In order to answer the inequality , you must give all of the values of x that will satisfy the inequality. That means you must list every number that can be plugged in to x and make the inequality true. You figured this out by making your number line. Because the section to the left and to the right were colored in, those are the x values to will make the inequality true. Therefore, the answer to this inequality is:
x<– 4 or x>1.
Example 2:
STEP 1:
(x + 1)(x + 5)
x + 1 = 0 x + 5 = 0
x = – 1 x = – 5
STEP 2:
STEP 3: f(-6) =
36 – 36 + 5 < 0
5 < 0
Because 5 is not less than zero the section to the left is not colored in.
f(-2) =
4 – 12 < 0
– 8 < 0
Because – 8 is less than 0 the middle section is colored in.
f(0) =
0 + 0 + 5 < 0
5 < 0
Because 5 is not less than 0 the section to the right is not colored in.
STEP 4: The solution to the inequality is -5 < x < 3.
Example 3:
STEP 1:
(x – 5)(x – 2)
x – 5 = 0 x – 2 = 0
x = 5 x = 2
STEP 2:
Notice that in this picture the circles are colored in. That is because the inequality is 0.
STEP 3: f(0) =
0 – 0 + 10 0
10 0
Because 10 is greater than or equal to 0 the left section is colored in.
f(3) =
9 – 21 + 10 0
– 2 0
Because – 2 is not greater than or equal to 0 the middle section is not colored in.
f(6) =
36 – 42 +10 0
4 0
Because 4 is greater than or equal to 0 the right section is colored in.
STEP 4: The solution to the inequality is x 2 or x 5.
Example 4:
When you have a negative a value, you must divide everything by – 1. When you divide by – 1 on both sides of an inequality, you must also switch the inequality sign:
=
– 1
Now solve this new equality like the examples before.
STEP 1:
(x + 3)(x + 5)
x + 3 = 0 x + 5 = 0
– 3 – 3 – 5 – 5
x = – 3 x = – 5
STEP 2:
STEP 3: f(-6)=
36 – 48 + 15 0
3 0
Because 3 is not less than or equal to 0, the section on the left is not colored in.
f(-4)=
16 – 32 + 15 0
–1 0
Because – 1 is less than 0, the section in the middle is colored in.
f(0)=
0 + 0 + 15 0
15 0
Because 15 is not less than or equal to 0, the section on the right is not colored in.
STEP 4: The solution to the inequality is -5 x -3.
Now you try a few.
1) 2)
3) -x2 - 2x + 15 < 0 4) 2x2 - x - 1 £ 0
Math 2: Quadratic Inequalities LS 1 Name: ______
1. 2x2-11x-60 2. x2 + 8x - 90 3. 2x2-3x - 90
4. -6x2+5x+110 5. 8x2+10x-30 6. x2 + 5x + 40
7. 7x2+12x-40 8. -x2+3x+40 9. x2 – 7x - 300
10. 12x2-x-60 11. x2 + x - 420 12. -6x2-x+20