Lillianmelin
Lillianmelin Lillianmelin
  • 04-05-2016
  • Mathematics
contestada

I am confused on number 15, the instructions are at the top.

I am confused on number 15 the instructions are at the top class=

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huntertpham
huntertpham huntertpham
  • 04-05-2016
You don't have #15 included.  If you meant to put #25, I already answered that.  Here is the solution again though.

#25 is fairly simple.  Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
[tex] \sqrt{12-(-4)} = \sqrt{16} = \frac{+}{}4[/tex]
Extraneous root in this case is positive four since +4≠-4
[tex] \sqrt{12-3} = \sqrt{9} = \frac{+}{}3[/tex]
In this case it's negative 3, since -3≠3
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