How do you solve 0=p^2-p-110=p2−p−11?

2 Answers
George C. · Arunraju Naspuri
Jun 13, 2015

Use the quadratic formula to find p = (1+-3sqrt(5))/2p=1±3√52

Explanation:

If ap^2+bp+c = 0ap2+bp+c=0 then

p = (-b+-sqrt(b^2-4ac))/(2a)p=−b±√b2−4ac2a

We have a=1a=1, b=-1b=−1, c=-11c=−11

So:

p = (1+-sqrt(1+44))/2 = (1+-sqrt(45))/2p=1±√1+442=1±√452

= (1+-sqrt(3^2*5))/2 = (1+-3sqrt(5))/2=1±√32⋅52=1±3√52

Hence solutions are
p = (1+-3sqrt(5))/2p=1±3√52

Mir Tareq
Jun 13, 2015

p=(1+3sqrt5)/2, (1-3sqrt5)/2p=1+3√52,1−3√52

Explanation:

p^2−2p(1/2)+(1/4)-(11+1/4)=0 or, (p−1/2)^2=45/4=5(3/2)^2 or, [option 1] p−1/2 = 3sqrt5/2 or, p=(1+3sqrt5)/2 or, [option 2] p−1/2 = -3sqrt5/2 or, p=(1-3sqrt5)/2

Related questions
See all questions in Comparing Methods for Solving Quadratics
Impact of this question
2211 views around the world
You can reuse this answer
Creative Commons License