Completing the Square

Completing the square is a technique used to take a quadratic equation and put it in a form called the vertex form.  This is an important technique that does not stop just in algebra.  We will revisit this in the calculus post for integration techniques.

The technique is very easy.

Lets look at a quadratic equation in the standard form.

Y=X²+BX+C  

To write this equation in vertex form, we take 1/2 of the B value and write our equation like this:

Y=(X+1/2B)²+k

Both of these equations are equivalent so
We expand (X+1/2B)², we have (X+1/2B)²=X²+BX+1/4B²

We know that (X+1/2B)²-k will expand to: X²+BX+1/4B²+k=X²+BX+C 

So C=1/4B² + k or if we want to find k we do the following:
k=C-1/4B²

Not too bad right?  We are going to now do an example.

Y=X²+6X+8
B=6
1/2B=3

C=8

(X+1/2B)²=(X+3)²

k=8-3²=-1

So our equation becomes:

Y=(X+3)²-1
Vertex form is (X-h)²+k
The vertex is at x=-3 and y=-1 or (-3,-1)

Lets look at:

Y=X²-4X-3

B=-4

1/2B=-2

C=-3

(X+1/2B)²=(X-4)²
k=-3-(-2)²=-3-4=-7

So our Vertex form becomes:

Y=(X-2)²-7
Our vertex is at: (-2,-7)

Note that:   Even though our 1/2B term was negative we still ended up subtracting it from the C term.

Homework:

Factor the following into the vertex form by completing the square:

Y=X²+12X-7
Y=X²-6X+6

Y=X²+11X-1
Y=X²-14X-31
Y=X²-9x+4

The solutions are posted here.

In future posts, we will be looking at completing the square on equations with multipliers on the leading coefficient, a synthetic way of completing the square, and an explanation of why completing the square works. 

The Derivative of x ^x

^{When taking the derivative of} x ^{x it often looks very hard but this is one of the functions you can always understand if you can understand how it is derived.}

^y=x x

^{The trick here is to take the ln of both sides which allows us to use the logarithmic power rule.}

lny=lnx ^x

^{Power rule of logs:}

lny=xlnx

Take the derivative of both sides and on the right side we will use the quotient rule and on the left side we will use implicit differentiation and the chain rule.

y '(1/y)=lnx +x(1/x)=lnx+1

y '=y(lnx+1)

We know y=x ^x  

                          ^so

^{y '=}x ^x(lnx+1)

Pascal's Triangle

1

1 1      1

2 1   2    1

3 1   3    3    1

4 1    4   6    4    1

5 1   5   10   10   5   1

6 1   6  15   20   15   6    1

7  1  7  21  35  35   21     7   1

On the last post we looked at binomial expansion.  It involved calculating combinations as well as carrying large numbers and was not very easy.  With Pascal's triangle we, there is no need to calculate combinations because they are given on the table.

So lets look at the table.  The numbers in red are the order of polynomial ad the number go in front of the variable from left to right.  To show this lets say that we wanted to see what (x+1)⁴:

This equation is 4th order.  So we use: 1       4       6        4         1

This becomes:   (x+1)⁴=1 x⁴   +  4 x³   +   6 x²   +    4x    +     1= x⁴+4x³+x²+4x+1

That's pretty nifty isn't it?

If we wanted (x+y)⁴=x⁴+4x³y+x²y²+4xy³+y⁴

 So you might ask if you need to memorize this table of does this table only go down to 7th order equations.  The answer is no to both questions.  You can construct this table and it goes down to as large a order as you want.

How do we construct it.

Start with                                                                                                  1

Put 1 on outside of the 1                                                                        1    1

Put one on the outside of the 1 and add the middle numbers:             1    2    1      (2=1+1)

Repeat                                                                                                1    3    3   1    (3=1+1)
Repeat                                                                                             1    4     6   4   1   (4=3+1) (6=3+3)

So if you are observing, you will see the number in the triangle is between two numbers and it is the sum of those 2 numbers.

The order of the equation starts at 1  at the two 1's ie                               1   1    1

The numbers descend from 1 from there:                                                2   1   2    1

                                                                                                              3    1    3   3     1

                                                                                                         ...................................

So just to show how easy it is if we had (x+1)⁸=x⁸+8x⁷+28x⁶+56x⁵+70x⁴ +56x³+28x²+8x+1

Homework:  Construct the 8 the through 10 order of pascals triangle.  I will be posting a large table soon so check back. 

                                                                                                            

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Binomial Theory Using Polynomial Expansion

If we expand:
 (x+a)(x+b)=x²+(a+b)x+ab
(x+a)(x+b)(x+c)=x³+(a+b+c)x²+(ab+ac+bc)x+abc
(x+a)(x+b)(x+c)(x+d)=x⁴+(a+b+c+d)x³+(ab+ac+ad+bc+bd+cd)x²+(abc+abd +cd+bcd)x+abcd

Lets say that we want to have 2 numbers or a binomial we would change the expressions above to:
(x+1)², (x+1)³, and (x+1)⁴
They equal:
(x+1)²=x²+(1+1)x+1X1=x²+2x+1
(x+1)³=x³+(1+1+1)x²+(1X1+1X1+1X1)x+1X1X1=x³+3x²+3x=1
When when we expand a binomial, the number preceding each x is just the number of combinations of a,b,c variables preceding our x in our polynomial expansion.

(x+1)⁴=x⁴+4x³+6x²+4x+1
(x+1)ⁿ=xⁿ+[nC(n-1)]xⁿ⁻¹+[nC(n-2)]xⁿ⁻²+[nC(n-3)]xⁿ⁻³...........

Now lets lets see what happens if we use a variable like y instead of 1.
(x+y)²=x²+(y+y)x+y.y=x²+2yx+y²
(x+1)³=x³+(y+y+y)x²+(y.y+y.y+y.y)x+y.y.y=x³+3yx²+3y²x+y³
(x+y)⁴=x⁴+4yx³+6x²y²+4xy³+y⁴

Our expansion becomes
(x+y)ⁿ=xⁿ+[nC(n-1)]xⁿ⁻¹y+[nC(n-2)]xⁿ⁻²y²+[nC(n-3)]xⁿ⁻³y³...........yⁿ

Lets say we have (x+y)⁶=x⁶+6x⁵y+15x⁴y²+20x³y³+15x²y⁴+6xy⁵+y⁶

If we had a number like 3 instead of y, we would substitute 3 where we have y and it would look like this:

(x+3)⁶=x⁶+6x⁵(3)+15x⁴(9)+20x³(27)+15x²(81)+6x(243)+y⁶(729)=x⁶+18x⁵+135x⁴+540x³+ 1215x²+1458x+729y⁶

As you can see there is a lot of calculation involved in expanding binomial series.  There is a way that you can just plug a number in from a table. This table is called Pascal's Triangle.  We will do post on this next.  Many people skip binomial expansion and just use Pascal's Triangle.

Polynomial Expansion 

(x+a)(x+b)=x²+(a+b)x+ab
Is the first and many times the last way we learn to expand polynomial equations.  The equation on the left side of the equal sign is known as the factored form and the equation on the right side is known as the general form.  This is very simple.  Third, forth, fifth, and nth order polynomials are very easy to understand also even though most people are not taught how to do it.

Lets look at the expansion of a third order polynomial:

(x+a)(x+b)(x+c)=x³+(a+b+c)x²+(ab+ac+bc)x+abc

And lets look at a forth order expansion:

(x+a)(x+b)(x+c)(x+d)=x⁴+(a+b+c+d)x³+(ab+ac+ad+bc+bd+cd)x²+(abc+abd+acd+bcd)x+abcd

How do we arrive at these expansion solutions?  Lets multiply through and see.

(x+a)(x+b)(x+c)

Lets multiply (x+a)(x+b)=x²+(a+b)x+ab

Lets multiply  (x+a)(x+b)(x+c)=(x+c)(x+a)(x+b)=(x+c)[x²+(a+b)x+ab]

Multiplying x =x³+(a+b)x²+abx
Multiplying c=cx²+c(a+b)x+abc=cx²+(ca+cb)x+abc=

(x+c)[x²+(a+b)x+ab]=x³+(a+b)x²+abx+cx²+(ca+cb)x+abc
Combine terms:  x³+(a+b)x²+cx²+(ca+cb)x+abx+abc=x³+(a+b+c)x²+(ab+ac+bc)x+abc

This is a little tedious but there is a pattern:

• The first term is going to be an exponent of the order of the polynomial.
• The second term is going to be the addition of all the roots times xⁿ⁻¹
• The last term will be the multiplication of all the roots.  This is consistent among all polynomials.
• The middle terms are just permutations of the roots.

So lets look at (x+a)(x+b)(x+c)(x+d)(x+e)(x+f)=

• First lets look at the second term:  That will be (a+b+c+d+e+f)x⁵
• The last term is: abcdef
• The third term is ab+ac+ad+ae+af+bc+bd+be+bf+cd+ce+cf+de+df+ef
• The forth term is abc+abd+abe+abf+acd+ace+acf+ade+adf+aef+bcd+bce+bcf+bde+bdf+bef+cde+cdf +cef+def
• The fifth term: abcd+abce+abcf+acde+acdf+adef+bcde+bcdf+bdef+cdef
• The sixth term is abcde+abcdf+acdef

So (x+a)(x+b)(x+c)(x+d)(x+e)(x+f)=x⁶+(a+b+c+d+e+f)x⁵+(ab+ac+ad+ae+af+bc+bd+be+bf+cd+ce+cf+de+df+ef)x⁴+(abc+abd+abe+abf+acd+ace+acf+ade+adf+aef+bcd+bce+bcf+bde+bdf+bef+cde+cdf

+cef+def )x³+(abcd+abce+abcf+acde+acdf+adef+bcde+bcdf+bdef+cdef)x²+(abcde+abcdf+acdef)x + abcdef

That is pretty long.  The main idea is that is the second term and the last term are always very easy to ascertain.  While we have formulas for the quadratic equation, we do not have similar equations for finding roots of higher order polynomials but there are many things we can use to solve this equation like multiple similar roots, even and odd roots as well as other roots.   We will look at those in the subsequent posts; We will also look at expansion of the equations with the leading or first terms being numbers other than 1;   We will also use these equations to explain the binomial expansion theory; We will also look at algorithms for solving and expanding polynomials.

Lots of good things to follow.

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