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anonymous
 4 years ago
how do i solve x2 – 2x = 8 by completing the square?
anonymous
 4 years ago
how do i solve x2 – 2x = 8 by completing the square?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(x1)^2=8+1=9\] \[x1=\pm3\] \[x1=3\implies x=4\] \[x1=3\implies x=2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i need to show the work, can you explain how you got this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Could you @satelite try to answer my question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We generally don't complete the square to solve for a variable. We usually use it to place a quadratic equation into vertex form in order to graph or solve a word problem. \[x ^{2}2x=8\] Bring everything to one side \[x ^{2}2x8=0\] Now we place brackets around the first two terms: \[(x ^{2}2x)8=0\] Now we need to create a nice perfect square in the brackets. We do this by taking half of x's coefficient and squaring it. In this case (2/2)^2 = 1 We then add and subtract this number, thereby not "really" changing the equation. \[(x ^{2}2x+11)8=0\] We bring the subtracted term out of the brackets: \[(x ^{2}2x+1)18=0\] Then we factor the bracket: \[(x1)^{2}18=0\] Simplify: \[(x1)^{2}9=0\] That would be vertex form (if the 0 was a y) So if we want to solve: \[(x1)^{2}=9\] Square root both sides, this actually gives us a plus and minus version: \[x1= +/3\] so we have two solutions: \[x=2\] and \[x=4\]
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