How do you solve #2p ^ { 2} - 1= 161#?

1 Answer
smendyka
Jan 26, 2018

See a solution process below:

Explanation:

First, add #color(red)(1)# to each side of the equation to isolate the #p# term while keeping the equation balanced:

#2p^2 - 1 + color(red)(1) = 161 + color(red)(1)#

#2p^2 - 0 = 162#

#2p^2 = 162#

Now, divide each side of the equation by #color(red)(2)# to isolate the #p^2# while keeping the equation balanced:

#(2p^2)/color(red)(2) = 162/color(red)(2)#

#(color(red)(cancel(color(black)(2)))p^2)/cancel(color(red)(2)) = 81#

#p^2 = 81#

Now, take the square root of each side of the equation to solve for #p# while keeping the equation balanced. Remember, the square root of a number provides both a positive and negative result:

#sqrt(p^2) = +-sqrt(81)#

#p = +-9#

Or

#p = {-9, 9}#

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