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anonymous

  • 4 years ago

how do i solve x2 – 2x = 8 by completing the square?

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  1. anonymous
    • 4 years ago
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    \[(x-1)^2=8+1=9\] \[x-1=\pm3\] \[x-1=3\implies x=4\] \[x-1=-3\implies x=-2\]

  2. anonymous
    • 4 years ago
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    i need to show the work, can you explain how you got this?

  3. anonymous
    • 4 years ago
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    Could you @satelite try to answer my question

  4. anonymous
    • 4 years ago
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    Ummm

  5. anonymous
    • 4 years ago
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    We 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 \[x ^{2}-2x-8=0\] 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. \[(x ^{2}-2x+1-1)-8=0\] We bring the subtracted term out of the brackets: \[(x ^{2}-2x+1)-1-8=0\] Then we factor the bracket: \[(x-1)^{2}-1-8=0\] Simplify: \[(x-1)^{2}-9=0\] That would be vertex form (if the 0 was a y) So if we want to solve: \[(x-1)^{2}=9\] Square root both sides, this actually gives us a plus and minus version: \[x-1= +/-3\] so we have two solutions: \[x=-2\] and \[x=4\]

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