How do you solve #p^ { 2} - 12p - 73= 0#?
2 Answers
Explanation:
We can solve this by completing the square and using the difference of squares identity:
#A^2-B^2 = (A-B)(A+B)#
with
#0 = p^2-12p-73#
#color(white)(0) = p^2-2p(6)+36 -36-73#
#color(white)(0) = p^2-2p(6)+36-109#
#color(white)(0) = (p-6)^2-(sqrt(109))^2#
#color(white)(0) = ((p-6)-sqrt(109))((p-6)+sqrt(109))#
#color(white)(0) = (p-6-sqrt(109))(p-6+sqrt(109))#
Hence:
#p = 6+-sqrt(109)#
Explanation:
Use the quadratic formula.
where
When you plug everything in, you get two answers: