Posted by Micha on Monday, October 1, 2007 at 12:35am.
How do I work this correctly? I know the answer is 7 but I am doing something wrong.Here is what I have...
x+ã(2x5)=10
x+ã(2x5)^2=10^2
x+2x5=100
3x5+5=100+5
3x=105
x=35
PLEASE HELP!

Algebra  Nick, Monday, October 1, 2007 at 12:55am
what is the a~ all about? If it is a square root symbol then the solution looks like this:
x + (2x5)^(1/2) = 10 : move x to other side
(2x5)^(1/2) = 10  x : square both sides
2x5 = (10x)^2 : expand the squared term
2x5 = 10010x10x+x^2 : combine terms
0 = 105  22x + x^2 : use quadratic formula to solve for x
x =[(22) +/ sqrt(22^2 4*1*105)]/(2*1)
x= [22 +/ sqrt(484  420)]/2
x= [22 +/ 8] / 2
x= 7 or 15
Plugging back into our initial equation shows that 7 is a solution, but 15 is not. This is because we started with a square root, which is invalid for negative numbers.
Regards.
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