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AR15.COM
12/6/2009 8:45:47 PM EDT
I'm working on some math homework and just need some help with the answers here.


It calls for using the completing the square method for solving these quadratic equations.
x^2-2x=24


next one

x^2+8x-12=0



next one


x^2+5x=4


next

x(x-3)=10

last one

4x^2=8x+5

12/6/2009 8:46:35 PM EDT
[#1]
Pie is the answer.... in more ways then one

12/6/2009 8:54:46 PM EDT
[#2]
If you're looking for the solutions use WolframAlpha (copy the question into the box and click the compute button).
12/6/2009 8:56:02 PM EDT
[#3]
87

12/6/2009 8:57:01 PM EDT
[#4]
all i know is 2+2= 87
12/6/2009 8:57:07 PM EDT
[#5]
Let me Google that for you



12/6/2009 9:01:58 PM EDT
[#6]
Just FOIL that shit.

x^2-2x=24

set it to zero x^2-2x-24=0

(x-6)(X+4)=0

-6,4 are your possible answers.

plug them back in to check the answer.
12/6/2009 9:04:26 PM EDT
[#7]
Quoted:
I'm working on some math homework and just need some help with the answers here.


It calls for using the completing the square method for solving these quadratic equations.
x^2-2x=24
x^2-2x-24=0
(x+4)(x-6)=0
x=-4,x=6
next one

x^2+8x-12=0



next one


x^2+5x=4
x^2-5x-4=0

next

x(x-3)=10
x^2-3x=10
x^2-3x-10=0
(x-5)(x+3)=0
x=5.x=-3

last one

4x^2=8x+5



There are  the easy ones. I haven't done any kind of these shenanigans for a while.
Basically, you find the multiple of the digit of the number without an x. Then, you add or subtract them to match the x number, not the the x^2. Once you have that figured out, you lay it out like so, (x+y)(x-z). Make each set of numbers in the parenthesis equal zero and solve for x.
12/6/2009 9:06:36 PM EDT
[#8]
http://www.purplemath.com/modules/sqrquad.htm



This explains the process pretty well.