Now that you know how to solve quadratic equations, you are ready to learn how to solve quadratic inequalities.
A quadratic inequality is an inequality of the form
a
x
2
+
b
x
+
c
>
0
a
x
2
+
b
x
+
c
≥
0
a
x
2
+
b
x
+
c
<
0
a
x
2
+
b
x
+
c
≤
0
a
x
2
+
b
x
+
c
>
0
a
x
2
+
b
x
+
c
≥
0
a
x
2
+
b
x
+
c
<
0
a
x
2
+
b
x
+
c
≤
0
(1)
Solving a quadratic inequality corresponds to working out in what region the graph of a quadratic function lies above or below the xx-axis.
Solve the inequality 4x2-4x+1≤04x2-4x+1≤0 and interpret the solution graphically.
- Step 1. Factorise the quadratic :
Let f(x)=4x2-4x+1f(x)=4x2-4x+1. Factorising this quadratic function gives f(x)=(2x-1)2f(x)=(2x-1)2.
- Step 2. Re-write the original equation with factors :
(
2
x
-
1
)
2
≤
0
(
2
x
-
1
)
2
≤
0
(2)
- Step 3. Solve the equation :
f(x)=0f(x)=0 only when x=12x=12.
- Step 4. Write the final answer :
This means that the graph of f(x)=4x2-4x+1f(x)=4x2-4x+1 touches the xx-axis at x=12x=12, but there are no regions where the graph is below the xx-axis.
- Step 5. Graphical interpretation of solution :
Find all the solutions to the inequality x2-5x+6≥0x2-5x+6≥0.
- Step 1. Factorise the quadratic :
The factors of x2-5x+6x2-5x+6 are (x-3)(x-2)(x-3)(x-2).
- Step 2. Write the inequality with the factors :
x
2
-
5
x
+
6
≥
0
(
x
-
3
)
(
x
-
2
)
≥
0
x
2
-
5
x
+
6
≥
0
(
x
-
3
)
(
x
-
2
)
≥
0
(3)
- Step 3. Determine which ranges correspond to the inequality :
We need to figure out which values of xx satisfy the inequality. From the answers we have five regions to consider.
- Step 4. Determine whether the function is negative or positive in each of the regions :
Let f(x)=x2-5x+6f(x)=x2-5x+6. For each region, choose any point in the region and evaluate the function.
Table 1
| |
|
f
(
x
)
f
(
x
)
|
sign of f(x)f(x) |
| Region A |
x
<
2
x
<
2
|
f
(
1
)
=
2
f
(
1
)
=
2
|
+ |
| Region B |
x
=
2
x
=
2
|
f
(
2
)
=
0
f
(
2
)
=
0
|
+ |
| Region C |
2
<
x
<
3
2
<
x
<
3
|
f
(
2
,
5
)
=
-
2
,
5
f
(
2
,
5
)
=
-
2
,
5
|
- |
| Region D |
x
=
3
x
=
3
|
f
(
3
)
=
0
f
(
3
)
=
0
|
+ |
| Region E |
x
>
3
x
>
3
|
f
(
4
)
=
2
f
(
4
)
=
2
|
+ |
We see that the function is positive for x≤2x≤2 and x≥3x≥3.
- Step 5. Write the final answer and represent on a number line :
We see that x2-5x+6≥0x2-5x+6≥0 is true for x≤2x≤2 and x≥3x≥3.
Solve the quadratic inequality -x2-3x+5>0-x2-3x+5>0.
- Step 1. Determine how to approach the problem :
Let f(x)=-x2-3x+5f(x)=-x2-3x+5. f(x)f(x) cannot be factorised so, use the quadratic formula to determine the roots of f(x)f(x). The xx-intercepts are solutions to the quadratic equation
-
x
2
-
3
x
+
5
=
0
x
2
+
3
x
-
5
=
0
∴
x
=
-
3
±
(
3
)
2
-
4
(
1
)
(
-
5
)
2
(
1
)
=
-
3
±
29
2
x
1
=
-
3
-
29
2
x
2
=
-
3
+
29
2
-
x
2
-
3
x
+
5
=
0
x
2
+
3
x
-
5
=
0
∴
x
=
-
3
±
(
3
)
2
-
4
(
1
)
(
-
5
)
2
(
1
)
=
-
3
±
29
2
x
1
=
-
3
-
29
2
x
2
=
-
3
+
29
2
(4)
- Step 2. Determine which ranges correspond to the inequality :
We need to figure out which values of xx satisfy the inequality. From the answers we have five regions to consider.
- Step 3. Determine whether the function is negative or positive in each of the regions :
We can use another method to determine the sign of the function over different regions, by drawing a rough sketch of the graph of the function. We know that the roots of the function correspond to the xx-intercepts of the graph. Let g(x)=-x2-3x+5g(x)=-x2-3x+5. We can see that this is a parabola with a maximum turning point that intersects the xx-axis at x1x1 and x2x2.
It is clear that g(x)>0g(x)>0 for x1<x<x2x1<x<x2
- Step 4. Write the final answer and represent the solution graphically :
-x2-3x+5>0-x2-3x+5>0 for x1<x<x2x1<x<x2
When working with an inequality where the variable is in the denominator, a different approach is needed.
- Step 1. Subtract 1x-31x-3 from both sides :
2
x
+
3
-
1
x
-
3
≤
0
2
x
+
3
-
1
x
-
3
≤
0
(5)
- Step 2. Simplify the fraction by finding LCD :
2
(
x
-
3
)
-
(
x
+
3
)
(
x
+
3
)
(
x
-
3
)
≤
0
x
-
9
(
x
+
3
)
(
x
-
3
)
≤
0
2
(
x
-
3
)
-
(
x
+
3
)
(
x
+
3
)
(
x
-
3
)
≤
0
x
-
9
(
x
+
3
)
(
x
-
3
)
≤
0
(6)
- Step 3. Draw a number line for the inequality :
We see that the expression is negative for x<-3x<-3 or 3<x≤93<x≤9.
- Step 4. Write the final answer :
x
<
-
3
o
r
3
<
x
≤
9
x
<
-
3
o
r
3
<
x
≤
9
(7)
Solve the following inequalities and show your answer on a number line.
- Solve: x2-x<12x2-x<12.
- Solve: 3x2>-x+43x2>-x+4
- Solve: y2<-y-2y2<-y-2
- Solve: -t2+2t>-3-t2+2t>-3
- Solve: s2-4s>-6s2-4s>-6
- Solve: 0≥7x2-x+80≥7x2-x+8
- Solve: 0≥-4x2-x0≥-4x2-x
- Solve: 0≥6x20≥6x2
- Solve: 2x2+x+6≤02x2+x+6≤0
- Solve for xx if: xx-3<2xx-3<2 and x≠3x≠3.
- Solve for xx if: 4x-3≤14x-3≤1.
- Solve for xx if: 4(x-3)2<14(x-3)2<1.
- Solve for xx: 2x-2x-3>32x-2x-3>3
- Solve for xx: -3(x-3)(x+1)<0-3(x-3)(x+1)<0
- Solve: (2x-3)2<4(2x-3)2<4
- Solve: 2x≤15-xx2x≤15-xx
- Solve for xx: x2+33x-2≤0x2+33x-2≤0
- Solve: x-2≥3xx-2≥3x
- Solve for xx: x2+3x-45+x4≤0x2+3x-45+x4≤0
- Determine all real solutions: x-23-x≥1x-23-x≥1
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