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help2013
 one year ago
Best ResponseYou've already chosen the best response.0can u show me how plzzzzz! an dthanks

bii17
 one year ago
Best ResponseYou've already chosen the best response.0\[(x4)^2 = x+8\] \[x^2 8x +16 = x+8\] combine similar terms it will become \[x^2 9x+8 =0\] factor (x+9) (x1) = 0 to check, substitute x=9 or x=1

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.1\(x4 = \sqrt{x+8}\) First, before ANYTHING else, think about the Domain. \(x+8\ge 0\;or\;x \ge 8\) You tell me why? \(x4\ge 0\;or\;x \ge 4\) You tell me why? After that, if we get ANY result for x that is less than 4, we will simply discard it as extraneous. It should not even be considered as a solution. x = 1 is not correct. x = 9 is not correct.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.1Seriously. Go try them in the original equation. They fail miserably.

help2013
 one year ago
Best ResponseYou've already chosen the best response.0well then is the answer zero

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.1No, x = 0 is also incorrect. The only suspected results are x = 1 and x = 8. (The factoring shown above was incorrect.) x = 1 is immediately discarded as not in the Domain. x = 8 is a solution. x = 9 could not have been a solutions, as 9 < 1. x = 0 could not be a solution, as 0 < 1 If you rule out ALL your suspects, there is NO solution. In this case x = 8 is the winner.

help2013
 one year ago
Best ResponseYou've already chosen the best response.0ohh well thats not one of my answer choices?:/

help2013
 one year ago
Best ResponseYou've already chosen the best response.0opps nv thanks for the help
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