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Transcript of © Boardworks Ltd 2005 1 of 50 © Boardworks Ltd 2005 1 of 50 AS-Level Maths: Core 1 for Edexcel...
© Boardworks Ltd 20051 of 50 © Boardworks Ltd 20051 of 50
AS-Level Maths: Core 1for Edexcel
C1.2 Algebra and functions 2
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© Boardworks Ltd 20052 of 50
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© Boardworks Ltd 20052 of 50
Quadratic expressions
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
© Boardworks Ltd 20053 of 50
Quadratic expressions
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:
© Boardworks Ltd 20054 of 50
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© Boardworks Ltd 20054 of 50
Factorizing quadratics
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
© Boardworks Ltd 20055 of 50
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.
© Boardworks Ltd 20056 of 50
Factorizing quadratic expressions
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)
© Boardworks Ltd 20057 of 50
Factorizing quadratic expressions
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
© Boardworks Ltd 20058 of 50
Factorizing quadratic expressions
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
© Boardworks Ltd 20059 of 50
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© Boardworks Ltd 20059 of 50
Completing the square
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
© Boardworks Ltd 200510 of 50
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
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Completing the square
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
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Completing the square
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= ( + )
© Boardworks Ltd 200513 of 50
2 22
2 2
b bx bx c x c
Completing the square
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:
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Completing the square
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= ( + )
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Completing the square
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:
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Completing the square
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)
© Boardworks Ltd 200517 of 50
Complete the square
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© Boardworks Ltd 200518 of 50
Solving quadratic equations
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
© Boardworks Ltd 200519 of 50
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.
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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.
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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
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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
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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
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Solving quadratics by completing the square
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
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Solving quadratics by completing the square
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
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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
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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
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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=
© Boardworks Ltd 200529 of 50
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
© Boardworks Ltd 200530 of 50
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© Boardworks Ltd 200530 of 50
The discriminant
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
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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
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The discriminant
We can demonstrate each of these possibilities graphically.
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© Boardworks Ltd 200533 of 50
Graphs of quadratic functions
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
© Boardworks Ltd 200534 of 50
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:
© Boardworks Ltd 200535 of 50
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.
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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.
© Boardworks Ltd 200537 of 50
Exploring graphs of the form y = ax2 + bx + c
© Boardworks Ltd 200538 of 50
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
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Sketching graphs of quadratic functions
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:
© Boardworks Ltd 200540 of 50
Sketching graphs of quadratic functions
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.
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Sketching graphs of quadratic functions
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:
© Boardworks Ltd 200542 of 50
Exploring graphs of the form y = a(x – α)(x – β)
© Boardworks Ltd 200543 of 50
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.
© Boardworks Ltd 200544 of 50
Sketching graphs by completing the square
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)
© Boardworks Ltd 200545 of 50
Sketching graphs by completing the square
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.
© Boardworks Ltd 200546 of 50
Exploring graphs of the form y = a(x + p)2 + q
© Boardworks Ltd 200547 of 50
Co
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© Boardworks Ltd 200547 of 50
Examination-style questions
Quadratic expressions
Factorizing quadratics
Completing the square
Solving quadratic equations
The discriminant
Graphs of quadratic functions
Examination-style questions
© Boardworks Ltd 200548 of 50
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
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Examination-style question
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+
© Boardworks Ltd 200550 of 50
Examination-style question
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