SOLUTION: square root of p^2-3p+16 (the square root ends on top of 16)=p+1

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Question 242786: square root of p^2-3p+16 (the square root ends on top of 16)=p+1
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28p%5E2-3p%2B16%29=p%2B1 Start with the given equation.


p%5E2-3p%2B16=%28p%2B1%29%5E2 Square both sides.


p%5E2-3p%2B16=p%5E2%2B2p%2B1 FOIL


-3p%2B16=2p%2B1 Subtract p%5E2 from both sides.


-3p=2p%2B1-16 Subtract 16 from both sides.


-3p-2p=1-16 Subtract 2p from both sides.


-5p=1-16 Combine like terms on the left side.


-5p=-15 Combine like terms on the right side.


p=%28-15%29%2F%28-5%29 Divide both sides by -5 to isolate p.


p=3 Reduce.


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Answer:

So the solution is p=3