anonymous
  • anonymous
complete the square z^2=16z-64
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
try putting everything on one side. \[z^2 -16z + 64 = 0\]
anonymous
  • anonymous
can you see it now?
anonymous
  • anonymous
i got that part

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anonymous
  • anonymous
You need to add half of the coefficient of z and square it (do it to both sides).\[z^2-16z=-64 \rightarrow z^2-16z+(-\frac{16}{2})^2=-64+(-\frac{16}{2})^2\]
anonymous
  • anonymous
\[z^2-16z+(-8)^2=-64+(-8)^2\]You can read off the left now as:\[(z-8)^2\]and the right is simply 0, so your equation becomes,\[(z-8)^2=0\]
anonymous
  • anonymous
Which means one solution for z, z=8.
anonymous
  • anonymous
its suppose to be two answers so would it be 8,-8??
anonymous
  • anonymous
Well, not in this case, since the right-hand side ended up being zero. When you take the square root of both sides, z-8 = +/- 0 = 0 so z = 8. If you'd had, (z-8)^2 = 7, for example, THEN you'd have two solutions:\[(z-8)^2=7 \rightarrow z-8=\pm \sqrt{7} \rightarrow z=8 \pm \sqrt{7}\]
anonymous
  • anonymous
thank u
anonymous
  • anonymous
welcome :) hope you pass!

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