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help2013Best ResponseYou've already chosen the best response.0
can u show me how plzzzzz! an dthanks
 5 months ago

bii17Best 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
 5 months ago

tkhunnyBest 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.
 5 months ago

tkhunnyBest ResponseYou've already chosen the best response.1
Seriously. Go try them in the original equation. They fail miserably.
 5 months ago

help2013Best ResponseYou've already chosen the best response.0
well then is the answer zero
 5 months ago

tkhunnyBest ResponseYou've already chosen the best response.1
No, 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.
 5 months ago

help2013Best ResponseYou've already chosen the best response.0
ohh well thats not one of my answer choices?:/
 5 months ago

help2013Best ResponseYou've already chosen the best response.0
opps nv thanks for the help
 5 months ago
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