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AS-Level Maths: Core 1for Edexcel

C1.2 Algebra and functions 2

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Quadratic expressions


Factorizing quadratics

Completing the square

Solving quadratic equations

The discriminant

Graphs of quadratic functions

Examination-style questions



A quadratic expression is an expression in which the highest power of the variable is 2. For example:

x2 – 2 w2 + 3w + 1 4 – 5g2 t2

2

x is a variable.

a is the coefficient of x2.

b is the coefficient of x.

c is a constant term.

ax2 + bx + c (where a ≠ 0)The general form of a quadratic expression in x is:


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The discriminant




Factorizing quadratic expressions

Factorizing an expression is the inverse of expanding it.

Expanding or multiplying out

Factorizing

When we expand an expression we multiply out the brackets.

(x + 1)(x + 2) x2 + 3x + 2

When we factorize an expression we write it with brackets.



Quadratic expressions of the form ax2 + bx can always be factorized by taking out the common factor x. For example:

No constant term

3x2 – 5x = x(3x – 5)

When a quadratic has no term in x and the other two terms can be written as the difference between two squares, we can use the identity

The difference between two squares

a2 – b2 = (a + b)(a – b)

to factorize it. For example:

9x2 – 49 = (3x + 7)(3x – 7)



Quadratic expressions of the form x2 + bx + c can be factorized if they can be written using brackets as

(x + d)(x + e)

where d and e are integers.

If we expand (x + d)(x + e), we have

(x + d)(x + e) = x2 + dx + ex + de

= x2 + (d + e)x + de

Quadratic expressions with a = 1



Quadratic expressions of the general form ax2 + bx + c can be factorized if they can be written using brackets as

(dx + e)(fx + g)

where d, e, f and g are integers.

If we expand (dx + e)(fx + g), we have

(dx + e)(fx + g)= dfx2 + dgx + efx + eg

= dfx2 + (dg + ef)x + eg

The general form


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The discriminant




Perfect squares

Some quadratic expressions can be written as perfect squares. For example:

x2 + 2x + 1 = (x + 1)2

x2 + 4x + 4 = (x + 2)2

x2 + 6x + 9 = (x + 3)2

x2 – 2x + 1 = (x – 1)2

x2 – 4x + 4 = (x – 2)2

x2 – 6x + 9 = (x – 3)2

How could the quadratic expression x2 + 8x be made into a perfect square?

We could add 16 to it.

In general:

x2 + 2ax + a2 = (x + a)2 or x2 – 2ax + a2 = (x – a)2 x2 + 2ax + a2 = (x + a)2 or x2 – 2ax + a2 = (x – a)2



Adding 16 to the expression x2 + 8x to make it into a perfect square is called completing the square.

x2 + 8x = x2 + 8x + 16 – 16We can write

If we add 16 we then have to subtract 16 so that both sides are still equal.

By writing x2 + 8x + 16 we have completed the square and so we can write this as

x2 + 8x = (x + 4)2 – 16In general:

2 22

2 2

b bx bx x



Complete the square for x2 – 10x.

Compare this expression to (x – 5)2 = x2 – 10x + 25

= (x – 5)2 – 25

x2 – 10x = x2 – 10x + 25 – 25

Complete the square for x2 + 3x.

Compare this expression to x x x2 23 92 4( + ) = + 3 +

x xx x 9 9244

2 + 3 ++ 3 =

x 232

94= ( + )


2 22

2 2

b bx bx c x c


How can we complete the square for x2 – 8x + 7?

= (x – 4)2 – 9

x2 – 8x + 7 = x2 – 8x + 16 – 16 + 7

Look at the coefficient of x.

This is –8 so compare the expression to (x – 4)2 = x2 – 8x + 16.

In general:



Complete the square for x2 + 12x – 5.

Compare this expression to (x + 6)2 = x2 + 12x + 36

= (x + 6)2 – 41

x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5

Complete the square for x2 – 5x + 7.

Compare this expression to x x x2 25 252 4( ) = 5 +

xx x x 2 22 254

5455 + 7 7+=

x 252

34= ( + )



When the coefficient of x2 is not 1, quadratic equations in the form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by completing the square.

Complete the square for 2x2 + 8x + 3.

2x2 + 8x + 3 = 2(x2 + 4x) + 3

By completing the square, x2 + 4x = (x + 2)2 – 4 so

2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3

= 2(x + 2)2 – 8 + 3

= 2(x + 2)2 – 5

Take out the coefficient of x2 as a factor from the terms in x:



Complete the square for 5 + 6x – 3x2.

By completing the square, x2 – 2x = (x – 1)2 – 1 so

5 + 6x – 3x2 = 5 – 3((x – 1)2 – 1)

= 5 – 3(x – 1)2 + 3

= 8 – 3(x – 1)2

Take out the coefficient of x2 as a factor from the terms in x:

5 + 6x – 3x2 = 5 – 3(–2x + x2)

= 5 – 3(x2 – 2x)


Complete the square


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The discriminant




Quadratic equations

Quadratic equations can be solved by:

completing the square, or

factorization

using the quadratic formula.

ax2 + bx + c = 0 (where a ≠ 0)The general form of a quadratic equation in x is:

The solutions to a quadratic equation are called the roots of the equation.

A quadratic equation may have:

one repeated root, or

two real distinct roots

no real roots.


The roots of a quadratic equation

If we sketch the graph of a quadratic function y = ax2 + bx + c the roots of the equation coincide with the points where the function cuts the x-axis.

As can be seen here, this can happen twice, once or not at all.


Solving quadratic equations by factorization

Start by rearranging the equation so that the terms are on the left-hand side:

Factorizing the left-hand side gives us

Solve the equation 5x2 = 3x

5x2 – 3x = 0

x(5x – 3) = 0

or 5x – 3 = 0

5x = 3

x = 0So

Don’t divide through by x!

35=x


Solving quadratic equations by factorization

Start by rearranging the equation so that the terms are on the left-hand side.

We need to find two integers that add together to make –5 and multiply together to make 4.

Factorizing the left-hand side gives us

Solve the equation x2 – 5x = –4 by factorization.

x2 – 5x + 4 = 0

(x – 1)(x – 4) = 0

x – 1 = 0 or x – 4 = 0

x = 4

Because 4 is positive and –5 is negative, both the integers must be negative. These are –1 and –4.

x = 1


Solving quadratics by completing the square

Quadratic equations that cannot be solved by factorization can be solved by completing the square.

For example, the quadratic equation

x2 – 4x – 3 = 0

can be solved by completing the square as follows:

x = 4.65 x = –0.646 (to 3 s.f.)

(x – 2)2 – 7 = 0

(x – 2)2 = 7

x – 2 = 7

x = 2 + 7 or x = 2 – 7



Solve the equation 2x2 – 4x + 1 = 0 by completing the square. Write the answer to 3 significant figures.

= 2((x – 1)2 – 1) + 1

= 2(x – 1)2 – 2 + 1

= 2(x – 1)2 – 1

Start by completing the square for 2x2 – 4x + 1:

2x2 – 4x + 1 = 2(x2 – 2x) + 1



2(x – 1)2 = 1

2(x – 1)2 – 1 = 0

x = 1.71 x = 0.293 (to 3 s.f.)

Now solving the equation 2x2 – 4x + 1 = 0:

x 2( 1) = 12

x 1= 12

x =1+ 12 x =1 1

2or


Using the quadratic equation formula

Any quadratic equation of the form

can be solved by substituting the values of a, b and c into the formula

ax2 + bx + c = 0ax2 + bx + c = 0

This formula can be derived by completing the square on the general form of the quadratic equation.

a

ax

b cb 2± 4=

2


Using the quadratic formula

Use the quadratic formula to solve 2x2 + 5x – 1 = 0.

2x2 + 5x – 1 = 0

x = 0.186 x = –2.69 (to 3 s.f.)

x 2± (4× × )

=25 15

2×2

x 5 ± 25 + 8

=4

x 5 + 33

=4

x5 33

=4

or

a

ax

b cb 2± 4=

2


Using the quadratic formula

Use the quadratic formula to solve 9x2 – 12x + 4 = 0.

9x2 – 12x + 4 = 0

a

ax

b cb 2± 4=

2

x 2( ) ± ( ) (4× ×1 42 1 9

=2 )

2×9

x12 ± 144 144

=18

x12 ± 0

=18

There is one repeated root: x 23=


Equations that reduce to a quadratic form

Some equations, although not quadratic, can be written in quadratic form by using a substitution. For example:

Solve the equation t4 – 5t2 + 6 = 0.

This is an example of a quartic equation in t.

Let’s substitute x for t2:x2 – 5x + 6 = 0

This gives us a quadratic equation that can be solved by factorization:

(x – 2)(x – 3) = 0

x = 2

So t2 = 2

t = 2

or x = 3

t2 = 3

3t =or


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The discriminant





The discriminant




The discriminant

By solving quadratic equations using the formula

we can see that we can use the expression under the square root sign, b2 – 4ac, to decide how many roots there are.

When b2 – 4ac > 0, there are two real distinct roots.

2

2

4ab

a

bx

c

When b2 – 4ac = 0, there is one repeated root: .2

bx

a

When b2 – 4ac < 0, there are no real roots.

Also, when b2 – 4ac is a perfect square, the roots of the equation will be rational and the quadratic will factorize.

b2 – 4ac is called the discriminant of ax2 + bx + cb2 – 4ac is called the discriminant of ax2 + bx + c


The discriminant

We can demonstrate each of these possibilities graphically.


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The discriminant




Plotting graphs of quadratic functions

Plot the graph of y = x2 – 4x + 2 for –1 < x < 5.

We can plot the graph of a quadratic function using a table of values. For example:

x

x2

– 4x

+ 2

y = x2 – 4x + 2

–1 0 1 2 3 4 5

1 0 1 4 9 16 25

+ 4 + 0 – 4 – 8 – 12 – 16 – 20

+ 2 + 2 + 2 + 2 + 2 + 2 + 2

7 2 –1 –2 –1 2 7

y = ax2 + bx + c (where a ≠ 0)A quadratic function in x can be written in the form:


x

1

0 1–1 3 4 5–1

2

3

4

5

6

y

2

Plotting graphs of quadratic functions

x

y = x2 – 4x + 2

–1 0 1 2 3 4 5

7 2 –1 –2 –1 2 7

The points given in the table are plotted …

… and the points are then joined together with a smooth curve.

The shape of this curve is called a parabola.

It is characteristic of a quadratic function.


Parabolas

When the coefficient of x2 is positive the vertex is a minimum point and the graph is -shaped.

When the coefficient of x2 is negative the vertex is a maximum point and the graph is -shaped.

Parabolas have a vertical axis of symmetry …

…and a turning point called the vertex.


Exploring graphs of the form y = ax2 + bx + c


Sketching graphs of quadratic functions

When a quadratic function factorizes we can use its factorized form to find where it crosses the x-axis. For example:

Sketch the graph of the function y = x2 – 2x – 3.

The function crosses the x-axis when y = 0.

x2 – 2x – 3 = 0

(x + 1)(x – 3) = 0

x + 1 = 0 or x – 3 = 0

x = 3

The function crosses the x-axis at the points (–1, 0) and (3, 0).

x = –1



By putting x = 0 in y = 2x2 – 5x – 3 we can also find where the function crosses the y-axis.

y = 2(0)2 – 5(0) – 3

y = – 3

So the function crosses the y-axis at the point (0, –3).

The quadratic function y = ax2 + bx + c will cross the y-axis at the point (0, c).

The quadratic function y = ax2 + bx + c will cross the y-axis at the point (0, c).

We now know that the function y = x2 – 2x – 3 passes through the points (–1, 0), (3, 0) and (0, –3) and so we can place these points on our sketch.

In general:



0

y

x

(–1, 0) (3, 0)

(0, –3)

(1, –4)

We can also use the fact that a parabola is symmetrical to find the coordinates of the vertex.

The x coordinate of the vertex is half-way between –1 and 3.

1+ 3= =1

2x

When x = 1, y = (1)2 – 2(1) – 3

y = –4

So the coordinates of the vertex are (1, –4).

We can now sketch the graph.



When a quadratic function is written in the form y = a(x – α)(x – β), it will cut the x-axis at

the points (α, 0) and (β, 0).α and β are the roots of the quadratic function.

When a quadratic function is written in the form y = a(x – α)(x – β), it will cut the x-axis at

the points (α, 0) and (β, 0).α and β are the roots of the quadratic function.

For example, write the quadratic function y = 3x2 + 4x – 4 in the form y = a(x – α)(x – β) and hence find the roots of the function.

This function can be factorized as follows,

y = (3x – 2)(x + 2)

It can be written in the form y = a(x – p)(x – q) as23= 3( ) 2( )y x x

Therefore, the roots are 23 and .2

In general:


Exploring graphs of the form y = a(x – α)(x – β)


Sketching graphs by completing the square

When a function does not factorize we can write it in completed square form to find the coordinates of the vertex. For example:

Sketch the graph of y = x2 + 4x – 1 by writing it in completed square form.

x2 + 4x – 1 = (x + 2)2 – 5

The least value that (x + 2)2 can have is 0 because the square of a number cannot be negative.

(x + 2)2 ≥ 0

(x + 2)2 – 5 ≥ – 5 Therefore

The minimum value of the function y = x2 + 4x – 1 is therefore y = –5.



When y = –5, we have,(x + 2)2 – 5 = –5

(x + 2)2 = 0x = –2

The coordinates of the vertex are therefore (–2, –5).The equation of the axis of symmetry is x = –2.Also, when x = 0 we have

y = –1 So the curve cuts the y-axis at the point (–1, 0).Using symmetry we can now sketch the graph.

y

x0(–1, 0)

x = –2

y = x2 + 4x – 1

(–2, –5)



In general, when the quadratic function y = ax2 + bx + c is written in completed square form as

a(x + p)2 + q

The coordinates of the vertex will be (–p, q).

The axis of symmetry will have the equation x = –p.

Also:

If a > 0 (–p, q) will be the minimum point.

If a < 0 (–p, q) will be the maximum point.

Plotting the y-intercept, (0, c) will allow the curve to be sketched using symmetry.


Exploring graphs of the form y = a(x + p)2 + q


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The discriminant




Examination-style question

a) Write 2x2 – 8x + 7 in the form a(x + b)2 + c.

b) Write down the minimum value of f(x) = 2x2 – 8x + 7 and state the minimum value of x where this occurs.

c) Solve the equation 2x2 – 8x + 7 = 0 leaving your answer in surd form.

d) Sketch the graph of y = 2x2 – 8x + 7.

a) 2x2 – 8x + 7 = 2(x2 – 4x) + 7

= 2((x – 2)2 – 4) + 7

= 2(x – 2)2 – 8 + 7

= 2(x – 2)2 – 1



b)

From this we can see that the minimum value of f(x) is –1.

f(x) can be written as f(x) = 2(x – 2)2 – 1

This occurs when x = 2.

c) 2x2 – 8x + 7 = 0

2(x – 2)2 – 1 = 0

2(x – 2)2 = 1

(x – 2)2 = 12

x – 2 = 12±

x = 2 12±

x = 2 12 or x = 2 1

2+



d) When y = 0, x = 2 – or x = 2 +12

12

When x = 0, y = 7

So the graph cuts the coordinate axes at (2 + , 0), (2 – , 0) and (0, 7).

12

12

The parabola has a minimum at the point (2, –1).y

x

7

–1 2 + 122 – 1

2

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