C1.3 Algebra and functions 3

AS-Level Maths:

Core 1

for Edexcel

C1.3 Algebra and functions 3

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Linear simultaneous equations


Simultaneous equations involving one linear and one quadratic equation

Linear inequalities

Quadratic inequalities

Polynomials

Examination-style questions


An equation with two unknowns has an infinite number of solution pairs. For example:

x

+

y

= 3 is true when

x

= 1 and

y

= 2

x

=

–4 and

y

= 7

x

= 6.4 and

y

=

–3.4 and so on.

We can represent the set of solutions graphically.

The coordinates of every point on the line satisfy the equation.

3

y x

+

y

= 3

0

3

x

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Similarly, an infinite number of solution pairs exist for the equation

y

–

x

= 1

Again, we can represent the set of solutions graphically.

y

There is one pair of values that satisfies

both

these equations simultaneously.

1

y

–

x

= 1

This pair of values corresponds to the point where the lines and

y

–

x

= 1 intersect:

x

+

y

= 3

-1

0

x

+

y

= 3

This is the point (1, 2). At this point

x

= 1 and

y

= 2.

x

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Two linear equations with two unknowns, such as

x

and

y

, can be solved simultaneously to give a single pair of solutions.

When will a pair of linear simultaneous equations have no solutions?

In the case where the lines corresponding to the equations are parallel, they will never intersect and so there are no solutions.

Linear simultaneous equations can be solved algebraically using:

The elimination method,

or

The substitution method.

The solution to the equations can be illustrated graphically by finding the points where the two lines representing the equations intersect.

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The elimination method

If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example:

Solve the simultaneous equations

3

x

+ 7

y

= 22 and 3

x

+ 4

y

= 10.

Subtracting gives:

–

3

x

+ 7

y

= 22

3

x

+ 4

y

= 10

3

y y

= 12

= 4

The terms in

x

have been eliminated.

Substituting

y

= 4 into the first equation gives:

3

x

+ 28 = 22

3

x

=

–6

x

=

–2

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We can check whether

x

=

–2 and

y

= 4 solves both

3

x

+ 7

y

= 22

3

x

+ 4

y

= 10 by substituting them into the second equation.

LHS = 3

× –2 + 4 × 4

=

–6 + 16

= 10

= RHS

So the solution is

x

=

–2

,

y

= 4 .

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Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example:

Solve: 5

x

–

2

y

= 31

4

x

+ 3

y

= 11

1

2

We need to have the same number in front of either the

x

or the

y

terms before adding or subtracting the equations.

Call these equations

1

and

2

.

3

×

2

×

3

1

2

+

4

+

15

x

– 6

y

= 93

8

x

+ 6

y

= 22

23

x

=

115

x

= 5

3

4

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Substitute

x

= 5 in

1 :

5

× 5 – 2

y

= 31

25

– 2

y

= 31

–2

y

= 6

x

=

–3

Check by substituting

x

= 5 and

y

=

–3 into

LHS = 4

× 5 + 3 × –3

= 20

– 9

2

:

= 11

= RHS

So the solution is

x

= 5 ,

y

=

–3

.

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The substitution method

Two simultaneous equations can also be solved by substituting one equation into the other. For example:

Solve:

= 2

x

2

x

+ 3

y

= 23


1

and

2

.

Substitute

1

into

2

:

2

x

+ 3( 2

x

– 3

) = 23

2

x

+ 6

x

– 9 = 23

8

x

– 9 = 23

8

x

= 32

x

= 4

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1

2



Substituting

x

= 4 into

1

gives

y

= 2

× 4 – 3

y

= 5

Check by substituting

x

= 4 and

y

= 5 into

2

:

LHS = 2

× 4 + 3 × 5

= 8 + 15

= 23

= RHS

So the solution is

x

= 4 ,

y

= 5 .

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Solve:

3

x

–

y

= 9

8

x

+ 5

y

= 1

1

2


1

and

2

.

One of the equations needs to be arranged in the form

x

or

y

= …

= … before it can be substituted into the other equation.

Rearrange equation

1

:

3

x

–

y

= 9

–

y

= 9

– 3

x y

= 3

x

– 9

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Now substitute

y

= 3

x

– 9 into equation

2

:

8

x

8

x

+ 5( 3

x

+ 15

x

– 9

) = 1

– 45 = 1

23

x

– 45 = 1

23

x

= 46

x

= 2

Substitute

x

= 2 into equation

1

to find the value of

y

:

6

–

y

–

y y

= 9

= 3

=

–3

So the solution is

x

= 2 ,

y

=

–3

.

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One linear and one quadratic equation





Polynomials



Suppose one of the equations in a pair of simultaneous equations is linear and the other is a quadratic of the form

y

=

ax

2

+

bx

+

c

.

By considering the points where the graphs of the two equations might intersect we can see that there could be two, one or no pairs of solutions.

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If the second equation contains terms in

xy

or

y

2 the shape of the corresponding graph will not be a parabola but a circle, a hyperbola or an ellipse:

A line and a circle

A line and a hyperbola

A line and an ellipse

Again we can have two, one or no pairs of solutions.

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When a pair of simultaneous equations contains one linear and one quadratic equation, we usually solve them by substitution. For example:

Solve:

y

=

x

2

+ 1

y

=

x

+ 3

1

2

Substituting equation

1

into equation

2

gives

x

2

+ 1 =

x

+ 3

Rearranging to give a quadratic equation of the form

ax

2

+

bx

+

c

= 0 gives

x

2

–

x

– 2 = 0

(

x

+ 1)(

x

– 2) = 0

x

=

–1 or

x

= 2

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We can substitute these values of

x

into one of the equations

y

=

x

2

+ 1

y

=

x

+ 3 to find the corresponding values of

y

.

1

2

It is easiest to substitute into equation

2

because it is linear.

When

x

=

–1:

y y

=

–1 + 3

= 2

When

x

= 2:

y

= 2 + 3

y

= 5

The solutions are

x

=

–1

,

y

= 2 and

x

= 2 ,

y

= 5 .

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y

y

=

x

2

+ 1

y

=

x

+ 3 We can demonstrate the solutions to

y

=

x

2

+ 1

y

=

x

+ 3 using a graph.

(

–1, 2)

(2, 5)

0

x

It is difficult to sketch a parabola accurately. For this reason, it is difficult to solve simultaneous equations with quadratic terms using graphs, particularly when the solutions are not integers.

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Look at this pair of simultaneous equations:

y

–

x

= 1

x

2

+

y

2

= 13

1

2

What shape is the graph given by

x

2

+

y

2

= 13?

The graph of

x

2

+

y

2

= 13 is a circular graph with its centre at

We can solve this pair of simultaneous equations algebraically using substitution.

We can then sketch the graphs of the equations to demonstrate where they intersect.

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y

–

x

= 1

x

2

+

y

2

= 13

1

2

Rearranging

1

gives

y

=

x

+ 1

Substituting into

2

gives

x

2

+ (

x

+ 1)

2

= 13

x

2

+

x

2

+ 2

x

+ 1 = 13

2

x

2

+ 2

x

– 12 = 0

x

2

+

x

– 6 = 0

(

x

+ 3)(

x

– 2) = 0

x

=

–3 or

x

= 2

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We can substitute these values of

x

into one of the equations

y

=

x

+ 1

x

2

+

y

2

= 13 to find the corresponding values of

y

.

1

2

It is easiest to substitute into equation

1

because it is linear.

When

x

=

–3:

When

x

= 2:

y

=

–3 + 1

y

=

–2

y

= 2 + 1

y

= 3

The solutions are

x

=

–3

,

y

=

–2 and

x

= 2 ,

y

= 3 .

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Demonstrating these solutions graphically gives:

x

2

+

y

2

= 13

y

(2, 3)

y

=

x

+ 1

0

x

(

–3, –2)

The graphs intersect at the points (

–3, –2) and (2, 3).

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Using the discriminant

In summary, to solve a pair of simultaneous equations where one equation is linear and the other is quadratic:

Rearrange the linear equation so that one of the variables is written in terms of the other.

Substitute the linear equation into the quadratic equation to give a single equation of the form

ax

2

+

bx

+

c

= 0.

We can find the determinant of this equation to find how many times the line and the curve will intersect. When

b

2

– 4

ac

> 0, there are two distinct points of intersection.

b

2

– 4

ac

= 0, there is one point of intersection (or two coincident points). The line is a

tangent

to the curve.

b

2

– 4

ac

< 0, there are no points of intersection.

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Using the discriminant

Show that the line

y

– 4 tangent to the curve

y x

=

x

+ 7 = 0 is a

2

– 2

x

+ 2.


1

and

2

.

y

– 4

x

+ 7 = 0

y

=

x

2

– 2

x

+ 2

1

2

Rearranging

1

gives

Substituting into

2

gives

4

x y

= 4

x

– 7

– 7 =

x x

2

– 6

x

+ 9 = 0

2

– 2

x

+ 2

The discriminant =

b

2

– 4

ac

= (

–6)

2

– 4(9)

= 36

– 36

= 0

b

2

– 4

ac

= 0 and so the line is a tangent to the curve.

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Polynomials



An

inequality

the symbols: links two or more expressions with

<, >,

≤ or ≥.

Inequalities are linear if the expressions they contain can be written in the form

ax

+

b

where

a

and

b

are constants.

For example:

3

x

+ 2 > 5

Solving this inequality involves finding the values of

x

that make the inequality true.

In this example, the inequality is true when

x

> 1.

The solution can be illustrated using a number line as follows:

–3

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–2 –1

0 1 2 3 4 5


Solving linear inequalities

Like an equation, we can solve an inequality by adding or subtracting the same value to both sides of the inequality sign.

We can also multiply or divide both sides of the inequality by a

positive

value. For example:

Add 7 to both sides:

Add 2

x

to both sides:

Divide both sides by 6:

Solve 4

x

– 7 > 11 – 2

x.

4

x

> 18

– 2

x

6

x

> 18

x

> 3

How could we check this solution?

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Checking solutions

To verify that is the solution to

x

> 3

4

x

– 7 > 11 – 2

x

substitute a value just above 3 into the inequality and then substitute a value just below 3.

Substituting

x

= 4 into the inequality gives

4

× 4 – 7 > 11 – 2 × 4

16

– 7 > 11 – 8

9 > 3 This is true.

Substituting

x

= 2 into the inequality gives

4

× 2 – 7 > 11 – 2 × 2

8

– 7 > 11 – 4

1 > 7 This is not true.

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Multiplying or dividing by negatives

Although most inequalities can be solved like equations we have to take great care when multiplying or dividing both sides of an inequality by a negative value.

The following simple inequality is true:

–3 < 5

Look what happens if we multiply both sides by

–1:

–3 × –1

< 5

× –1

3 <

–5

This is not true.

To keep the inequality true we have to reverse the inequality sign:

3 >

–5

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Multiplying or dividing by negatives

Remember:

when both sides of an inequality are multiplied or divided by a negative number the inequality is reversed.

For example:

4

– 3

x

≤ 10

–3

x

≤ 6

x

≥ –2

The inequality sign is reversed.

We could also solve this type of inequality by collecting

x

terms on the right and reversing the inequality sign at the end.

4

– 3

x

≤ 10

4

≤ 10 + 3

x

–6 ≤ 3

x

–2 ≤

x x

≥ –2

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Solving combined linear inequalities

The two inequalities 4

x

+ 3

≥ 5 and 4

x

as a single combined inequality:

+ 3 < 15 can be written

5

≤ 4

x

+ 3 < 15

We can solve this inequality as follows:

Subtract 3 from each part:

2

≤ 4

x

< 12

Divide each part by 4:

0.5

≤

x

< 3

We can illustrate this solution on a number line as follows:

–1 –0.5

0 0.5

1 1.5

2 2.5

3 3.5

4

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Solving combined linear inequalities

Some combined inequalities contain variables in more than one part. For example:

x

– 2 ≤ 3

x

+ 2

≤ 2

x

+ 7

Treat this as two separate inequalities:

x

– 2 ≤ 3

x

+ 2 and 3

x

– 2 ≤ 2

x

– 4 ≤ 2

x

+ 2

x

+ 2

x

≤ 2

+ 2

≤ 7

≤ 5

x

+ 7

– 2 ≤

x

We can write the complete solution as

–2 ≤

x

≤ 5 and illustrate it on a number line as follows:

–3 –2 –1

0 1 2 3 4 5 6 7

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Overlapping solutions

Solve the following inequality and illustrate the solution on a number line:

2

x

– 1 ≤

x

+ 2 < 7

Treating as two separate inequalities gives

2

x

– 1 ≤

x

+ 2 and

x

+ 2 < 7

x

– 1 ≤ 2

x

≤ 3

x

< 5

If

x

< 5 then it is also

≤ 3. The whole solution set is therefore given by

x

< 5. This can be seen on the number line:

–3 –2 –1

0 1 2 3 4 5 6 7

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Solutions in two parts

Solve the following inequality and illustrate the solution on a number line:

4

x

+ 5 < 3

x

+ 5

≤ 4

x

+ 3

Treating as two separate inequalities gives

4

x

+ 5 < 3

4

x x

< 3

< 0

x x

+ 5 and 3

x

+ 5

≤ 4

x

+ 3

5

≤

2

≤

x x x

≥ 2

+ 3

We cannot write these solutions as a single combined inequality. The solution has two parts.

–3 –2 –1

0 1 2 3 4 5 6 7

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Polynomials




contain terms in both

x

2 example:

x

2

+

x

– 6 ≥ 0 and

x

. For

Factorizing gives

(

x

+ 3)(

x

– 2) ≥ 0

x

2

+

x

– 6 is equal to 0 when:

x

+ 3 = 0

x

=

–3 and

x

– 2 = 0

x

= 2

These values give the end points of the solution set:

–5 –4 –3 –2 –1

0 1 2 3 4 5

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To find the solution set we can substitute a value from each of the following three regions:

region 2 region 3

–5 –4 –3 –2 –1

0 1 into the original inequality

x

2

+

x

– 6 ≥ 0.

2 3 4

When

x

=

–4: –4

2

+

–4 – 6 ≥ 0

16

– 4 – 6 ≥ 0

6

≥ 0

This is true and so values in region 1 satisfy the inequality.

5

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When

x

= 0 0

2

+ 0

– 6 ≥ 0

–6 ≥ 0

This is not true and so values in region 2 do not satisfy the inequality.

region 1

–5 –4

When

x

= 3

–3 –2 –1

0 1

3

2

+ 3

–

6

≥ 0

9 + 3

–

6

≥ 0

6

≥ 0

2 3 4

This is true and so values in region 3 satisfy the inequality.

5

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We have shown that values in

region 1

inequality

x

2

+

x

–

6

≥ 0.

and

region 3

satisfy the

region 1 region 2 region 3

–5 –4 –3 –2 –1

0 1 2 3

We can show the complete solution set as follows:

4 5

–5 –4 –3 –2 –1

0

So the solution to

x

2

+

x

– 6 ≥ 0 is:

x

≤ –3 or

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1

x

≥ 2

2 3 4


5


An alternative method for solving inequalities involves using graphs. For example:

Solve

x

2

+

x

– 3 > 4

x

+ 1.

The first step is to rearrange the inequality so that all the terms are on one side and 0 is on the other.

x

2

– 3

x

– 4 > 0

Sketching the graph of

y

inequality.

=

x

2

– 3

x

– 4 will help us to solve this

The coefficient of

x

2

> 0 and so the graph will be



-shaped.

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Next, we find the roots by solving

x

2

– 3

x

– 4 = 0.

Factorizing gives (

x

+ 1)(

x

– 4) = 0

x

=

–1 or

x

= 4

We can now sketch the graph.

The inequality

x

2

– 3

x

– 4 > 0 is true for the parts of the curve that lie above the

x

-axis.

( (

–1, 0)

So, the solution to

x

2

+

x

– 3 > 4

x

+ 1 is

x

<

–1 or

x

> 4

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Solving quadratic inequalities using graphs

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Polynomials





Polynomials


Polynomials

A

polynomial in

x

ax n

is an expression of the form

+

bx n



1

+

cx n



2

+...+

px

2

+ where

a

,

b

,

c

, … are constant coefficients and

n

is a positive integer.

The value of

a

is called the

leading coefficient

.

Examples of polynomials include:

3

x

7

+ 4

x

3

–

x

+ 8

x

11

– 2

x

8

+ 9

x

and 5 + 3

x

2

– 2

x

3

.

Polynomials are usually written in

descending

powers of

x

.

They can also be written in

ascending

powers of

x

, especially when the leading coefficient is negative, as in the last example.


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Polynomials

The

degree

, or

order

, of a polynomial is given by the highest power of the variable.

A polynomial of degree 1 is called

linear

and has the general form

ax

+

b

.


quadratic

general form

ax

2

+

bx

+

c

.

and has the


cubic

general form

ax

3

+

bx

2

+

cx

+

d

.

and has the


quartic

general form

ax

4

+

bx

3

+

cx

2

+

dx

+

e

.

and has the

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Using function notation

Polynomials are often expressed using function notation.

For example, consider the polynomial function:

f

(

x

) = 2

x

2

– 7

We can use this notation to substitute given values of

x

.

For example:

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Find

f

(

x

) when a)

f

(

–2) = 2(–2)

2

– 7

= 8

– 7

= 1 a)

x

=

–2 b)

x

=

t

+ 1 b)

f

(

t

+ 1) = 2(

t

+ 1)

2

– 7

= 2(

t

2

+ 2

t

+ 1)

– 7

= 2

t

2

+ 4

t

+ 2

– 7

= 2

t

2

+ 4

t

– 5


Adding and subtracting polynomials

When two or more polynomials are added, subtracted or multiplied, the result is another polynomial.

Polynomials are added and subtracted by

collecting like terms

.

For example:

f

(

x

) = 2

x

3

– 5

x

+ 4 and

g

(

x

) = 2

x

– 4

Find a)

f

(

x

) +

g

(

x

) b)

f

(

x

)

–

g

(

x

) a)

f

(

x

) +

= 2

x

3

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g

(

x

– 3

)

= 2

x

3

– 5

x x

+ 4 + 2

x

– 4 b)

f

(

x

)

–

g

(

x

)

= 2

x

3

– 5

x

+ 4

– (2

x

– 4)

= 2

x

3

– 5

x

+ 4

– 2

x

+ 4

= 2

x

3

– 7

x

+ 8


Multiplying polynomials

When two polynomials are multiplied together every term in the first polynomial must by multiplied by every term in the second polynomial. For example:

f

(

x

) = 3

x

3

– 2 and

g

(

x

) =

x

3

+ 5

x

– 1

f

(

x

)

g

(

x

) = (3

x

3

– 2)(

x

3

+ 5

x

– 1)

= 3

x

6

+ 15

x

4

– 3

x

3

– 2

x

3

– 10

x

+ 2

= 3

x

6

+ 15

x

4

– 5

x

3

– 10

x

+ 2

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Multiplying polynomials

Sometimes we only need to find the coefficient of a single term.

For example:

Find the coefficient of

x

2 is multiplied by 2

x

3 when

x

3

+ 5

x

2

–

x

– 4

x

2

– 6.

+ 2

x

We don’t need to multiply this out in full. We only need to decide which terms will multiply together to give terms in

x

2

.

(

x

3

– 4

x

2

+ 2

x

)(2

x

3

+ 5

x

2

–

x

– 6)

We have: 24

x

2

– 2

x

2

= 22

x

2

So, the coefficient of

x

2 is 22 .

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Polynomials


Examination-style question 1

a) Solve the simultaneous equations

x

– 2

y

= 2

x

2

+ 4

y

2

= 100 b) Interpret your solution geometrically.

a) Label the equations

x

– 2

y

= 2

x

2

+ 4

y

2

= 100

1

2

Rearranging equation

1

Substituting into equation

2

x

= 2 + 2

y

(2 + 2

y

)

2

+ 4

y

2

= 100

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4 + 8

y

+ 4

y

2

+ 4

y

2

= 100

8

y

2

+ 8

y

– 96 = 0

y

2

+

y

– 12 = 0

(

y

+ 4)(

y

– 3) = 0

y

=

–4 or

y

= 3

Substituting into equation

1

When

y

=

–4,

x

=

–6

When

y

= 3,

x

= 8 b) The line

x

– 2

y

= 2 crosses the curve

x

2 points (

–4, –6) and (3, 8).

+ 4

y

2

= 100 at the

54 of 56



a) Write an expression for the area

A

of the following rectangle:

x

– 2

x

+ 2 b) If the area satisfies the inequality

5 <

A

< 12 find the range of possible values for

x

.

a)

A

= (

x

=

x

2

+ 2)(

x

– 4

– 2) b) The range of possible values for

x

is given by

5 <

x

2

– 4 < 12

55 of 56



We have to solve 5 <

x

2

– 4 and then solve

x

2

– 4 < 12

5 <

x

2

– 4

x

2

– 9 > 0

x x

2

2

– 4 < 12

– 16 < 0

(

x

+ 3)(

x

– 3) > 0

Sketching

y

=

x

2

– 9

(

x

+ 4)(

x

– 4) < 0

Sketching

y

=

x

2

– 16

y y

(

–3, 0)

0

(3, 0)

x

(

–4, 0)

0

(4, 0)

x

x

2

– 9 > 0 when

x

<

–3 or

x

> 3

x

2

– 16 < 0 when –4 <

x

So the range of possible value for

x

is

< 4

3 <

x

< 4 (ignoring negative solutions)

56 of 56


C1.3 Algebra and functions 3

AS-Level Maths:

Core 1