Can someone help solve for P: p=(p+20)^(1/2)?

Answer 1

p = (p +20)^1/2

*square both sides,
p^2 = p + 20

*transpose,
p^2 -p - 20 = 0

*think of factors of -20 whose difference is -1...
*factors of -20 are:
1 / -20
-1/20
2 / -10
-2 / 10
4/ -5
-4 / 5

*factors whose difference is -1 is -5/4
*factor out

(p-5) (p+4) = 0
p = 5 and -4

Answer 2

square both sides; p^2=p+20, then set the eqn equal to zero;
p^2-p-20 =0, factor: (p+4)(p-5), and solve for p
p+4 =0;p-5=0, so p = -4, 5

Answer 3

p=(p+20)^(1/2)
p^2= p+20
p^2 -p -20 = 0
(p-5)(p+4)=0
p-5=0 --> p= 5
p+4=0 --> p=-4

Answer 4

p=(p+20)^(1/2) square both sides
p^2=p+20 subtract p+20 from both sides
p^2-p-20=0
(p-5)(p+4)=0
p-5=0
p=5
p+4=0
p=-4
p=-4, 5

Answer 5

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