How do you solve #p^ { 2} - 10p + 7= - 2#?

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Answer 1 · 2017-11-09T14:47:03

See a solution process below:

First, add #color(red)(2)# to each side of the equation to put it into standard quadratic format:

#p^2 - 10p + 7 + color(red)(2) = -2 + color(red)(2)#

#p^2 - 10p + 9 = 0#

We can next factor the left side of the equation as:

#(p - 9)(p - 1) = 0#

Now, we can solve by equating each term on the left to #0# and solving for #p#:

Solution 1:

#p - 9 = 0#

#p - 9 + color(red)(9) = 0 + color(red)(9)#

#p - 0 = 9#

#p = 9#

Solution 2:

#p - 1 = 0#

#p - 1 + color(red)(1) = 0 + color(red)(1)#

#p - 0 = 1#

#p = 1#

The Solution Is: #p = {1, 9}#