SOLUTION: Hi I need help in understanding one of the steps in "completing the square" for a quadratic equation.Here is my problem x^2+9x+8=0 IN the 2ND step my homework tells me that I

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hi I need help in understanding one of the steps in "completing the square" for a quadratic equation.Here is my problem x^2+9x+8=0 IN the 2ND step my homework tells me that I      Log On


   

Question 141840: Hi I need help in understanding one of the steps in "completing the square" for a quadratic equation.Here is my problem
x^2+9x+8=0
IN the 2ND step my homework tells me that I have to multiply the equation by A, whatever value it has.Now everytime they show a model there is a different value for "A" everytime.How do I find the value of "A" for my own homework equations.Here is one of my models from my school...


Solve x^2 + 7x + 12 = 0 by completing the square.
1. Write the quadratic equation in general for and identify A and B.
A is 1 and B is 7.
2. Multiply the terms of the equation by 4A.

4A is 4: 4[x^2 + 7x + 12]= 4[0]
4x^2 + 28x + 48 = 0
3.Isolate the constant term on the right side of the equation.
4x^2 + 28x = -48
4.Add B^2 to each side of the equation.
B^2 is 49: [4x^2 + 28x] +49 = [-48]+49
4x^2 + 28x + 49 = 1
5.Factor the left side of the equation to the square of a binomial.
(2x + 7)(2x + 7) = 1
(2x + 7)^2 = 1
Now solve:
1^2 is 1, so (-1)^2 is 1, so
2x + 7 = 1 2x + 7 = -1
2x = -6 2x = -8
x = -3 x = -4


The solution set is {-4, -3}.
What I dont understand is step 2 how did they decide to multiply the equation by 4? Why dont they use any other number like 2 or 6?

I hope I gave you enough information.Thanks for your help...
Found 2 solutions by nabla, solver91311:
Answer by nabla(475) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+9x+8=0
To complete the square, we need the digit in front of the x^2 term to be 1. All you have to do to complete the square is take half of the number in front of the x term, then square it and add/subtract it:
x%5E2%2B9x%2B9%5E2%2F%282%5E2%29-9%5E2%2F%282%5E2%29%2B8=0
x%5E2%2B9x%2B81%2F4-81%2F4%2B32%2F4=0
x%5E2%2B9x%2B81%2F4-49%2F4=0
x%5E2%2B9x%2B81%2F4=49%2F4
Now, the left hand side can be factored:
%28x%2B9%2F2%29%5E2=49%2F4
x%2B9%2F2=0+%2B-+sqrt%2849%2F4%29
x=-9%2F2+%2B-+sqrt%2849%2F4%29
x=-9%2F2+%2B-+7%2F2
Which gives x=-1 or x=-8.

It is important to note that the factoring of the square will always be half of the original term in front of the x. E-mail me at enabla@gmail.com if you still have difficulty or would like it explained differently.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Well, the obvious answer is that 4 is a perfect square, while 2 and 6 are not. Having said that, the process you showed is not the one that I use.

Here's my process given ax%5E2%2Bbx%2Bc=0

Move the constant term to the right:
ax%5E2%2Bbx=-c

Divide both sides by the lead coefficient (if it is already 1, you don't have to do anything)
x%5E2%2B%28b%2Fa%29x=%28%28-c%29%2Fa%29

Divide the coefficient on the 1st degree term by 2, and then square the result. Add that result to both sides of the equation.
x%5E2%2B%28b%2Fa%29x%2B%28b%5E2%2F4a%5E2%29=%28b%5E2%2F4a%5E2%29%2B%28%28-c%29%2Fa%29

Factor the left now that it is a perfect square.
%28x%2B%28b%2F2a%29%29%5E2=%28b%5E2%2F4a%5E2%29%2B%28%28-c%29%2Fa%29

Take the square root of both sides.
x%2B%28b%2F2a%29=sqrt%28%28b%5E2%2F4a%5E2%29%2B%28%28-c%29%2Fa%29%29 or x%2B%28b%2F2a%29=-sqrt%28%28b%5E2%2F4a%5E2%29%2B%28%28-c%29%2Fa%29%29

Add -b%2F2a to both sides.
x=-b%2F2a%2B-sqrt%28%28b%5E2%2F4a%5E2%29%2B%28%28-c%29%2Fa%29%29

Simplify.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

Hope that helps.