anonymous
  • anonymous
Solve by completing the square x^2-14x=10
Mathematics
chestercat
  • chestercat
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chestercat
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anonymous
  • anonymous
take half the coefficent of -14, it should be -7. The we have (x-7)^2 which equals=x^2-14x+49. That means you get: (x-7)^2=59
anonymous
  • anonymous
or (x-7)^2-59=0
jim_thompson5910
  • jim_thompson5910
\[\large x^2-14x=10\] \[\large x^2-14x-10=0\] \[\large x^2-14x+49-49-10=0\] \[\large (x-7)^2-49-10=0\] \[\large (x-7)^2-59=0\] \[\large (x-7)^2=59\] \[\large x-7=\pm\sqrt{59}\] \[\large x=7\pm\sqrt{59}\]

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anonymous
  • anonymous
x=7plus or mius sqrt59
anonymous
  • anonymous
\[x^2-14x=10\] \[(x-7)^2=10+7^2=59\] \[x-7=\pm\sqrt{59}\] \[x=7\pm\sqrt{59}\]
nilankshi
  • nilankshi
good

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