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n0toriousB

  • 2 years ago

4 = the square root of p, -2?

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  1. zepdrix
    • 2 years ago
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    \[\huge 4=\sqrt p\]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

  2. zepdrix
    • 2 years ago
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    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

  3. n0toriousB
    • 2 years ago
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    Yes there is a minus 2 . So does it just become p = 2^2 ?

  4. zepdrix
    • 2 years ago
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    \[\huge 4=\sqrt p - 2\]So this is the problem? :D

  5. zepdrix
    • 2 years ago
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    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.\[\huge 6^2=\sqrt p ^2 \qquad \rightarrow \qquad 6^2=p\]

  6. n0toriousB
    • 2 years ago
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    Thanks so much !!

  7. n0toriousB
    • 2 years ago
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    I understand how to do the rest of the problems now (:

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