Download Completing the square

Completing the square
This is another way of solving a quadratic equation if the equation will not
factorise.
It's often convenient to write an algebraic expression as a square plus another
term. The other term is found by dividing the coefficient of x by 2, and
squaring it.
Any quadratic equation can be rearranged so that it can be solved this way.
Have a look at this example.
Example 1
Re-write x2 + 6x as a square plus another term.
The coeffient of x is 6. Dividing 6 by 2 and squaring it gives 9.
x2 + 6x = (x2 + 6x + 9) - 9
= (x + 3)2 - 9
Have a look at another example before trying one for yourself.
Example 2
We have seen in the previous example that x2 + 6x = (x + 3)2 - 9
So work out x2 + 6x - 2
x2 + 6x - 2 = ( x2 + 6x + 9 ) - 9 - 2
= (x + 3)2 - 11
Question 1
Now solve x2 + 6x - 2 = 0
The Answer
From the previous examples we know that x2 + 6x - 2 = 0 can be written as (x
+ 3)2 - 11 = 0
So to solve the equation, take the square root of both sides
So (x + 3)2 = 11
x+3=+
or x + 3 = x=-3+
or x = - 3 x = - 3 + 3.317 or x = - 3 - 3.317 (
is 3.317.)
x = - 0.317 (3 s.f) or x = - 6.317 (3 s.f)
Example 3
Re-write 2x2 + 20x + 3
Re-write to get x2 on its own.
2( x2 + 10x ) + 3
The coefficient of x is 10. Divide 10 by 2 and square to get 25.
= 2 ( ( x + 5)2 - 25) + 3
= 2 (x + 5)2 - 50 + 3
= 2 (x + 5)2 - 47
Question 2
Now use the previous example to solve 2x2 + 20x + 3 = 0
The Answer
From the previous example, we know that 2x2 + 20x + 3 can be rewritten as:
2 (x + 5)2 - 47
Therefore, we can rewrite the equation as:
2(x + 5 )2 - 47 = 0
2(x + 5 )2 = 47
(x + 5 )2 = 23.5 (Dividing both sides by 2)
Take the square root of both sides
x+5=
or x + 5 = x=-5+
or x = - 5 x=-5+
or x = - 5 x = - 0.152 (3 s.f) or x = - 9.85 (3 s.f)
Using the quadratic formula
Using the technique of completing the square we can make a formula which
will work for all quadratic equations.
The most general way you can write a quadratic equation is:
ax2 + bx + c = 0
Here a, b and c are numbers that vary for different equations.
So if the equation was:
2x2 + 7x + 11 = 0
then a = 2, b = 7, c = 11.
The formula for the solution is then
This formula will work for all equations that can be solved.
Always try to factorise first, if the equation factorises this is the easier method.
In an exam any question that asks for an answer to a quadratic equation
correct to x decimal places should be solved using this formula.
Have a look at the example below:
Solve 2x2 - 5x - 6 = 0
Here a = 2, b = -5 c = -6
Substituting these values in the formula you have
Simultaneous equations with quadratics
If you have two equations to solve, one linear and the other quadratic, you will
usually have two pairs of numbers as solutions.
Pairs of equations like this can be shown on graphs as a line and a curve.
To solve the equations, make one of the variables the subject of the linear
equation and substitute into the other equation.
Here's an example.
Example 1
Solve
Equation 1 : x + y = 2 and
Equation 2: y = x 2 - 4
Using equation 1,
y=2-x
Substitute y = 2 - x into equation 2.
2 - x = x2 - 4
0=x2-4-2+x
This is a quadratic equation.
0 = x2 + x - 6
Now factorise this equation.
(x - 2) (x + 3) = 0
So either x = 2 or x = -3
Substituting these 2 values back into the equation x + y = 2
When x = 2
2+y=2
y=0
When x = -3
-3 + y = 2
y=2+3
y=5
So the solutions are in pairs
x = 2, y = 0 or
x = -3, y = 5
Now try this question below.
Question 1
Solve the simultaneous equations:
Equation 1 : x + 2y = 3 and
Equation 2 :x2 + y2 = 5
The Answer
Did you get the solutions
Well done. If you didn't get this answer, follow the steps below.
Using equation 1, make x the subject.
x = 3 - 2y
Substitute x = 3 - 2y into equation 2:
(3 - 2y)2 + y2 = 5
9 - 12y + 4y2 + y2 = 5
5y2 - 12y + 9 - 5 = 0
5y2 - 12y + 4 = 0
Again, you factorise the equation so that you can solve it in the usual way.
(5y - 2) (y - 2) = 0
Either 5y - 2 = 0 or y - 2 = 0
If 5y - 2 = 0
Substitute y = 2/5 back into the equation x + 2y = 3
x = 3 - 2 × 2/5
Now find the values if y - 2 = 0
If y - 2 = 0
y=2
Substitute y = 2 into the equation x +2y = 3
x = 3 -2 × 2
x = -1
So the values when (x + 3) = 0 are x = -1, y = 2
Solving simultaneous equations using graphs
Solving simultaneous equations using a graph is a lot easier than you might
think. First you need to draw the lines of the equations, and the points where
the lines cross is the solution.
The equations can be linear, which means they will give straight lines.
Have a look at the example below.
Question 1
Solve the simultaneous equations by drawing graphs.
2x + 3y = 6
4x - 6y = - 4
The Answer
You can also have one straight line and a curve.
Question 2
Solve the simultaneous equations by drawing graphs.
y - 2x = 1
y = x2 - 2
The Answer
In an exam you should only use this method if you are prompted to by a
question. It is usually quicker to use algebra if you are not asked to use
graphs.
Perpendicular lines
Remember that perpendicular lines will always cross at right angles.
In this diagram the lines y = 2x - 3 and y = -1/2 x -1 cross at right angles.
The gradients of these lines are 2 and -1/2 and the product of the gradients is 1.
You can work out whether 2 lines are perpendicular by multiplying their
gradients. The product of the gradient of perpendicular lines will always be -1.
In exams you will often be asked to find the equation of a line perpendicular to
a given line. To do this, you will need to work out the gradient of one line
before finding the gradient and equation of the other.
Have a look at the examples below.
Question 1
Find the perpendicular line to 4y - 3x = 8 through the point (0, 2)
The Answer
Re-arrange the equation 4y - 3x = 8 in the form y = mx + c.
y = 3/4x +2
Now write down the gradient of the line 4y - 3x = 8.
The gradient is 3/4.
Now we need to work out the gradient of the 2nd line. Remember that when 2
lines are perpendicular, the product of their gradients is -1. Let's call the
gradient of the second line m.
3/ m = -1
4
m = -4/3
We also know that the line passes through the point (0, 2). This means that
the line crosses the y axis at + 2.
So the equation of the line perpendicular to 4y - 3x = 8 is y = - 4/3x + 2.
Question 2
Line A goes through the points (4, 9 ) and (1, 3). Find the perpendicular line
through the point (0, 1).
The Answer
First, find the gradient of line A. We know that the line passes through the
points (1, 3) and (4, 9). To calculate the gradient, find the difference in the y
co-ordinates and the difference in the x co-ordinates and then divide the y
value by the x value.
9 - 3/
4 - 1 = 2. The gradient of line A is 2.
We know that the product of the gradients of perpendicular lines is -1. If we
call the gradient of line B m then: m × 2 = -1
m = -1/2
Line B passes through the point (2, 0). To find out the value of c in the
equation y = mx + c, we substitute the values x = 2 and y = 0 into the
equation.
0 = -1/2 × 2 + c
0=-1+c
c=1
The equation of line B is therefore y = -1/2x + 1
Cubic graphs
The simplest cubic graph is the graph of y = x3.
Any graph where the highest power is x3 will have a similar shape. In your
exam you will have to plot points to draw the graph accurately.
Question 1
Calculate the points and then plot the graph for the equation y = x3 - x + 8
The Answer
You could be asked to solve equations using a graph.
Question 2
Solve the equation x3 + x2 - 12 using a graph
The Answer
Remember that if the term in x3 is negative the graph will be the other way up.
The graph below shows the equation y = -x3.
Graphs and equations of circles
A circle on a graph will join up all the points at a distance from a given point.
The diagram shows a circle radius 5 and centre at the origin. Any point on the
circle must be a distance 5 from the origin.
An equation for the circle can be found using Pythagoras' Theorem.
The equation for a circle, centre the origin and radius r is x2 + y2 = r2
For the circle below, r = 5, x = 4 and y = 3
so the equation of the circle is x2 + y2 = 52
You can also find the equation of circles when the centre isn't at the origin.
Question 1
Find the equation of the circle with centre at (3, 2) and radius 5.
The Answer
Mark a point P (x, y) on the circle, and draw in a right angled triangle.
The lengths of the sides of the triangle are r (y - 2) and (x - 3).
Remember to use Pythagoras' Theorem, x2 + y2 = r2 to find the equation of the
circle.
(x - 3)2 + (y - 2)2 = 52
Graphs and translations
Have a look at the graph y = x2.
Now, have a look at the graphs of:
y = x2 + 2
y = (x - 3)2
y = (x - 3)2 + 2
You can see that y = (x - 3)2 + 2 is the same shape as y = x2 but it has been
moved 2 up the y axis and 3 along the x axis. This is called a translation and
we show it with the vector:
You can often find a transformation by completing the square. Have a look at
the example below.
Question 1
Complete the square and sketch the graph of y = x2 - 2x + 5
The Answer
y = x2 - 2x + 5
y = (x2 - 2x + 1) - 1 + 5
y = (x - 1)2 + 4
Now from the equation written in this format, we know that the lowest value of
y = 4.
The lowest value will be when the bracket adds up to 0. This occurs when x =
1.
So the lowest point on the curve is at the point (1 , 4). You can then easily
sketch the curve.
Question 2
The graph of y = x3 is drawn in the diagram.
On the same diagram sketch the graph of y = (x + 2)3.
The Answer
Stretches
Compare the graphs y = x2 with y = 3x2.
You can see that the graph of y = 3x2 has been stretched compared to y = x2.
The stretch is parallel to the y axis and has a scale factor of 3. Each point has
moved up by 3.
If the scale factor is 1/3 the graph y = x2 is transformed into y = (1/3)x2. The new
graph looks like this.
Because the scale factor is less than one the stretch flattens out the graph.
You can stretch graphs parallel to the x axis in a very similar way.
Look at these transformations of y = x2 into y = (3x)2 and y = (1/3 x)2
Notice that y = (3x)2 represents a stretch parallel to the x axis of scale factor
1/ .
3
y = (1/3 x)2 represents a stretch parallel to the x axis of scale factor 3.
This will work for any graph. If y = f(x) is any function of x then
y = 3f(x) is a stretch of scale factor 3 parallel to the y axis.
y = f(3x) is a stretch of scale factor 1/3 parallel to the x axis.
Reflections
When you draw the graphs of y = x3 + 5 and y = -( x3 + 5) you see that the
second graph is a reflection of the first in the x axis.
This is true for any function. y = - f(x) is the reflection of y = f(x) in the x axis.
If you compare the graphs of y = x3 + 5 and y = (-x)3 + 5 you see that the
second graph is a reflection of the first in the y axis.
Again there is a general result that for any function.
y = f(- x) is the reflection of y = f(x) in the y axis.
Combine the results in the last three sections for translations, stretches and
rotations and try this question.
The graph of y = f(x) is shown. On the same axes sketch the graph of y = f(x) - 4.
Direct proportion
If two quantities are in direct proportion, it means that as one increases, the
other increases by the same percentage.
If y is directly proportional to x we write y x.
A simple example of two things in the same proportion is the amount of
apples bought and the amount you pay.
If you buy twice as many apples as your friend, you pay twice as much.
We can write the connection between cost and amount as an equation:
Cost of apples = price per apple × number of apples bought.
This can also be written as y = kx where k is the cost e.g. the price per apple.
This means that, for some constant k, y = kx for all values of x and y.
k is called the constant of proportionality.
Question 1
y is directly proportional to x.
y = 3 when x = 12
Find the constant of proportionality and the value of x when y = 8.
The Answer
y x
So y = kx
Substitute the values y = 3 and x = 12
3 = k × 12
So k = 1/4
When y = 8
Substitute y = 8 into y = kx.
8 = (1/4) x
So x = 32
These questions can involve squares, cubes or other powers.
Always form an equation with k, a constant multiplier (the constant of
proportionality) at the start.
Question 2
y x 3.
If y = 1 when x = 2, find the value of y when x = 4.
The Answer
y x3
So y = kx3
Substitute the value y = 1 and x = 2 into y = kx3 to find the value of k.
1 = k × 23
So k = 1/8
Now use the values k = 1/8 and x = 4.
y = 1/8 × 64
y=8
Inverse proportion
Inverse proportion is when one value increase as the other value decreases.
A simple example of inversely proportional quantities is the lengths and widths
of rectangles with the same area. As the length of one side doubles, the width
has to be halved in order for the product to stay the same.
Question 1
y is inversely proportional to x . When y = 3 x = 12.
Find the constant of proportionality and the value of x when y = 8.
The Answer
y 1/x
So xy = k
Substitute the values x = 12 and y = 3 into xy = k.
3 × 12 = 36
So k = 36
To find the value of x when y = 8, substitute k = 36 and y = 8 into xy = k
8x = 36
So x = 4.5
Again you can have questions involving squares, cubes or other powers of the
variables.
Question 2
v is inversely proportional to r3. When r = 2 v = 25.
Find r when v = 60.
The Answer
so
Re-arrange the above to get k on its own.
k = vr3
k = 25 × 23
So k = 200
When v = 60
60r3 = 200
r3 = 200/60
r3 = 3.333
So r = 1.494
Graphs and proportion
If y is proportional to a power of x we can draw a graph that shows the
relationship. The shape of the graph will depend on the power and on the
constant of proportionality.
Here are some graphs that you should be able to recognise.
y x
y = kx and we have a straight line. The gradient of the line is k.
y x2
y = kx2 and we have a curve. The larger the value of k the steeper the graph.
y x3
y = kx 3 and we have a curve. The larger the value of k the steeper the graph.
You have similar shaped curve for any powers between 0 and 1. Again
increasing k will make the graph steeper.
y 1/x
y = k /x
Inverse proportion leads to curved graphs.