Download The suare of a binomial

The square of a binomial
Benjamin Baumslag gives a brief revision lesson for more experienced pupils which shows the
value of one of the earliest results learnt in Algebra
By the square of a number we mean the product of the number by itself. For instance, 3 squared
is 33 or 9. We remind the reader that we express this by writing 2 as a superscript, so we write
32 to denote the square of 3. If x denotes an arbitrary number, then x.2 denotes the square of x.
Geometrically the interpretation of x.2 is that of the area of a square of side x. .
The square of 2 is 4, the square of 3 is 9. What is the square of 2 + 3 = 5? A guess might be the
square of 2 plus the square of 3, but this gives a total of 13, and the square of 5 is in fact 25. If a
and b represent any two numbers, then the correct formula is
(a + b)2 = a2 + b2 + 2ab
(1)
In other words, if we take the square of a sum, we add not only the squares of the original
numbers, we also add twice the product of the numbers. Thus (2 + 3)2 = 22 + 32 + 2(23) = 25,
which we know is the correct result.
Equation (1) is a rather dull looking result, one of the first one comes across in school algebra.
But it turns out to be a very valuable result. Below we give five interesting consequences.
We first prove Equation (1) with a geometrical proof. If you prefer there is an alternative
algebraic proof below.
Figure 1 Proof of Equation (1)
To prove the result geometrically consider (see Figure 1) a square of side a + b. Inside this
square we draw in the top left-hand corner a square of side a, which we denote by the symbol Sa.
In the bottom right hand-corner we draw a square Sb of side b. Two rectangles, each having sides
a and b ,fill the rest of the square of side a + b. Thus the area of the square of side a + b is
(a + b)2 and is equal to the area a2 of Sa together with the area b2 of Sb together with the area of
the two rectangles, each of area ab. Hence Equation (1).
The algebraic proof is based on the distributive law. This says that if you have the product of
two sums, multiply each of the numbers one from the first sum and one from the second, and add.
Thus for instance,
(3 + 4) (5 + 6) = 35 + 3 6 + 4 5 + 46.
In general, if a, b, c and d represent any four numbers,
(a + b)(c + d) = ac + bc + ad + bd.
(2)
Note that here as is customary we have left out the  sign from for example ac and simply
written ac.
If we take Equation (2) and put c = a and d = b, we get
(a + b)2 = aa + ab + ba + bb. In other words, we get Equation (1).
Five applications of Equation (1)
1. Calculating squares
We can use Equation (1) to calculate a table of squares by taking the case b = 1.
The equation then becomes
(a + 1) 2 = a2 + 2a + 1. If we have calculated a2then we can calculate (a + 1)2
by simply adding 2a + 1 to a2. This is considerably easier than calculating the square all over
again from scratch.
The idea is we calculate first the square of 1, then the square of 2, then the square of 3 etc. to
make our table of squares.
To go from a2 to (a + 1)2 we simply add 2a + 1.
However, what is the importance of calculating a table of squares? The Babylonians did so, and
there are tables of squares one of which goes up to 59 squared, which are preserved on their clay
tablets. We are talking about 4,000 years ago.
In our system of mathematics we use a base of 10, so that a number like 345 means 3
hundreds, 4 tens and 5 units. The Babylonians used a base of 60. How do we know this?
From one of their tables of squares.
This table lists the squares, and for the square of 7 we get 49 marks. The next square which
must be 8 squared is represented by a single mark in the first position and 4 marks in the
second.. In other words, they used a base of 60.
Why did the ancient Babylonians need tables of squares? The answer is they used them to
multiply.
2. The Babylonian’s method of multiplying
To multiply two numbers the Babylonians added the numbers and took their difference, squaring
these two numbers and then subtracting them. By dividing by 4 they calculated the product.
(They had other methods as well.)
For instance, 7 times 3 is calculated by taking 10 the sum of 7 and 3, and 4 their difference.
Squaring these two and subtracting gives 100 – 16 = 84, and if we divide by 4 we get 21, the
product of 7 and 3.
Compared with the method we learnt at school, this seems very long winded. But the
Babylonians did not have the method we have learnt at school. And since they had a system
based on 60, they would have needed to know off by hearts not only the produces up to 9 times 9,
but would need the products up to 59 times 59. Their system with their tables of squares
therefore worked very well.
To see how the Babylonian method works, consider
(a + b)2 - (a - b)2 .. Equation (1) tells us how to write (a + b)2
The second square (a - b)2 comes from the same equation, but now 2ab is replaced by – 2ab.
Since we are subtracting one from the other, the terms a2 and b2 cancel, and the only ones left
are 2ab from the first bracket squared, minus the – 2ab from the second. The total then becomes
4ab, as required.
Thus we have the equation
(a + b)2 - (a - b)2 .= 4ab………..
(3)
3. Pythagorean triples
4.
You no doubt remember that Pythagoras’s theorem concerns a right-angled triangle. If the sides
in order of magnitude are x, y and z then Pythagoras’s theorem says that
x2+ y 2 = z2.
The Babylonians were well aware of Pythagoras’s theorem something like 1,000 years before
Pythagoras. They also had a method of finding whole numbers x, y and z satisfying the same
equation as Pythagoras theorem, namely
x2 + y2 = z2. ….
An easy example is 3, 4, 5, since 32 + 42 = 52. This fact is still used nowadays to mark out a
right-angle when building.
Whole numbers x, y, z which satisfy the above equation are called Pythagorean triples.
One example of that the Babylonians knew was
46012 + 48002 = 66492.
We do not know how they managed to find such a triple.
However we have a good method to find such triples. We take any two differing whole numbers
m and n.
Putting
x = 2mn, y = m2 - n2, and z = m2 + n2 then
x2 + y2 = z2 .
An example will make this clearer.
Example:
Take m = 2 and n = 1. Then
x = 2mn = 4, y = m2 - n2 = 4 – 1 = 3, and z = m2 + n2 = 5.
So with this choice of m and n we get the easiest example which we have already mentioned of a
Pythagorean Triple.
As another example, take m = 7 and n = 5, to get
x = 70, y = 24 and z = 74
To see why this method gives us Pythagorean triples, consider
Equation (3)
(a + b)2 - (a - b)2 .= 4ab……….
And with a = m2 and b = n2 .
we see that the right-hand side of the above equation becomes 4m2 n2 or (2mn) 2 .
Put x =(2mn) so that
x2 = 4ab. Put
z = (a + b) = m2 + n2 and y = a – b = m2 – n2 , then
z2 ..- y2 = x2
which means that
x2 + y2 = z2
5. Newton’s Cradle
Newton’s cradle consists of a number of identical round steel balls hanging from two parallel
beams. Each ball is supported by two strings, one on each side, the one supported by the one
beam, the other supported by the other beam. This ensures that each ball when it strikes its
neighbour, does so with a direct blow, and not a glancing one. . Five balls is quite a good
number to have. At rest, each ball touches its neighbour.When the left-most ball is pulled to
the side and allowed to swing on its own, it will swing until it knocks into the adjacent ball.
Then there is a surprise. The first ball stops instantly, and all the other balls with the
exception of the last one, stay still. The last ball swings to the right with the same speed as
the initial ball hits the stationary ball. Intuitively this seems wrong.
To see why this is the case we need to use two principles from Physics. The first is that the
energy of the system after the balls have collided is the same as the energy before the
collision. To simplify the discussion we will discuss only the case with two balls. Let the
left hand ball strike the stationary ball with a speed of u. After the collision let it have a speed
of v, and let the other ball have a speed of w. We assume both balls are the same, and hence
have the same mass say m. Then the energy according to Physics is the mass m of the ball
moving at speed u, is ½mu2 The energy of the system after the collision is
½mv2 + ½mw2..
Since the energy is conserved,
½mu2 = ½mv2 + ½mw2.
Dividing by ½m we find that
u2 = v2 + w2…..
(4)
The second principle is that the momentum after the collision is equal to the momentum
before the collision.
The momentum is defined to be the product of mass and the speed.
Before the collision, the momentum is mu. After the collision it is mv + mw. Thus
mu = mv + mw. Or dividing by m,
u =v +w
Hence u2 = v2 + w2+ 2vw. as a consequence of Equation (1).
But from Equation (4)
2
u = v2 + w2 and so 2vw = 0. So either v or w is 0, and if w is 0, then since
u = v + w, we see that u = v, and so that v is positive. This means the first ball is moving to the
right with a positive speed while being impeded by a ball which is moving at a speed of 0, i.e. not
moving at all. This is a contradiction and so v must be 0 as claimed.
- 5. Solving quadratic equations
As our final example of what Equation (1) can be used for, we will show how it can be used
to solve quadratic equations. This is a method which always appears in the school books, and
we shall illustrate the method with an example. A quadratic equation in the unknown x is a
sum of an x2 term, an x term and a constant, e.g.
3x2 + 6x - 27 = 0.
3 is called the coefficient of x2 and 6 is called the coefficient of x, while -27 is called the
constant of the equation.
The first step is to convert this equation into one which has 1 as the coefficient of x2. To do
this we divide by 3, the coefficient of x2, to get
x2 + 2x – 9 = 0
(5)
The next step is where we use Equation (1) above. We divide the coefficient of x by 2. In
this case we get 1. By Equation (1) above,
(x + 1)2 = x2 + 2x + 1. So x2 + 2x = (x + 1)2 – 1. This means that
x2 + 2x – 9 = (x + 1)2 – 1 - 9= 0..
As a consequence,
(x + 1)2 = 10, so that x + 1 is the square root of 10, that is +10.
Thus x = - 1 + 10, and we have the answer.
If we use the method above and apply it to a general quadratic equation we get the standard
formula for the solution of quadratic equations which appears in the school books,
As an afterthought, it is interesting to note that the Babylonians could also solve quadratic
equations. In our discussion we have made use of Algebra, which was only developed in
about 600 AD. Some of the ancient Babylonians really must have been genii to handle these
problems without the benefit of Algebra.
Acknowlegement:I thank Natalie Kehr for improving the article.