anonymous
  • anonymous
4 = the square root of p, -2?
Algebra
chestercat
  • chestercat
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zepdrix
  • zepdrix
\[\huge 4=\sqrt p\]To solve for p, square both sides.\[\huge 4^2=\sqrt{p}^2\]\[\huge p=4^2\]Is this what you were trying to do? Solve for p? :D
zepdrix
  • zepdrix
os was there a minus 2 in the problem also? :O i couldn't tell based on the comma, i thought maybe that was a guess :D
anonymous
  • anonymous
Yes there is a minus 2 . So does it just become p = 2^2 ?

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zepdrix
  • zepdrix
\[\huge 4=\sqrt p - 2\]So this is the problem? :D
zepdrix
  • zepdrix
We'll add 2 to both sides, giving us:\[\huge 6=\sqrt p\]Then we'll do the same step we did before, square both sides, that will eliminate the square root.\[\huge 6^2=\sqrt p ^2 \qquad \rightarrow \qquad 6^2=p\]
anonymous
  • anonymous
Thanks so much !!
anonymous
  • anonymous
I understand how to do the rest of the problems now (:

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