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Inequalities
Introduction
β’ You will be familiar with solving Inequalities from GCSE maths and C1
β’ In this chapter you will see how to solve some more complicated Inequalities
β’ You will also see how to avoid making a very common error!
β’ You will see how to use diagrams to help identify the correct regions for a question
Teachings for Exercise 1A
InequalitiesYou can manipulate Inequalities in
order to solve them
Remember that solving an Inequality is very similar to solving an equation:
1A
5 π₯+8=23 5 π₯+8>23
5 π₯=15 5 π₯>15
π₯=3 π₯>3
Subtract 8
Divide by 5
Subtract 8
Divide by 5
So the value of x in this case is 3
So the value of x in this case is anything greater
than 3
The steps are effectively the same. However, there is one special situation when solving Inequalities that you need to
be aware ofβ¦
InequalitiesYou can manipulate Inequalities in
order to solve them
Remember that solving an Inequality is very similar to solving an equation:
1A
6β2π₯=2
β2 π₯=β4
βπ₯=β4
Subtract 6
Divide by 2
If you multiply or divide by a negative in an Inequality, you must reverse the direction of the
signβ¦
(you can check by substituting values back into the first step if you like!)
Multiply by -1
π₯=4
6β2π₯<2
β2 π₯<β4
βπ₯<β4
Subtract 6
Divide by 2
Multiply by -1. This REVERSES the
sign!π₯>4
InequalitiesYou can manipulate Inequalities in
order to solve them
Solve the Inequality below:
1A
2 π₯2<π₯+3
2 π₯2<π₯+3
2 π₯2βπ₯β3<0Subtract x and subtract 3
Factorise(2 π₯β3)(π₯+1)<0
So the βcritical valuesβ are x = 3/2 and x = -1 Now draw a sketch. Use the critical values and the fact this is a positive quadraticβ¦
-1 3/2
Consider the Inequality β we want the range of values where the graph is below 0
So therefore:
β1<π₯<32
x
y
InequalitiesYou can manipulate Inequalities in order
to solve them
Solve the Inequality below:
You MUST be careful in this situation.
The normal process would be to multiply each side by (x β 2)
However, this could be negative, there is no way to know for sure at this stage
What you can do is multiply by (x β 2)2, as this will definitely be positive (as it has
been squared)
Then you rearrange and solve as in the previous exampleβ¦
You will need to use the βclever factorisationβ technique from FP1! 1A
π₯2
π₯β2<π₯+1 π₯β 2
π₯2
π₯β2<π₯+1
π₯2(π₯β2)2
π₯β2<(π₯+1)(π₯β2)2
π₯2(π₯β2)<(π₯+1)(π₯β2)2
π₯2 (π₯β2 )β(π₯+1)(π₯β2)2<0
(π₯β2 )ΒΏ ΒΏπ₯2β(π₯+1)(π₯β2)ΒΏ0
(π₯β2 )π₯2β(π₯2βπ₯β2)ΒΏ ΒΏΒΏ0(π₯β2 ) (π₯+2 )<0
So the critical values of x are 2 and -2 Now sketch a graph to help with solving the inequality
Multiply by (x β 2)2
Cancel an (x β 2) on the left
Rearrange terms to one side
Take out (x β 2) as a factor
Multiply out the inner bracket
Simplify
InequalitiesYou can manipulate Inequalities in order
to solve them
Solve the Inequality below:
You MUST be careful in this situation.
The normal process would be to multiply each side by (x β 2)
However, this could be negative, there is no way to know for sure at this stage
What you can do is multiply by (x β 2)2, as this will definitely be positive (as it has
been squared)
Then you rearrange and solve as in the previous exampleβ¦
You will need to use the βclever factorisationβ technique from FP1! 1A
π₯2
π₯β2<π₯+1 π₯β 2
π₯2
π₯β2<π₯+1
(π₯β2 ) (π₯+2 )<0We have shown that this Inequality is
equivalent
-2 2
Plot a graph
The shape is a positive quadratic
The x-intercepts are 2 and -2
We want the region below the x-axis (< 0)
Write this as an Inequality
β2<π₯<2
InequalitiesYou can manipulate Inequalities in
order to solve them
Solve the Inequality below:
Sometimes you need to multiply by two different denominators in order to cancel
them both!
As before, they both need to be squared to ensure they arenβt negativeβ¦
1A
π₯π₯+1
β€2
π₯+3 π₯β β1 ,π₯ β β3
π₯π₯+1
β€2
π₯+3
π₯ (π₯+1)2(π₯+3)2
π₯+1β€2(π₯+1)2(π₯+3)2
π₯+3
π₯ (π₯+1) (π₯+3 )2β€2 (π₯+1 )2(π₯+3)
π₯ (π₯+1 ) (π₯+3 )2β2 (π₯+1 )2(π₯+3)β€0
(π₯+1)(π₯+3)π₯ (π₯+3 )β2(π₯+1)ΒΏ ΒΏβ€0(π₯+1)(π₯+3)π₯2+3π₯β2 π₯β2ΒΏ ΒΏβ€0(π₯+1)(π₯+3)π₯2+π₯β2ΒΏ ΒΏβ€0(π₯+1)(π₯+3)(π₯+2)(π₯β1)β€0
So the critical values of x are -1, -3, -2 and 1 Now sketch a graph to help with solving the inequality!
Multiply by (x+1)2(x+3)2
Cancel terms where
appropriate
Rearrange and set equal to 0
βClever Factorisationβ
Multiply out terms
Simplify
Factorise the expression in the squared bracket
InequalitiesYou can manipulate Inequalities in
order to solve them
Solve the Inequality below:
Sometimes you need to multiply by two different denominators in order to cancel
them both!
As before, they both need to be squared to ensure they arenβt negativeβ¦
1A
π₯π₯+1
β€2
π₯+3 π₯β β1 ,π₯ β β3
π₯π₯+1
β€2
π₯+3
(π₯+1)(π₯+3)(π₯+2)(π₯β1)β€0We have shown that this Inequality is
equivalent
Plot a graph
The shape is a positive quartic (same basic shape as a
quadratic β βUβ, just with more changes of direction!)
The x-intercepts are -1, -3, -2, 1
We want the region below the x-axis (< 0)
Write this using Inequalities
-3
-2 -1
1
β3<π₯<β2
β1<π₯<β1ππ
Teachings for Exercise 1B
InequalitiesYou can use graphs to help solve
Inequalities
If you are solving an Inequality you can also find answers by drawing
graphs of each side and looking for the region(s) where one graph is
above/beneath the otherβ¦
a) On the same axes sketch the graphs of the curves with
equations:
b) Find the points of intersection of the two graphs
c) Solve the following equation:
1B
π¦=7 π₯3 π₯+1 π¦=4βπ₯
7 π₯3π₯+1
<4βπ₯
π¦=7 π₯3 π₯+1
π¦=4βπ₯The sketch for this graph is simple Downward sloping
graph x and y intercepts at
(4,0) and (0,4)
This one is more difficultSubbing in x = 0 or y = 0 will yield the intercept
(0,0)
π¦=7 π₯3 π₯+1
3 π₯β1=0π₯=β
13
Vertical asymptote at x =
-1/3
An asymptote will be at the value for x that makes the denominator 0 (as this is
not possible)
Solve
4
4
InequalitiesYou can use graphs to help solve
Inequalities
If you are solving an Inequality you can also find answers by drawing
graphs of each side and looking for the region(s) where one graph is
above/beneath the otherβ¦
a) On the same axes sketch the graphs of the curves with
equations:
b) Find the points of intersection of the two graphs
c) Solve the following equation:
1B
π¦=7 π₯3 π₯+1 π¦=4βπ₯
7 π₯3π₯+1
<4βπ₯
π¦=7 π₯3 π₯+1
π¦=4βπ₯The sketch for this graph is simple Downward sloping
graph x and y intercepts at
(4,0) and (0,4)
This one is more difficultSubbing in x = 0 or y = 0 will yield the intercept
(0,0)Vertical asymptote at x = -1/3
π¦=7 π₯3 π₯+1 Rearrange to write in terms of
x Multiply by (3x + 1)
4
4
π¦ (3 π₯+1 )=7 π₯
3 π₯π¦+π¦=7 π₯π¦=7 π₯β3π₯π¦π¦=π₯ (7β3 π¦)
π¦7β3 π¦
=π₯
Multiply out the bracket
Subtract 3xy
Factorise
Divide by (7 β 3y)
InequalitiesYou can use graphs to help solve
Inequalities
If you are solving an Inequality you can also find answers by drawing
graphs of each side and looking for the region(s) where one graph is
above/beneath the otherβ¦
a) On the same axes sketch the graphs of the curves with
equations:
b) Find the points of intersection of the two graphs
c) Solve the following equation:
1B
π¦=7 π₯3 π₯+1 π¦=4βπ₯
7 π₯3π₯+1
<4βπ₯
π¦=7 π₯3 π₯+1
π¦=4βπ₯The sketch for this graph is simple Downward sloping
graph x and y intercepts at
(4,0) and (0,4)
This one is more difficultSubbing in x = 0 or y = 0 will yield the intercept
(0,0)Vertical asymptote at x = -1/3
4
4
π¦7β3 π¦
=π₯
7β3 π¦=0
7=3 π¦73=π¦
Find the value that would make the denominator 0 (which isnβt
possible)
Divide by 3
Add 3y
Horizontal asymptote at y =
7/3
InequalitiesYou can use graphs to help solve
Inequalities
If you are solving an Inequality you can also find answers by drawing
graphs of each side and looking for the region(s) where one graph is
above/beneath the otherβ¦
a) On the same axes sketch the graphs of the curves with
equations:
b) Find the points of intersection of the two graphs
c) Solve the following equation:
1B
π¦=7 π₯3 π₯+1 π¦=4βπ₯
7 π₯3π₯+1
<4βπ₯
π¦=7 π₯3 π₯+1
π¦=4βπ₯The sketch for this graph is simple Downward sloping
graph x and y intercepts at
(4,0) and (0,4)
This one is more difficultSubbing in x = 0 or y = 0 will yield the intercept
(0,0)Vertical asymptote at x = -1/3
Horizontal asymptote at y = 7/3
4
4
-1/3
7/3
(0,0)
InequalitiesYou can use graphs to help solve
Inequalities
If you are solving an Inequality you can also find answers by drawing
graphs of each side and looking for the region(s) where one graph is
above/beneath the otherβ¦
a) On the same axes sketch the graphs of the curves with
equations:
b) Find the points of intersection of the two graphs
c) Solve the following equation:
1B
π¦=7 π₯3 π₯+1 π¦=4βπ₯
7 π₯3π₯+1
<4βπ₯
π¦=7 π₯3 π₯+1
π¦=4βπ₯
4
4
-1/3
7/3
b) The points of intersection will be where the two
equations are set equal to each other
7 π₯3π₯+1
=4βπ₯
7 π₯=(4βπ₯)(3 π₯+1)
7 π₯=β3π₯2+11π₯+4
3 π₯2β4 π₯β4=0
(3 π₯+2 ) (π₯β2 )=0
π₯=β23ππ π₯=2
Multiply by (3x + 1)
Expand brackets
Rearrange and set equal to 0
Factorise
Now you know the intersections
(0,0)
InequalitiesYou can use graphs to help solve
Inequalities
If you are solving an Inequality you can also find answers by drawing
graphs of each side and looking for the region(s) where one graph is
above/beneath the otherβ¦
a) On the same axes sketch the graphs of the curves with
equations:
b) Find the points of intersection of the two graphs
c) Solve the following equation:
1B
π¦=7 π₯3 π₯+1 π¦=4βπ₯
7 π₯3π₯+1
<4βπ₯
4
4
7 π₯3π₯+1
<4βπ₯
7 π₯3π₯+1
<4βπ₯
-2/3
2
Consider the colours (in this case)
So we want to know where the blue line is below the red lineβ¦
-1/3
7/3
The blue line is below the red line for x-values below
-2/3
The blue line is below the red line for x-
values between -1/3 and 2
π₯<β23β13<π₯<2
(0,0)
InequalitiesYou can use graphs to help solve
Inequalities
Solve the Inequality:
Start by sketching a graph of each side
Remember for the modulus side, think about what the graph would
look like without the modulus partβ¦
So the lowest value will be when x = 2 (so the minimum point will have a
value of -4)
This is important as when we reflect the lower part for the modulus, the peak will be above the y = 3 line
1B
|π₯2β4 π₯|<3
3
y =βx2 β 4xβ
π₯2β4 π₯=0π₯ (π₯β4)=0π₯=0ππ 4
We can see visually where the modulus graph is
below y = 3, but we need the critical pointsβ¦
The original red line has equation y = x2 β 4x
The reflected part has equation y = -(x2 β 4x)
π₯2β4 π₯=3
(π₯β2)2β4=3
(π₯β2)2=7
π₯β2=Β±β7π₯=2Β±β7
Use completing the square (or the quadratic formula β this wonβt factorise
nicely!)
Add 4
Square root
Add 2
2-β7 2+β70 4
(2,-4)
(2,4)
Intersection of y = 3 on the original graph
InequalitiesYou can use graphs to help solve
Inequalities
Solve the Inequality:
Start by sketching a graph of each side
Remember for the modulus side, think about what the graph would
look like without the modulus partβ¦
So the lowest value will be when x = 2 (so the minimum point will have a
value of -4)
This is important as when we reflect the lower part for the modulus, the peak will be above the y = 3 line
1B
|π₯2β4 π₯|<3
3
y =βx2 β 4xβ
π₯2β4 π₯=0π₯ (π₯β4)=0π₯=0ππ 4
We can see visually where the modulus graph is
below y = 3, but we need the critical pointsβ¦
The original red line has equation y = x2 β 4x
The reflected part has equation y = -(x2 β 4x)
β(π₯2β4 π₯)=3βExpandβ the
bracket
2-β7 2+β7
(2,4)
Intersection of y = 3 on the reflected graph
βπ₯2+4 π₯=3
π₯2β4 π₯+3=0
(π₯β3 ) (π₯β1 )=0
Rearrange and set equal to 0
Factorise
π₯=1ππ π₯=3
1 3
InequalitiesYou can use graphs to help solve
Inequalities
Solve the Inequality:
Start by sketching a graph of each side
Remember for the modulus side, think about what the graph would
look like without the modulus partβ¦
So the lowest value will be when x = 2 (so the minimum point will have a
value of -4)
This is important as when we reflect the lower part for the modulus, the peak will be above the y = 3 line
1B
|π₯2β4 π₯|<3
3
y =βx2 β 4xβ
π₯2β4 π₯=0π₯ (π₯β4)=0π₯=0ππ 4
We can see visually where the modulus graph is
below y = 3, but we need the critical pointsβ¦
The original red line has equation y = x2 β 4x
The reflected part has equation y = -(x2 β 4x)
2-β7 2+β7
(2,4)
1 3
|π₯2β4 π₯|<3|π₯2β4 π₯|<3
We need the ranges where the red graph is below the blue graph
2ββ7<π₯<1 3<π₯<2+β7ππ
InequalitiesYou can use graphs to help
solve Inequalities
Sometimes rearranging the equation can make the sketch far
easier to draw!
Remember to be wary of whether you might by
multiplying or dividing by a negative though!
Solve:
Now it is easier to sketch them both!
1B
|3 π₯|+π₯β€2|3 π₯|β€2β π₯
y =β3xβ
y = 2 - x
2
2
Find the critical values.
Remember to use y = 3x for the original red graph and y = -(3x)
for the reflected partβ¦
3 π₯=2βπ₯4 π₯=2π₯=0.5
Intersection on the original red
line β(3 π₯)=2β π₯β2 π₯=2π₯=β1
Intersection on the reflected red
lineAdd x
Solve
Add x
Solve
-1 0.5
y = 3x
InequalitiesYou can use graphs to help
solve Inequalities
Sometimes rearranging the equation can make the sketch far
easier to draw!
Remember to be wary of whether you might by
multiplying or dividing by a negative though!
Solve:
Now it is easier to sketch them both!
1B
|3 π₯|+π₯β€2|3 π₯|β€2β π₯
y =β3xβ
y = 2 - x
2
2-1 0.5
|3 π₯|β€2β π₯|3 π₯|β€2β π₯
We want where the red line is below the
blue line
β1β€ π₯β€0.5
Summary
β’ We have seen how to solve more complicated Inequalities
β’ We have seen how to avoid multiplying or dividing by a negative
β’ We have also seen how to use graphs to help answer questions!