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Solve the equation
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what is it
\[4=\sqrt{p}-2\]
\[4 = \sqrt{p} - 2\] transpose -2 to the left. Combine first then square both sides
\[4+2 = \sqrt{p}\]
\[6 = \sqrt{p}\] or \[\sqrt{p} = 6\] square both sides just like what happened last time
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\[(\sqrt p)^2 = 6^2\]
so you add 4=2 then square root
opps 4+2
yep. in the equation \[4 = \sqrt p - 2\] you will first transpose -2 to the left or in other terms, add 2 to both sides of the equation \[4 + 2 = \sqrt p -2 + 2\] \[6 = \sqrt p \] square both sides of the equation \[6^2 = (\sqrt p)^2\] \[p=?\]
i want to say 36 but i feel worng
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its 36 i think
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