How do you solve #p^ { 2} - 12p - 73= 0#?

2 Answers
George C. · EZ as pi
Mar 15, 2018

#p = 6+-sqrt(109)#

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 #A=(p-6)# and #B=sqrt(109)# as follows:

#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)#

KPaukstis · EZ as pi · Jacobi J.
Mar 15, 2018

#p=6+-sqrt109#

Explanation:

Use the quadratic formula. #x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#

where #a = 1," "b = -12 and c = -73#

When you plug everything in, you get two answers:

#p=6+sqrt109# and #p=6-sqrt109#.

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