How do you solve #16n ^ { 2} - 10n + 129= 8n ^ { 2} - 8# by completing the square?

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Answer 1 · 2017-03-27T16:22:12

No real solutions

Move the #8n^2# to the other side:

#16n^2-10n+129-8n^2=color(red)cancel(color(black)(8n^2))-8color(red)cancel(color(black)(-8n^2))#

#8n^2-10n+129=-8#

Now move 129 to the other side:

#8n^2-10ncolor(red)cancel(color(black)(+129))color(red)cancel(color(black)(-129))=8-129#

#8n^2-10n=-121#

Now add #(b/(2a))^2# to both sides where #b= -10# and #a=8#

#8n^2-10n+(-10/16)^2=-121+(-10/16)^2#

#(8n-10/16)^2=-121+(-10/16)^2#

Now if we square root, there will be a negative inside the square root so we will end up getting imaginary solutions. So there are no real solutions