Transcript GCSE Mathematics - Energy E

Analytical Toolbox
Higher order Equations and Solutions
By
Dr J.P.M. Whitty
Learning objectives
•
After the session you will be able to:
•
•
•
•
•
Find the solution to Quadratic equations using
graphical means.
Find algebraic solutions to quadratic equations.
Solve simultaneous equations one quadratic and
one linear.
Use graphical techniques to estimate the solution
of higher order equations
Use math software to solve systems of equations.
2
Learning Check:
Solve simultaneous equations
• Solve the following simultaneous equations
 y=2x+4 & 3x=7-5y using three different
mathematical techniques, i.e:
• Algebraically
• Graphically
• MATLAB matrix method
3
Solution
 -2x+y=4
 3x+5y=7
 -6x+3y=12
 6x+10y=14
 13y=26; y=2
 -2x+2=4
 x=(4-2)/-2=-1
Commands
>> ezplot('-2*x+y=4')
>> hold on
>> ezplot('3*x+5*y=7')
>> grid on
>> ginput
4
Solution cont…
•
The MATLAB matrix method is as follows:
1.
2.
3.
4.
A=[-2 1; 3 5]
b=[4; 7]
A\b
ans=[-1 2]
5
Recap: Linear Functions
 We have already learned about linear functions, for example:

f(x) =y= 4x + 1
 g(x) =y= -6x -2
 h(x) =y= (-2/3)x + 2
The graphs of these functions
are straight lines and each
equation gives us “clues” as to
what each line looks like:
Slope of 4
f(x) = 4x + 1
6
Quadratic Functions
 What do the functions
 f(x) = 2x2 + 4x – 3


g(x) = x2 – 4x
h(x) = -x2 – 5x + 6
have in common?
They are all QUADRATIC FUNCTIONS that can be
written in standard form:
f(x) = ax2 +bx + c
where a ≠ 0
Pay close attention… these coefficients a, b and c will
be clues that help us to know what the graphs of these
quadratic functions look like…
7
Graphs of Quadratic Functions
• Quadratic Functions graph into a shape called a
Parabola
If the coefficient of the
If the coefficient of the
x2 term, “a”, is negative,
the parabola opens
downward (Sad)
f(x) = -x2 + 5
x2 term, “a”, is positive,
the parabola opens
upward (Happy)
Maximum Point
f(x) = x2 - 4
Vertex
Minimum Point
8
The parabola
y

ax
 The graph of
2
is a parabola with
 vertex (0, 0)
Example: Locate the vertex and find the axis of symmetry of
the following parabola. Does it open up or down? Graph the
parabola.
9
Quadratic equations

Any equation that can be written so that one side
is a quadratic and the other side is zero is called
a quadratic equation.
6x  4x 1  0
2
In fact, we can consider a “general form” of
quadratic equation. However a quadratic is given
to us, we can always rewrite it as:
ax  bx  c  0
2
10
Solving Quadratics
 As with all equations, solving them simply means finding
the values that make them true. The easiest quadratic
equation we could be asked to solve would be
something like:
x2  4
There are TWO values that make this true: x = 2 and x = -2
(in MATLAB roots([1 0 -4]); the value of the coefficients).
We call these the roots of the equation. In fact this idea of
two solutions extends to just about all
quadratic equations.
11
Solution by factorization


At GCSE level you would have spent a lot of time practising
factorization of quadratic expressions. Here’s why...
Consider the following equation:
•
We can factorise the LHS of the equation
2 as follows:
•
This leaves us with the equation looking like this:

x  3x  2  0
x 2  3x  2  ( x  2)( x  1)
( x  2)( x  1)  0
12
Sketching the curve
 Here the solution to
renders intercepts on
the x-axis, while putting
x=0; gives the intercept
on the y-axis, hence
generating the required
three points, thus:
( x  2)( x  1)  0
(x-2) (x-1)
6
5
4
3
2
1
0
-1
0
1
2
3
4
x
For a forth point let x=1.5; y=0.25
13
Try one:
 Solve the equation
hence or
otherwise sketch the curve:
x2  2x  3  0
y  x2  2x  3  0
Factorize: y  ( x  1)( x  3)
(x+1) (x-3)
12
10
8
6
4
2
0
x=0;y=-3
-2
-4
x=1;y=-4
-2
0
2
x
4
14
Formula solution to quadratic equations
The process of completing the square leads us to a
formula for solving any quadratic equation.
We can solve a quadratic in the form:
ax 2  bx  c  0
With the formula:
 b  b 2  4ac
x
2a
Those feeling powerful should try to derive this formula. The
starting point should be a general formula for completing
the square.
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The formula for quadratic equations
Examples
x2  4x  3  0
x2  x  6  0
x 2  8x  3  0
 b  b 2  4ac
x
2a
2 x 2  5x  1  0
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Example solutions:
x2  4x  3  0
Examples:
 4  16  12
2
42
x
2
 x  1 or x  3
x
x2  x  6  0
 1  1  (24)
2
1  5
x
2
 x  2 or x  3
x
8  76 8  2 19
x  8x  3  0  x 

 4  19
2
2
2
 5  33
2 x  5x  1  0  x 
4
2
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The formula for quadratic equations
Note that you MUST rearrange an equation into
ax2 + bx + c = 0
form before using the formula.
Also… remember that NOT every quadratic equation has
two
solutions, and indeed, some have none at all!!
We’ll look at the following examples with the formula.
x  6x  9  0
2
3x  2 x  20  0
2
18
Repeated roots
x  6x  9  0
2
6  36  36
x
2
60
x
2
x3
What’s going on here is that the
expression x2 – 6x + 9 is a perfect
square. Namely (x – 3)2.
So we only have one solution to
x2 – 6x + 9 = 0.
We say that 3 is a repeated root of
the equation x2 – 6x + 9 = 0 .
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No Solutions??
3 x  2 x  20  0
2
 2  4  240
x
6
 2   236
x
6
Our problem here is that we
CANNOT find the square root
of a negative number.
In situations like this, we simply
say that the equation has no
solutions.
That is, there are NO values of x
that make 3x2 + 2x + 20 equal 0.
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A general approach for solving quadratics
Rearrange into the form
ax 2  bx  c  0
Try to factorise. The idea is that you are so well practised in
factorisation that this shouldn’t take too long!
If you can’t factorise, use the formula.
You’ll note that completing the square isn’t mentioned. That’s
why we didn’t spend long talking about it!! The reasons I
bothered to include it at all are (a) the method leads to the
formula, and (b) so that if anyone else ever mentions it to you,
you will know what they mean, and have somewhere to come
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back to!
When is a quadratic not a quadratic?
We will often be faced with equations that don’t look very
quadratic, that turn out to be when rearranged.
x 3  2 x 2  x  1  x 3  3x 2  7 x  9
3 x4

x
2
x4  x2  6  0
All of these can be rearranged-to and/or tackled as
quadratics.
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When it’s a jar…
x  2 x  x  1  x  3x  7 x  9
x4  x2  6  0
 x  6x  8  0
 ( x  2)( x  4)  0
 ( x 2  2)( x 2  3)  0
 x 2  2 or
x2  3
 x  2 or
 no solutions
and
3
2
3
2
2
x4
x 3
3 x4

x
2
 6  x2  4x
 0  x2  4x  6
and using formula we have
x  2  10
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Summary
You should now
know what is meant by an equation
know what is meant by a quadratic equation
be able to solve a quadratic equation by factorisation
be able to rearrange and solve without losing a solution
be able to use the completing the square technique
know and be able to use the formula for quadratic
equations
be able to recognise (thanks to the formula) when a
quadratic
equation has no solutions
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Examination Type Question
• Using suitable math software or otherwise sketch the
graph of the following functions
• f(x)=3x2+2x-7 [6]
• g(x)=2x+3 [4]
• From the graph obtain approximate values of the
interception points [2]. Use an analytical method to
obtain better solutions [5]. How could you obtain
better approximations using math software [3]
• [20 marks]
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Solution:
 Use the MATLAB
ginput command to
obtain approximate
solutions
3 x 2+2 x-7
120
100
80
60
40
20
ans =
0
-1.7807 -0.6651
1.8386 6.9621
-6
-4
-2
0
x
2
4
6
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Analytical
solutions:
• f(x)=3x2+2x-7
• g(x)=2x+3
3x 2  2 x  7  2 x  3
 3 x 2  10 
10
10
2
x 
x 
 1.825
3
3
2
 10 
  3  6.65
y  2
 3 


Find y values :
 10 

y  2
 3   3  0.651


*Better solutions could be obtained from MATLAB
ginput command or alternatively as for the roots of the
quadratic
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Further Examination Type Q:
 The graph shows
two functions
namely: 4x2-2x+1 &
x+4. Explain which
curve is which giving
reasons for your
answer [5]; and find
the interception points
[5].
4 x 2-2 x+1
20
15
10
5
0
-5
-10
-15
-4
-2
0
2
4
6
x
[10 marks]
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Solution
 The first function is a parabolaB2 (only curveB1) due to the
polynomial being of second orderB1. The second graph is a is a
lineB1 as it has the form y=mx+cB1.
4x2  2x 1  x  4
 4 x 2  3x  3  0
x
x
  3 
3
M1
 32  44  3
24 
A1
M2
A1

 1.319
9  48
 8
  3 57  
8

 0.568
 8
3
57
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Another Exam Question
1.
Find the interception of the graphs:
y  3x 2  2 x  1
and
y  4x  1
[5]
3 x 2  2 x  1  4 x  1 M1
 3x 2  6 x  0
M1A1
2
 x (3 x  6)  0  x  
0
( 2,42   1)
Find points values : 
(0,40   1)
 ( 2,7) A1

(0,1)
A1
You could use the
formula here
also, with a=3,
b=-6 and c=0
M1A1
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