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

2 Answers
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)

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.