How do you solve #0=p^2-p-11#?

2 Answers

Answer:

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

Explanation:

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

#p = (-b+-sqrt(b^2-4ac))/(2a)#

We have #a=1#, #b=-1#, #c=-11#

So:

#p = (1+-sqrt(1+44))/2 = (1+-sqrt(45))/2#

#= (1+-sqrt(3^2*5))/2 = (1+-3sqrt(5))/2#

Hence solutions are
#p = (1+-3sqrt(5))/2#

Jun 13, 2015

Answer:

#p=(1+3sqrt5)/2, (1-3sqrt5)/2#

Explanation:

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