Introduction You will be familiar with solving Inequalities from GCSE maths and C1 In this chapter...

Inequalities

Introduction

• You will be familiar with solving Inequalities from GCSE maths and C1

• In this chapter you will see how to solve some more complicated Inequalities

• You will also see how to avoid making a very common error!

• 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 𝑥+8=23 5 𝑥+8>23

5 𝑥=15 5 𝑥>15

𝑥=3 𝑥>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 of…


order to solve them

Remember that solving an Inequality is very similar to solving an equation:

1A

6−2𝑥=2

−2 𝑥=−4

−𝑥=−4

Subtract 6

Divide by 2

If you multiply or divide by a negative in an Inequality, you must reverse the direction of the

sign…

(you can check by substituting values back into the first step if you like!)

Multiply by -1

𝑥=4

6−2𝑥<2

−2 𝑥<−4

−𝑥<−4

Subtract 6

Divide by 2

Multiply by -1. This REVERSES the

sign!𝑥>4


order to solve them

Solve the Inequality below:

1A

2 𝑥2<𝑥+3

2 𝑥2<𝑥+3

2 𝑥2−𝑥−3<0Subtract x and subtract 3

Factorise(2 𝑥−3)(𝑥+1)<0

So the ‘critical values’ are x = 3/2 and x = -1 Now draw a sketch. Use the critical values and the fact this is a positive quadratic…

-1 3/2

Consider the Inequality – we want the range of values where the graph is below 0

So therefore:

−1<𝑥<32

x

y

InequalitiesYou can manipulate Inequalities in order

to solve them


You MUST be careful in this situation.

The normal process would be to multiply each side by (x – 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 – 2)2, as this will definitely be positive (as it has

been squared)

Then you rearrange and solve as in the previous example…

You will need to use the ‘clever factorisation’ technique from FP1! 1A

𝑥2

𝑥−2<𝑥+1 𝑥≠2

𝑥2

𝑥−2<𝑥+1

𝑥2(𝑥−2)2

𝑥−2<(𝑥+1)(𝑥−2)2

𝑥2(𝑥−2)<(𝑥+1)(𝑥−2)2

𝑥2 (𝑥−2 )−(𝑥+1)(𝑥−2)2<0

(𝑥−2 )¿ ¿𝑥2−(𝑥+1)(𝑥−2)¿0

(𝑥−2 )𝑥2−(𝑥2−𝑥−2)¿ ¿¿0(𝑥−2 ) (𝑥+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 – 2)2

Cancel an (x – 2) on the left

Rearrange terms to one side

Take out (x – 2) as a factor

Multiply out the inner bracket

Simplify

InequalitiesYou can manipulate Inequalities in order

to solve them


You MUST be careful in this situation.

The normal process would be to multiply each side by (x – 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 – 2)2, as this will definitely be positive (as it has

been squared)

Then you rearrange and solve as in the previous example…

You will need to use the ‘clever factorisation’ technique from FP1! 1A

𝑥2

𝑥−2<𝑥+1 𝑥≠2

𝑥2

𝑥−2<𝑥+1

(𝑥−2 ) (𝑥+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

−2<𝑥<2


order to solve them


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 aren’t negative…

1A

𝑥𝑥+1

≤2

𝑥+3 𝑥≠−1 ,𝑥 ≠−3

𝑥𝑥+1

≤2

𝑥+3

𝑥 (𝑥+1)2(𝑥+3)2

𝑥+1≤2(𝑥+1)2(𝑥+3)2

𝑥+3

𝑥 (𝑥+1) (𝑥+3 )2≤2 (𝑥+1 )2(𝑥+3)

𝑥 (𝑥+1 ) (𝑥+3 )2−2 (𝑥+1 )2(𝑥+3)≤0

(𝑥+1)(𝑥+3)𝑥 (𝑥+3 )−2(𝑥+1)¿ ¿≤0(𝑥+1)(𝑥+3)𝑥2+3𝑥−2 𝑥−2¿ ¿≤0(𝑥+1)(𝑥+3)𝑥2+𝑥−2¿ ¿≤0(𝑥+1)(𝑥+3)(𝑥+2)(𝑥−1)≤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

‘Clever Factorisation’

Multiply out terms

Simplify

Factorise the expression in the squared bracket


order to solve them


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 aren’t negative…

1A

𝑥𝑥+1

≤2

𝑥+3 𝑥≠−1 ,𝑥 ≠−3

𝑥𝑥+1

≤2

𝑥+3

(𝑥+1)(𝑥+3)(𝑥+2)(𝑥−1)≤0We have shown that this Inequality is

equivalent

Plot a graph

The shape is a positive quartic (same basic shape as a

quadratic – ‘U’, 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

−3<𝑥<−2

−1<𝑥<−1𝑜𝑟

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 other…

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

𝑦=7 𝑥3 𝑥+1 𝑦=4−𝑥

7 𝑥3𝑥+1

<4−𝑥

𝑦=7 𝑥3 𝑥+1

𝑦=4−𝑥The 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)

𝑦=7 𝑥3 𝑥+1

3 𝑥−1=0𝑥=−

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


Inequalities





equations:



1B

𝑦=7 𝑥3 𝑥+1 𝑦=4−𝑥

7 𝑥3𝑥+1

<4−𝑥

𝑦=7 𝑥3 𝑥+1



(4,0) and (0,4)


(0,0)Vertical asymptote at x = -1/3

𝑦=7 𝑥3 𝑥+1 Rearrange to write in terms of

x Multiply by (3x + 1)

4

4

𝑦 (3 𝑥+1 )=7 𝑥

3 𝑥𝑦+𝑦=7 𝑥𝑦=7 𝑥−3𝑥𝑦𝑦=𝑥 (7−3 𝑦)

𝑦7−3 𝑦

=𝑥

Multiply out the bracket

Subtract 3xy

Factorise

Divide by (7 – 3y)


Inequalities





equations:



1B

𝑦=7 𝑥3 𝑥+1 𝑦=4−𝑥

7 𝑥3𝑥+1

<4−𝑥

𝑦=7 𝑥3 𝑥+1



(4,0) and (0,4)



4

4

𝑦7−3 𝑦

=𝑥

7−3 𝑦=0

7=3 𝑦73=𝑦

Find the value that would make the denominator 0 (which isn’t

possible)

Divide by 3

Add 3y

Horizontal asymptote at y =

7/3


Inequalities





equations:



1B

𝑦=7 𝑥3 𝑥+1 𝑦=4−𝑥

7 𝑥3𝑥+1

<4−𝑥

𝑦=7 𝑥3 𝑥+1



(4,0) and (0,4)



Horizontal asymptote at y = 7/3

4

4

-1/3

7/3

(0,0)


Inequalities





equations:



1B

𝑦=7 𝑥3 𝑥+1 𝑦=4−𝑥

7 𝑥3𝑥+1

<4−𝑥

𝑦=7 𝑥3 𝑥+1

𝑦=4−𝑥

4

4

-1/3

7/3

b) The points of intersection will be where the two

equations are set equal to each other

7 𝑥3𝑥+1

=4−𝑥

7 𝑥=(4−𝑥)(3 𝑥+1)

7 𝑥=−3𝑥2+11𝑥+4

3 𝑥2−4 𝑥−4=0

(3 𝑥+2 ) (𝑥−2 )=0

𝑥=−23𝑜𝑟 𝑥=2

Multiply by (3x + 1)

Expand brackets


Factorise

Now you know the intersections

(0,0)


Inequalities





equations:



1B

𝑦=7 𝑥3 𝑥+1 𝑦=4−𝑥

7 𝑥3𝑥+1

<4−𝑥

4

4

7 𝑥3𝑥+1

<4−𝑥

7 𝑥3𝑥+1

<4−𝑥

-2/3

2

Consider the colours (in this case)

So we want to know where the blue line is below the red line…

-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

𝑥<−23−13<𝑥<2

(0,0)


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 part…

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

|𝑥2−4 𝑥|<3

3

y =│x2 – 4x│

𝑥2−4 𝑥=0𝑥 (𝑥−4)=0𝑥=0𝑜𝑟 4

We can see visually where the modulus graph is

below y = 3, but we need the critical points…

The original red line has equation y = x2 – 4x

The reflected part has equation y = -(x2 – 4x)

𝑥2−4 𝑥=3

(𝑥−2)2−4=3

(𝑥−2)2=7

𝑥−2=±√7𝑥=2±√7

Use completing the square (or the quadratic formula – this won’t factorise

nicely!)

Add 4

Square root

Add 2

2-√7 2+√70 4

(2,-4)

(2,4)

Intersection of y = 3 on the original graph


Inequalities






value of -4)


1B

|𝑥2−4 𝑥|<3

3

y =│x2 – 4x│

𝑥2−4 𝑥=0𝑥 (𝑥−4)=0𝑥=0𝑜𝑟 4





−(𝑥2−4 𝑥)=3‘Expand’ the

bracket

2-√7 2+√7

(2,4)

Intersection of y = 3 on the reflected graph

−𝑥2+4 𝑥=3

𝑥2−4 𝑥+3=0

(𝑥−3 ) (𝑥−1 )=0


Factorise

𝑥=1𝑜𝑟 𝑥=3

1 3


Inequalities






value of -4)


1B

|𝑥2−4 𝑥|<3

3

y =│x2 – 4x│

𝑥2−4 𝑥=0𝑥 (𝑥−4)=0𝑥=0𝑜𝑟 4





2-√7 2+√7

(2,4)

1 3

|𝑥2−4 𝑥|<3|𝑥2−4 𝑥|<3

We need the ranges where the red graph is below the blue graph

2−√7<𝑥<1 3<𝑥<2+√7𝑜𝑟

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 𝑥|+𝑥≤2|3 𝑥|≤2− 𝑥

y =│3x│

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 part…

3 𝑥=2−𝑥4 𝑥=2𝑥=0.5

Intersection on the original red

line −(3 𝑥)=2− 𝑥−2 𝑥=2𝑥=−1

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 𝑥|+𝑥≤2|3 𝑥|≤2− 𝑥

y =│3x│

y = 2 - x

2

2-1 0.5

|3 𝑥|≤2− 𝑥|3 𝑥|≤2− 𝑥

We want where the red line is below the

blue line

−1≤ 𝑥≤0.5

Summary

• We have seen how to solve more complicated Inequalities

• We have seen how to avoid multiplying or dividing by a negative

• We have also seen how to use graphs to help answer questions!

Introduction You will be familiar with solving Inequalities from GCSE maths and C1 In this chapter...

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Transcript of Introduction You will be familiar with solving Inequalities from GCSE maths and C1 In this chapter...