if (4p+1)^2= 81 and p>0, what is a possible value of p?

(4p + 1)^2 = 81
4p + 1 = ±√81
4p + 1 = ±9
4p = -1 ±9

4p = -1 + 9
4p = 8
p = 8/4
p = 2

4p = -1 - 9
4p = -10
p = -10/4
p = -5/2 (-2.5)

∴ p = -5/2 (-2.5) , 2

(4p+1)^2 = 81
(4p+1)^2 - 81 = 0
(4p+1)^2 - 9^ = 0

(4p+1 -9)(4p+1 +9) = 0
p = (9-1)/4 or p = -(9+1)4
p = 2 or p = -5/2

But p > 0
ther fore
p = 2

First you are able to desire to distribute 7p + 40 9 - 36 - 4p = 12 then you incredibly combine like words 3p +13 = 12 then you incredibly subtract 13 from the two aspects 3p = -a million then you incredibly divide by applying 3 p = -a million/3 the respond is A. :)

First,
sqr((4p+1)^2)=sqr(81),
4p+1=9,
4p=9-1,
4p=8,
p=8/4=2.

4p + 1 = ± 9
4p = - 1 ± 9
4p = 8 , 4p = - 10
p = 2 , p = - 5/2

Answer;
p = 2

Solution:
(4p + 1)² = 81
4p + 1 = √81 <===get the square root of both sides
4p + 1 = 9
4p = 9 -1
4p = 8
p = 8/4
p = 2

(4p + 1) ^ 2 = 9 ^ 2

=> 4p + 1 = 9 or 4p + 1 = -9 since 9 and -9 both squared give 81
=>p=2 or p= -10/4

4p + 1 = 9

4p = 8

p = 2

2

p=2