Skip to 0 minutes and 12 secondsSo now we come to a fundamental algebraic manipulation that goes back to an Arab mathematician, Al-Khwarizmi, who lived around 780 to 850 AD. So he actually proposed a method of solving this particular equation, the equation x squared plus 10x equals 39. Faced with such an equation, how are you going to find x? Here's how he reasoned. He said, OK, we don't know what x is. Suppose that we make a square of side x. That represents x squared.

Skip to 0 minutes and 52 secondsSo now let's take the 10x and divide it into two rectangles. One rectangle will be x this way and 5 this way, and the other rectangle we'll put over here. It will be 5 this way and x this way. So the area here is 5x. The area here is 5x. So total, we have x squared plus 10x. And on the other side, we have 39. So now what he said is what we need to do is complete the square. There's a little chunk missing in here. And we can fix that by putting in a little extra area. How big must that area be?

Skip to 1 minute and 43 secondsWell, this is 5 this way and 5 this way, so this is a little 5 by 5 square that we're adding here. So we're adding 25 to the left hand side. So to keep the equality we ought to add 25 to the right hand side as well.

Skip to 2 minutes and 2 secondsAnd what is this now telling us? It's telling us, well, here we have, in fact, a square. It's a square of side x plus 5. So its area has to be x plus 5 squared. And on the right hand side, well, we get 39 plus 25, which is 64.

Skip to 2 minutes and 22 secondsSo the equation has been transformed into a simpler equation, which is just something squared equals something else. There's no intermediate term like the 10x. We've gotten rid of that. Now we can solve this. Because if the something squared equals 64, we can conclude that x plus 5 equals 8. Well, actually, there's also a negative value. So the other possibility is that x plus 5 equals minus 8. So therefore, we can conclude that x equals either 8 minus 5, which is 3, or x equals minus 8 minus 5, which is minus 13.

Skip to 3 minutes and 9 secondsThat's how Al-Khwarizmi completed the square. So this is a very useful technique which is really behind the standard quadratic formula. And from our point of view, it allows us to take a quadratic expression of form y equals a x squared plus bx plus c and to rewrite it in the form a times, say, x minus h, squared, plus k. When we've done that, then we've converted this general parabola into a form that we know precisely what it looks like, because it's exactly the same kind of thing that we talked about in the previous slide-- taking a x squared, shifting it over by h in the x direction, and then up by k in the y direction.

Skip to 4 minutes and 0 secondsSo let's have a look at an example. Here is a quadratic function given by the equation y equals x squared minus 6x plus 5. So what one can do is apply Al-Khwarizmi's completing the square to rewrite this as x minus 3 squared minus 4. It's basically the completing the square operation. This form of the parabola tells us immediately where it's situated. It has a vertex when x equals 3 and y equals minus 4. There it is right there. Now, there's something else that we can do-- another alternate way of thinking about rewriting this equation, which is also very familiar to you, I hope, which is the idea of factoring it.

Skip to 4 minutes and 49 secondsIt's a quadratic expression, so it factors into factors x minus 1 and x minus 5. And Descartes-- the same Descartes who gave us the Cartesian coordinate system-- had a very important realisation. That in order to find these factors, it's basically, geometrically we're talking about the places where the function takes on the value 0-- in other words, where it crosses the x-axis. So it's the analogue of the x-intercepts for a line. So these are the zeros of this function. So Descartes realised that these two points, 1 and 5, were the places where the function was equal to 0 and that they corresponded to factors of the actual expression.

Skip to 5 minutes and 42 secondsSo once we've got this value x equals 1, where this expression is equal to 0, we know there must be a factor x minus 1, and same for 5. In fact, that 5 gives us a 0 for the actual function. It tells us that x minus 5 is a factor. So in summary, we have basically three different but all useful ways of thinking about the same quadratic expression-- one simply as a polynomial, two in completed the square form, and three in factored form à la Descartes. Each of them gives us useful information about the shape, the position of this quadratic function.

# Zeroes, completing the square, and the quadratic formula II

Completing the square is an important algebraic procedure that was discovered by the Persian mathematician Muhammad al Khwarizmi (780-850 AD) that is the foundation for the quadratic formula. This technique really has a strong geometrical flavour.

## Completing the square

Al Khwarizmi’s completing the square operation has a strong geometrical flavour. Suppose we consider an expression of the form \(\normalsize{x^2+2bx}\) for some number \(\normalsize{b}\). This quadratic expression can be interpreted visually as the sum of a square of side \(\normalsize{x}\), and two rectangles of sides \(\normalsize{b}\) and \(\normalsize{x}\).

By adding the small remaining square of area \(\normalsize{b^2}\), we get a perfect square.

So we have rearranged the original expression, and added something to it, to make a perfect square. Algebraically we would write this as

\[\Large{x^2+2bx+b^2=(x+b)^2}.\]This is the algebraic essence behind the quadratic formula. In general, the rule is that to complete the square, we add the square of one-half of the coefficient of the linear term in \(\normalsize{x}\).

Q1(E): How much would we have to add to these expressions to complete the square i) \(\normalsize{x^2+8x}\) ii) \(\normalsize{x^2-22x}\) iii) \(\normalsize{x^2+5x}\) iv) \(\normalsize{x^2-17x}\)? What would we get in each case after having done that?

## Answer

A1.i) To \(\normalsize{x^2+8x}\) we would add \(\normalsize{4^2=16}\) to get \(\normalsize{x^2+8x+16=(x+4)^2}\).

ii) To \(\normalsize{x^2-22x}\) we would add \(\normalsize{11^2=121}\) to get \(\normalsize{x^2-22x+121=(x-11)^2}\).

iii) To \(\normalsize{x^2+5x}\) we would add \(\normalsize{(\frac{5}{2})^2=\frac{25}{4}}\) to get \(\normalsize{x^2+5x+\frac{25}{4}=(x+\frac{5}{2})^2}\).

iv) To \(\normalsize{x^2-17x}\) we would add \(\normalsize{(\frac{-17}{2})^2=\frac{289}{4}}\) to get \(\normalsize{x^2-17x+\frac{289}{4}=(x-\frac{17}{2})^2}\).

© UNSW Australia 2016