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Practical uses of completing the square
By Anonymous on Sunday, October 10, 1999 - 12:40 pm:

Can someone please give me a practical use of completing the square?!

By Chris Jefferson (Caj30) on Sunday, October 10, 1999 - 01:57 pm:

Hmm.. I assume you mean writing a2x2 + bx + c

As (ax + (b/2a))2 - (b/2a)2 + c

Well... you can look at c - (b/2a)2. As the first term is always positive, then if this is less than zero, you know no roots exist. If it is zero, the roots are the same and if it is more than one, then two roots exist (a root being where the equation is zero for a certain value of x)

Although you probably haven't done it yet, it can also be useful in very complicated integrals, by using it you can integrate ea quadratic equation quite easily. Without this, it is VERY hard...

Hmm.. It can make the graph easier to draw and finally, who says maths has to be useful... If we are honest, most of it is just there for the fun of it!!!

Bye for now, Chris. If anyone has any other uses, feel free to add them!

By Anonymous on Sunday, October 10, 1999 - 03:30 pm:

How do you integrate equadratic expression?

By Kerwin Hui (P1312) on Monday, October 25, 1999 - 08:58 pm:

This is how I would do it:

First, let the quadratic expression be ax2+bx+c.

Then, completing the square to give the exponent in the form au2+d, where u=x+(b/2a).

If a is negative, the answer would be an error function, in fact, I found that it is

edErf([x+b/2a]sqrt[|a|])/sqrt(|a|)+constant

but if a is positive, hm... I don't know, perhaps expressing the integrand as power series and integrate.

By Dan Goodman (Dfmg2) on Friday, November 19, 1999 - 10:04 pm:

Another integral that you can use completing the square for is ò (quadratic)-1dx or ò (quadratic)-0.5 which turn out to be things like sin-1 and tanh-1. This is something that is usually done at A Level. Another nice use for it is that you can use it to find the general solution to ax2+bx+c=0 by writing the equation as (x-d)^2-e=0 which gives the solutions x=d±sqrt(e) which you can expand as x=(-b±sqrt(b2-4ac))/2a which you will probably have seen before.