Quad. Equations Questions?

Question

Look here at work..shown
EXAMPLE A
x²+2/3x-3=0
(x+1/3)²-1/9=3
(x+1/3)²=28/9
x+1/3=2sqrt7/3
ect....

1) In this equation we don't times "2" in the denomenator of 1/3 in (x+1/3)² why???

EXAMPLE B
x²+1/4x-1/2=0
x²+1/4x=1/2
(x+1/8)²-1/64=1/2
ect...

2)In example B, we times "2" and squared it to get -1/64 why???

3)Why are example A and example B different, with *2 and squaring it and without it?

Answer 1

in part a we took the mid term 2/3x to mean 2 times 1/3 times x. so the one third is timesed by 2 in the form of 2/3.

in part b the 1/4x is really 2 times 1/8 times x and the 2 & 8 were reduced to 1/4x. a simple trick, look at the mid term as 2/8 x rather than 1/4x..so the 1/8 squared gives 1/64 to complete the square.

Answer 2

For the example 1,
You have something squared = something else. You are taking square root of both sides. If you take sqrt(28/9), what do you get?

sqrt(28/9) = sqrt28/sqrt9 = 2*sqrt7 / 3 correct? It is NOT 2 times per say. sqrt28 becomes sqrt(2*2*7). Since sqrt(2*2) = 2, it comes outside.

For this, you are using a method called 'completing the square'

For example 2, try not to skip so many steps.

x²+1/4x-1/2=0
x²+1/4x= 1/2
x²+1/4x + 1/64 = 1/2 + 1/64
(x+1/8)^2 = 32/64 + 1/64
(x+1/8)^2 = 33/64

I don't know what you mean by times 2 and squared, but you take half of the second term and square it to get the third term. It's exactly the same thing you did for the first example. It is the basis of completing the square method. If you didn't do this, it will never be (a+b)^2 form that is required at the end.

Answer 3

To complete the square, you take half of the bx term, square it, and add it to both sides. In example A, one half of 2/3 is 2/6 which reduces to 1/3 before it is squared.

In example B, one half of 1/4 is 1/8. 1/8 does not reduce, so it is then squared to get 1/64