Question

How do you solve `2p = sqrtp+5`?

Answer 1

Set `p=t^2>=0` hence we have that

#2t^2=sqrt(t^2)+5=>2t^2=t+5=>2t^2-t-5=0=> t=1/4(1+sqrt41)#

hence `p=t^2=1/8(21+sqrt41)`