could someone explain completeing the square and taking the derivative.?

Answer 1

I don't know how completing the square and taking the derivative have anything to do with each other, but anyways.

Completing the square is a strategy used to solve second degree quadratic equations and conic sections. It does exactly what its name says, it "completes" the statement into a perfect square, so you can take the square root of it and get an answer:

For example, x^2-3x-7=0
This is not a perfect square, so we complete the square,
X^2-3x-7
now, we need to find something to add to both sides that will make x^2-3x + something become a perfect square
To do that, recall that (a+b)^2 = a^2+2ab+b^2, meaning that the "b" term will have to be half of the middle term, squared
The middle term is 3, halving produces 3/2, squaring produces (3/2)^2

We need to add this to both sides
x^2-3x+(3/2)^2 = 7+9/4
(x-3/2)^2=37/4
x-3/2 = +/- root(37)/2
x=3 +/- root37/2


The derivative is a completely different thing. It is the slope of the tangent line to a function at a given point

The most common rule for differentiation (taking the derivative) is d/dx x^n = nx^(n-1). Of course, d/dx c where c is a constant = 0, and d/dx x = 1

Answer 2

completing the square is going to follow some pretty standard forms
ax^2 + bx + c=0 is usually what youll see

usually when completing the square a will be 1 so you wont have to worry about it

when b is positive c is (b^2)/4

you can also think of it as you need to find two numbers that multiply to c and add up to b

i dont know exactly what you want when you say taking the derivative because it has many many forms. The most basic is the power rule and it allows you to take the derivative of most basic functions such as x^2 or 5x^6 etc.

the derivative of a function can be found by multiplying the function by the exponent of x and then reducing the exponent of x by 1

for example:
x^3 -> 3x^2
4x^5 -> 20x^4
4x^1 -> 4 (x^0=1)

if you are doing more advanced derivatives than this make another post

Answer 3

Completing the square:

Suppose you have 4*x^2 + 6*x +1
You want to make it into a square.
Try to put it in the form:
y^2 +2*y*p + p^2 .
Take y=2*x . Then 4*x^2 + 6*x +1 = (2*x)^2 + 2* x*3 + 3^2 -3^2 +1 = (2*x)^2 + 2*x*3 + 3^2 -9+1
= (2*x + 3)^2 -8

So you have to add 8 on the left hand side to complete the square .
4*x^2 + 6*x +1 +8 = (2*x + 3)^2

Or put the equation 4*x^2 + 6*x +1 =0 in the form
4*x^2 + 6*x +9 = 8
or, (2*x + 3)^2 = (sqrt(8) )^2

2*x +3 = + sqrt(8) or -sqrt(8) ,
x= -3 + sqrt(8)
or
x= -3 - sqrt(8) ,
as solutions.


Link: http://www.purplemath.com/modules/sqrquad.htm
http://www.sosmath.com/algebra/quadraticeq/complsquare/complsquare.html


B. Taking derivative.
It is actually too long to explain here. Take a look at tghe links:

1.http://www.math.montana.edu/frankw/ccp/calculus/deriv/derlimit/learn.htm
2.http://www.nipissingu.ca/calculus/tutorials/derivatives.html

Cheers.

Answer 4

Jack,

Please see the previous question I answered. In response to your comments I put an addendum at the bottom about solving it without calculus. It does use completing the square.

Completing the square and taking a derivative really are not the same thing at all, but in the case of this particular problem you can solve it by completing the square instead of taking a derivative. That won't generally be true however.